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Page 1

29-38.

Determine the set of points at which the function is continuous.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}F(x,y) = {\sin{(xy)} \over e^x - y^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}e^x - y^2 \ne 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}e^x \ne y^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}e^x = y^2 is a pair of curves (\newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y=e^{x/2} and \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y=-e^{x/2}). \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}F is discontinuous only on those curves.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{(x,y)|e^x \ne y^2}

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}F(x,y) = {x - y \over 1 + x^2 + y^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}1 + x^2 + y^2 \ne 0 is always true:

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^2 \ge 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y^2 \ge 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^2 + y^2 \ge 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}1 + x^2 + y^2 \ge 1

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}F is continuous on the whole plane.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{(x,y)|(x,y)\in\mathbb{R}^2)}

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}F(x,y) = e^{x^2y} + \sqrt{x + y^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x + y^2 \ge 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x \ge -y^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x \ge -y^2 is a closed region bounded by the curve \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x = -y^2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}F is continuous on that region.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{(x,y)|x \ge -y^2}

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}G(x,y) = \ln{(x^2 + y^2 - 4)}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^2 + y^2 - 4 > 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^2 + y^2 > 4

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^2 + y^2 \le 4 is the closed region of the interior of a circle with center \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}(0,0) and radius \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}G is continuous outside that circle.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{(x,y)|x^2 + y^2 > 4}

Page 2

15-38.

Find the first partial derivatives of the function.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = y^5 - 3xy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = -3y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = 5y^4 - 3x

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = x^4y^3 + 8x^2y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = 4x^3y^3 + 16xy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = 3x^4y^2 + 8x^2

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,t) = e^{-t}\cos{\pi x}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = -\pi e^{-t}\sin{\pi x}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{t} = -e^{-t}\cos{\pi x}

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,t) = \sqrt x \ln t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = {1\over 2\sqrt x}\ln t
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{t} = \sqrt x {1\over t}

  5. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = {(2x + 3y)}^10

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{x} = 20(2x + 3y)^9
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{y} = 30(2x + 3y)^9

  6. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = \tan{xy}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{x} = y\sec^2{xy}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{y} = x\sec^2{xy}

  7. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = {x - y \over x + y}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = {(x + y) - (x - y) \over (x + y)^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {y^2 \over (x + y)^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = {(-1)(x + y) - (x - y) \over (x + y)^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {x^2 \over (x + y)^2}

  8. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = x^y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = yx^{y-1}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = x^y\ln x

Page 3

51-56

Find all the second partial derivatives.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = x^3y^5 + 2x^4y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = 3x^2y^5 + 8x^3y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = 5x^3y^4 + 2x^4

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{x}{x} = 6xy^5 + 24x^2y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{x} = 15x^2y^4 + 8x^3
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{x}{y} = 15x^2y^4 + 8x^3
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{y} = 20x^3y^3

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}w = \sqrt{u^2 + v^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{w}{u} = {u \over \sqrt{u^2 + v^2}}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{w}{v} = {v \over \sqrt{u^2 + v^2}}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{w}{u}{u} = {\sqrt{u^2 + v^2} - u(u / \sqrt{u^2 + v^2}) \over u^2 + v^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over \sqrt{u^2 + v^2}} - {u^2 \over (\sqrt{u^2 + v^2})^3}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{w}{v}{u} = {-uv \over (\sqrt{u^2 + v^2})^3}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{w}{u}{v} = {-uv \over (\sqrt{u^2 + v^2})^3}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{w}{v}{v} = {\sqrt{u^2 + v^2} - v(v / \sqrt{u^2 + v^2}) \over u^2 + v^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over \sqrt{u^2 + v^2}} - {v^2 \over (\sqrt{u^2 + v^2})^3}

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}v = {xy \over x - y}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{v}{x} = {y(x - y) - xy \over (x - y)^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {-y^2 \over (x - y)^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{v}{y} = {x(x - y) + xy \over (x - y)^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {x^2 \over (x - y)^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{v}{x}{x} = {2y^2 \over (x - y)^3}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{v}{y}{x} = {-2y(x - y)^2 - 2y^2(x - y) \over (x - y)^4}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {-2y(x - y) - 2y^2 \over (x - y)^3}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {-2xy \over (x - y)^3}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{v}{x}{y} = {2x(x - y)^2 - 2x^2(x - y) \over (x - y)^4}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2x(x - y) - 2x^2 \over (x - y)^3}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {-2xy \over (x - y)^3}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{v}{y}{y} = {-2x^2 \over (x - y)^3}

Note: 71, 72, and 75 use subscript notation for partials; see “notation” in section 5 of the notes.

  1. Verify that the function \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}u = e^{-\alpha^2k^2t}\sin{kx} is a solution of the heat conduction equation \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}u_t = \alpha^2u_{xx}.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{u}{t} = -\alpha^2k^2e^{-\alpha^2k^2t}\sin{kx}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{u}{x} = e^{-\alpha^2k^2t}k\cos{kx}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{u}{x}{x} = -e^{-\alpha^2k^2t}k^2\sin{kx}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}-\alpha^2k^2e^{-\alpha^2k^2t}\sin{kx} = -\alpha^2e^{-\alpha^2k^2t}k^2\sin{kx} ~~~~\checkmark

  2. Determine whether each of the following functions is a solution of Laplace’s equation \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}u_{xx} + u_{yy} = 0.

    a) \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}u = x^2 + y^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{u}{x} = 2x
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{u}{x}{x} = 2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{u}{y} = 2y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{u}{y}{y} = 2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}2 + 2 \ne 0 ~~~~\xmark

  1. If \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f and \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}g are twice differentiable functions of a single variable, show that the function

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}u(x,t) = f(x + at) + g(x - at)

    is a solution of the wave equation \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}u_{tt} = a^2u_{xx}.

    Note: Physically, this says a plane wave is a superposition of left and right traveling waves

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{u}{t} = a{df\over dt}(x + at) - a{dg\over dt}(x - at)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{u}{t}{t} = a^2{d^2f\over dt^2}(x + at) + a^2{d^2g\over dt^2}(x-at)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{u}{t} = {df\over dx}(x + at) + {dg\over dx}(x - at)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{u}{t}{t} = {d^2f\over dx^2}(x + at) + {d^2g\over dx^2}(x - at)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}a^2{d^2f\over dt^2}(x + at) + a^2{d^2g\over dt^2}(x-at) = a^2({d^2f\over dx^2}(x + at) + {d^2g\over dx^2}(x - at)) is not necessarily true due to the differences between what the derivatives are “in respect to.”

Page 4

  1. The total resistance R produced by three conductors with resistances \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}R_1, \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}R_2, \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}R_3 connected in a parallel electrical circuit is given by the formula \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{1\over R} = {1\over R_1} + {1\over R_2} + {1\over R_3}. Find \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\partial R/\partial R_1.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{R}{R_1} = \pder{}{R_1}{1\over 1/R_1 + 1/R_2 + 1/R_3}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over R_1^2(1/R_1 + 1/R_2 + 1/R_3)^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over R_1^2((R_2R_3 + R_1R_3 + R_1R_2)/(R_1R_2R_3))^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {R_2^2R_3^2 \over (R_2R_3 + R_1R_3 + R_1R_2)^2}

  2. The gas law for a fixed mass \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}m of an ideal gas at absolute temperature \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}T, pressure \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}P, and volume \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}V is \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}PV = mRT, where \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}R is the gas constant. Show that

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{P}{V}\pder{V}{T}\pder{T}{P} = -1


    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}P = {mRT \over V}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{P}{V} = -{mRT \over V^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}V = {mRT \over P}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{V}{T} = {mR \over P}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}T = {PV \over mR}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{T}{P} = {V \over mR}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}-{mRT \over V^2}{mR \over P}{V \over mR}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -{mRT \over PV}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -{mRT \over mRT} = -1 ~~~~\checkmark

Page 5

1-6.

Find an equation of the tangent plane to the given surface at the specified point.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = 4x^2 - y^2 + 2y,~(-1,2,4)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{x} = 8x
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{y} = 2 - 2y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z - 4 = 8(-1)(x + 1) + (2 - 2(2))(y - 2)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = -8x - 8 - 2y + 4 + 4

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = -8x - 2y

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = 3(x - 1)^2 + 2(y + 3)^2 + 7,~(2,-2,12)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{x} = 6(x - 1) \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{y} = 4(y + 3)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z - 12 = 6(2 - 1)(x - 2) + 4(-2 + 3)(y + 2)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = 6x - 12 + 4y + 8 + 12

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = 6x + 4y + 8

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = \sqrt{xy},~(1,1,1)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{x} = {y \over 2\sqrt{xy}}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{y} = {x \over 2\sqrt{xy}}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z - 1 = {1\over 2}(x - 1) + {1\over 2}(y - 1)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = {x + y \over 2}

17-18.

Verify the linear approximation at \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}(0,0).

Note: here “linear approximation” means the same as “tangent plane approximation” as defined in section 6 of the notes.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{2x + 3 \over 4y + 1} \approx 3 + 2x - 12y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{}{x} {2x + 3 \over 4y + 1} = {2 \over 4y + 1}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{}{y} {2x + 3 \over 4y + 1} = {-8x - 12 \over (4y + 1)^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{2x + 3 \over 4y + 1} \approx {2\over 1}x + {-12\over 1}y + {3\over 1}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 3 + 2x - 12y ~~~~\checkmark

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\sqrt{y + \cos^2 x} \approx 1 + {1\over 2}y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{}{x} \sqrt{y + \cos^2 x} = -\sin x (2\cos^2 x){1\over 2\sqrt{y + \cos^2 x}}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -\sin x \cos^2 x {1\over \sqrt{y + \cos^2 x}}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{}{y} \sqrt{y + \cos^2 x} = {1\over 2\sqrt{y + cos^2 x}}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\sqrt{y + \cos^2 x} \approx -\sin 0 \cos^2 0 {1\over \sqrt{0 + \cos^2 0}} x + {1\over 2\sqrt{0 + \cos^2 0}} y + \sqrt{0 + \cos^2 0}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 0 + {1\over 2(1)}y + 1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over 2}y + 1

Page 6

25-30.

Find the differential of the function.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = x^2\ln{(y^2)}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{x} = 2x\ln{(y^2)}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{y} = {2x^2y \over y^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2x^2 \over y}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dz = 2x\ln{(y^2)}dx + {2x^2 \over y}dy

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}v = y\cos{xy}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{v}{x} = -y^2\sin{xy}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{v}{y} = \cos{xy} - xy\sin{xy}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dv = -y^2\cos{(xy)}dx + (\cos{xy} - xy\sin{xy})dy

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}m = p^5q^3

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{m}{p} = 5p^4q^3
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{m}{q} = 3p^5q^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dm = 5p^4q^3dp + 3p^5q^2dq

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}T = {v \over 1 + uvw}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{T}{u} = {-v^2w \over (1 + uvw)^2} \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{T}{v} = {(1 + uvw) + vuw \over (1 + uvw)^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1 + 2uvw \over (1 + uvw)^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{T}{w} = {-uv^2 \over (1 + uvw)^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dT = {1\over (1 + uvw)^2}(-v^2wdu + (1 + 2uvw)dv - uv^2dw)

Page 7

1-6.

Use the Chain Rule to find \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dz/dt or \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dw/dt.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = x^2 + y^2 + xy,~ x = \sin t,~ y = e^t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dz = (2x + y)dx + (2y + x)dy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dx = \cos{(t)}dt
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dy = e^t dt

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{dz\over dt} = (2\sin t + e^t)\cos t + (2e^t + \sin t)e^t

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = \cos{(x + 4y)},~ x = 5t^4,~ y = 1/t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dz = -\sin{(x + 4y)}dx - 4\sin{(x + 4y)}dy

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dx = 20t^3 dt
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dy = {-1\over t^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{dz\over dt} = -20\sin{(5t^4 + {4\over t})} + {4\over t^2}\sin{(5t^4 + {4\over t})}

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = \sqrt{1 + x^2 + y^2},~ x = \ln t,~ y = \cos t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dz = {x \over \sqrt{1 + x^2 + y^2}}dx +{y \over \sqrt{1 + x^2 + y^2}}dy
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over \sqrt{1 + x^2 + y^2}}(xdx + ydy)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dx = {1\over t}dt
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dy = -\sin{(t)}dt

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{dz\over dt} = {1\over \sqrt{1 + \ln^2 t + \cos^2 t}}({\ln t \over t} - \cos t \sin t)

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}w = \ln\sqrt{x^2 + y^2 + z^2},~ x = \sin t,~ y = \cos t,~ z = \tan t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{w}{x} = {x \over x^2 + y^2 + z^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{w}{y} = {y \over x^2 + y^2 + z^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{w}{z} = {z \over x^2 + y^2 + z^2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dw = {1\over x^2 + y^2 + z^2}(xdx + ydy + zdz)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dx = \cos{(t)}dt
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dy = -\sin{(t)}dt
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}dz = \sec^2{(t)}dt

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{dw\over dt} = {1\over \sin^2 t + \cos^2 t + \tan^2 t}(\sin t\cos t - \sin t\cos t + \tan t\sec^2 t)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over 1 + \tan^2 t}(\tan t\sec^2 t)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {\tan t\sec^2 t \over \sec^2 t}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= \tan t

7-12.

Use the Chain Rule to find \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\partial z/\partial s and $\partial z/\partial t$.

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = x^2y^3,~ x = s\cos t,~ y = s\sin t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{s} = \pder{z}{x}\pder{x}{s} + \pder{z}{y}\pder{y}{s}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2xy^3\pder{x}{s} + 3x^2y^2\pder{y}{s}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2(s\cos t)(s^3\sin^3 t)(\cos t) + 3(s^2\cos^2 t)(s^2\sin^2 t)(\sin t)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2s^4\sin^3 t\cos^2 t + 3s^4\sin^3 t\cos^2 t
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 5s^4\sin^3 t\cos^2 t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{t} = 2xy^3\pder{x}{t} + 3x^2y^2\pder{y}{t}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2(s\cos t)(s^3\sin^3 t)(-s\sin t) + 3(s^2\cos^2 t)(s^2\sin^2 t)(s\cos t) \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -2s^5\sin^4 t\cos t + 3s^5\sin^2 t\cos^3 t \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= s^5\sin^2 t\cos t(-2\sin^2 t + 3\cos^2 t) \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= s^5\sin^2 t\cos t(-2 + 5\cos^2 t)

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = \arcsin{(x - y)},~ x = s^2 + t^2,~ y = 1 - 2st

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{s} = {1\over \sqrt{1 - (x - y)^2}}\pder{x}{s} - {1\over \sqrt{1 - (x - y)^2}}\pder{y}{s}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over \sqrt{1 - (x - y)^2}}(\pder{x}{s} - \pder{y}{s})
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over \sqrt{1 - (s^2 + t^2 - 1 + 2st)^2}}(2s + 2t)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2(s + t) \over \sqrt{1 - ((s + t)^2 - 1)^2}}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{t} = {1\over \sqrt{1 - (x - y)^2}}(\pder{x}{t} - \pder{y}{t})
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over \sqrt{1 - (s^2 + t^2 - 1 + 2st)^2}}(2t + 2s)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {2(s + t) \over \sqrt{1 - ((s + t)^2 - 1)^2}}

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = \sin\theta\cos\phi,~ \theta = st^2,~ \phi = s^2t

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{s} = \cos\theta\cos\phi\pder{\theta}{s} - \sin\theta\sin\phi\pder{\phi}{s} \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= \cos(st^2)\cos(s^2t)t^2 - 2\sin(st^2)\sin(s^2t)st

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{t} = \cos\theta\cos\phi\pder{\theta}{t} - \sin\theta\sin\phi\pder{\phi}{t} \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2\cos(st^2)\cos(s^2t)st - \sin(st^2)\sin(s^2t)s^2

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}z = e^{x + 2y},~ x = s/t,~ y = t/s

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{s} = e^{x + 2y}\pder{x}{s} + 2e^{x + 2y}\pder{y}{s}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= e^{x + 2y}(\pder{x}{s} + 2\pder{y}{s})
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= e^{s/t + 2t/s}({1\over t} - {2t \over s^2})

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{z}{t} = e^{x + 2y}(\pder{x}{t} + 2\pder{y}{t})
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= e^{s/t + 2t/s}({1\over t} - {2t \over s^2})

Page 8

  1. The voltage \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}R is slowly increasing as the resistor heats up. Use Ohm’s Law, \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}V = IR, to find out how the current \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}I is changing at the moment when \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}R = 400\:\mathrm{\Omega}, $I = 0.008\:\text{A}\newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}, dV/dt = -0.01\:\text{V}/\text{s}\newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}, and dR/dt = 0.03\:\mathrm{\Omega}/\text{s}$.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}V = 400\:\mathrm{\Omega} \times 0.008\:\text{A} = 3.2\:\text{V}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}I = {V \over R}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}{dI\over dt} = {dI\over dV}{dV\over dt} + {dI\over dR}{dR \over dt}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over R}{dV\over dt} - {V \over R^2}{dR\over dt}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {-0.01\:\text{V}/\text{s} \over 400\:\mathrm{\Omega}} - {3.2\:\text{V} \times 0.03\:\mathrm{\Omega}/\text{s} \over (400\:\mathrm{\Omega})^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -0.000025\:\text{A}/\text{s} - 0.00024\:\text{A}/\text{s}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -0.000265\:\text{A}/\text{s}

Page 9

5-18.

Find the local maximum and minimum values and saddle point(s) of the function. ~~If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.~~

  1. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = 9 - 2x + 4y - x^2 - 4y^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = -2 - 2x
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = 4 - 8y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}-2 - 2x = 0 \Rightarrow x = -1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}4 - 8y = 0 \Rightarrow y = {1\over 2}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{x}{x} = -2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{x} = 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{y} = -8

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}|H(-1,{1\over 2})| = (-2)(-8) - 0^2 = 16 > 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}-2 < 0 \Rightarrow (-1,{1\over 2}) is local max.

  2. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = x^3y + 12x^2 - 8y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = 3x^2y + 24x
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = x^3 - 8

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}x^3 - 8 = 0 \Rightarrow x = 2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}3(2)^2y + 24(2) = 0 \Rightarrow y = -4

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{x}{x} = 6xy + 24
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{x} = 3x^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{y} = 0

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}|H(2,-4)| = 0 - 3(2)^2 = -12 < 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow (2,-4) is a saddle point.

  3. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = (1 + xy)(x + y)

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = y(x + y) + (1 + xy)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2xy + y^2 + 1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = x(x + y) + (1 + xy)
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= 2xy + x^2 + 1

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}2xy + y^2 + 1 = 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow x = {-y^2 - 1 \over 2y}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}2y{-y^2 - 1 \over 2y} + ({-y^2 - 1 \over 2y})^2 + 1 = 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {(-y^2 - 1)^2 \over 4y^2} - y^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {y^4 + 2y^2 + 1 \over 4y^2} - y^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= {1\over 4}y^2 + {1\over 2} + {1\over 4y^2} - y^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~= -{3\over 4}y^2 + {1\over 2} + {1\over 4y^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow {3\over 4}y^2 = {1\over 2} + {1\over 4y^2}
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow y^2 = 1 \Rightarrow y = \pm 1

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y = -1 \Rightarrow x = {-(-1)^2 - 1 \over 2(-1)} = 1
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y = 1 \Rightarrow x = {-(1)^2 - 1 \over 2(1)} = -1

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{x}{x} = 2y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{x} = 2x + 2y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{y} = 2x

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}|H(-1,1)| = 4(-1)(1) + (2(-1) + 2(1))^2 = -4 < 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow (-1,1) is a saddle point.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}|H(1,-1)| = 4(1)(-1) + (2(1) + 2(-1))^2 = -4 < 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow (1,-1) is a saddle point.

  4. \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}f(x,y) = x^3 - 12xy + 8y^3

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{x} = 3x^2 - 12y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\pder{f}{y} = -12x + 24y^2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}-12x + 24y^2 = 0 \Rightarrow x = 2y^2
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}3(2y^2)^2 - 12y = 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow 12y^4 - 12y = 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow y^4 = y
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow y \in {0, 1}

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y = 0 \Rightarrow x = 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}y = 1 \Rightarrow x = 2

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{x}{x} = 6x \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{x} = -12 \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}\spder{f}{y}{y} = 48y

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}|H(0,0)| = 0 - (-12)^2 = -144 < 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}~~~~\Rightarrow (0,0) is a saddle point.

    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}|H(2,1)| = 6(2)(48)(1) - (-12)^2 > 0
    \newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}\newcommand{\spder}[3]{\frac{\partial^2#1}{\partial#2\partial#3}}\newcommand{\cmark}{\checkmark}\newcommand{\xmark}{\Chi}6(2) > 0 \Rightarrow (2,1) is a local min.


© Emberlynn McKinney