Homework 10

1. (30 points)
   Use the chain rule to compute:
   (a) \(\frac{\partial z}{\partial t}\), for \(z = f(x,y) = 4x^2 + 3y^2\), \(x = x(t) = \sin(t)\), \(y = y(t) = \cos(t)\).
   (b) \(\frac{\partial z}{\partial t}\), for \(z = f(x,y) = \sqrt{x^2 + y^2}\), \(x = x(t) = e^{2t}\), \(y = y(t) = e^{-t}\).

   (a) \(z = f(x,y) = f(t) = 4(\sin t)^2 + 3(\cos t)^2 = 3 + \sin^2 t\)
   ​\(\frac{\partial z}{\partial t} = 2 \sin t \cos t\)

   (b) \(z = f(x,y) = f(t) = \sqrt{e^{4t} + e^{-2t}} = (e^{4t} + e^{-2t})^{\frac{1}{2}}\)
   Then \(\frac{\partial z}{\partial t} = \frac{1}{2} (e^{4t} + e^{-2t})^{-\frac{1}{2}} \times (4e^{4t} - 2e^{-2t}) = \frac{4e^{4t} - 2e^{-2t}}{2\sqrt{e^{4t} + e^{-2t}}} = \frac{2e^{4t} - e^{-2t}}{\sqrt{e^{4t} + e^{-2t}}}\)

2. (40 points)
   Compute (if they exist) the second order partial derivative of the following functions at all points:
   (a) \(f(x, y) = \cos(2x) - x^2e^{5y} + 3y^2\).
   (b) \(g(x, y) = \begin{cases} \frac{x^2-y^2}{x^2+y^2} & \text{if } (x, y) \neq (0,0) \\ 0 & \text{if } (x, y) = (0,0). \end{cases}\)
   (c) \(h(x, y) = \begin{cases} \frac{x^2y^2}{x^2+y^2} & \text{if } (x, y) \neq (0,0) \\ 0 & \text{if } (x, y) = (0,0). \end{cases}\)

   (a)
   ​\(\frac{\partial f}{\partial x} = -2 \sin(2x) - 2e^{5y}x\)
   ​\(\frac{\partial f}{\partial y}=-5x^{2}e^{5y}+6y\)
   ​\(\frac{\partial^2 f}{\partial x^2} = -4 \cos(2x) - 2e^{5y}\)
   ​\(\frac{\partial^{2} f}{\partial y^{2}}= -25x^{2}e^{5y}+ 6\)
   ​\(\frac{\partial^2 f}{\partial x \partial y} = -10xe^{5y}\)
   ​\(\frac{\partial^2 f}{\partial y \partial x} = -10e^{5y}x\)

   (b) At \((0,0)\),\(\frac{\partial g}{\partial x}(0,0) = \lim_{x \to 0}\frac{g(x,0) - g(0,0)}{x}= \lim _{x \to 0}\frac{1}{x}= \nexists\)
   ​\(\frac{\partial g}{\partial y}(0,0)=\lim_{y\to0}\frac{g(0,y)-g(0,0)}{y}=\lim_{y\to0} \frac{-1}{y}=\nexists\)

   Thus, the second derivative of g at \((0,0)\) doesn't exist.

   Not at \((0,0)\):
   ​\(\frac{\partial g}{\partial x} = \frac{2x(x^2+y^2) - (x^2-y^2)(2x)}{(x^2+y^2)^2} = \frac{2x^3+2xy^2-2x^3+2xy^2}{(x^2+y^2)^2} = \frac{4xy^2}{(x^2+y^2)^2}\)
   ​\(\frac{\partial g}{\partial y} = \frac{-4x^2y}{(x^2+y^2)^2}\)

   \(\frac{\partial^2 g}{\partial x^2} = \frac{4y^2(x^2+y^2)^2 - 4xy^2 \cdot 2(x^2+y^2) \cdot 2x}{(x^2+y^2)^4} = \frac{(x^2+y^2)(4y^2(x^2+y^2)-16x^2y^2)}{(x^2+y^2)^4} = \frac{4x^2y^2+4y^4-16x^2y^2}{(x^2+y^2)^3} = \frac{-12x^2y^2+4y^4}{(x^2+y^2)^3}\)
   ​\(\frac{\partial^2 g}{\partial y^2} = \frac{-4x^2(x^2+y^2)^2 - (-4x^2y) \cdot 2(x^2+y^2) \cdot 2y}{(x^2+y^2)^4} = \frac{-4x^2(x^2+y^2)+16x^2y^2}{(x^2+y^2)^3} = \frac{-4x^4-4x^2y^2+16x^2y^2}{(x^2+y^2)^3} = \frac{-4x^4+12x^2y^2}{(x^2+y^2)^3}\)

   \(\frac{\partial^2 g}{\partial x \partial y} = \frac{8xy(x^2+y^2)^2 - 4xy^2 \cdot 2(x^2+y^2) \cdot 2y}{(x^2+y^2)^4} = \frac{(x^2+y^2)(8xy(x^2+y^2)-16xy^3)}{(x^2+y^2)^4} = \frac{8x^3y+8xy^3-16xy^3}{(x^2+y^2)^3} = \frac{8x^3y-8xy^3}{(x^2+y^2)^3}\)

   \(\frac{\partial^{2}g}{\partial y\,\partial x}= \frac{8x^{3}y - 8xy^{3}}{(x^{2}+y^{2})^{3}}\)

   (c) At \((0,0)\), \(\frac{\partial h}{\partial x}(0,0) = \lim_{x \to 0} \frac{h(x,0)-h(0,0)}{x} = \lim_{x \to 0} \frac{0-0}{x} = 0\)
   \(\frac{\partial h}{\partial y}(0,0) = \lim_{y \to 0} \frac{h(0,y)-h(0,0)}{y} = \lim_{y \to 0} \frac{0-0}{y} = 0\)

   \(\frac{\partial^{2}h}{\partial x^{2}}(0,0)=\lim_{x\to0}\frac{\frac{\partial h}{\partial x}(x,0)-\frac{\partial h}{\partial x}(0,0)}{x}=\lim_{x\to0}\frac{0-0}{x}=0\)
   Similarly \(\frac{\partial^2 h}{\partial y^2} (0, 0) = 0.\)
   ​\(\frac{\partial^{2}h}{\partial x\partial y}(0,0)=\lim_{x\to0}\frac{\frac{\partial h}{\partial y}(x,0)-\frac{\partial h}{\partial y}(0,0)}{x}=\lim_{x\to 0}\frac{ \frac{2(x)^4(0)}{(x^2+0^2)^2}-0}{x}=0\)​
   Similarly, \(\frac{\partial^{2}h}{\partial y\partial x}(0,0)=0.\)

   Not at \((0,0)\), \(\frac{\partial h}{\partial x} = \frac{2y^2x(x^2+y^2)-x^2y^2(2x)}{(x^2+y^2)^2} = \frac{2y^4x}{(x^2+y^2)^2}\)
   ​\(\frac{\partial h}{\partial y} = \frac{2x^2y(x^2+y^2)-x^2y^2(2y)}{(x^2+y^2)^2} = \frac{2x^4y}{(x^2+y^2)^2}\)

   \(\frac{\partial^2 h}{\partial x^2} = \frac{2y^4(x^2+y^2)^2 - 2y^4x \cdot 2(x^2+y^2) \cdot 2x}{(x^2+y^2)^4} = \frac{2y^4x^2+2y^6-8y^4x^2}{(x^2+y^2)^3} = \frac{-6x^2y^4+2y^6}{(x^2+y^2)^3}\)
   ​\(\frac{\partial^2 h}{\partial y^2} = \frac{-6x^4y^2+2x^6}{(x^2+y^2)^3}\)
   ​\(\frac{\partial^2 h}{\partial x \partial y} = \frac{8xy^3(x^2+y^2)^2-2xy^4 \cdot 2 \cdot (x^2+y^2) \cdot 2y}{(x^2+y^2)^4} = \frac{8x^3y^3}{(x^2+y^2)^3}\)
   ​\(\frac{\partial^2 h}{\partial y \partial x} = \frac{8x^3y^3}{(x^2+y^2)^3}\)

3. (30 points)

   (a) Compute \(\int_D (x^3y + \cos(x)) \,dA\), where \(D\) is the triangle in the \(XY\)-plane with vertices \((0,0)\), \((\frac{\pi}{2}, 0)\) and \((\frac{\pi}{2}, \frac{\pi}{2})\). Do it in two different ways.
   (b) Compute the volume of the solid bounded below by the \(XY\)-plane and above by the surface \(z = \ln(xy)\) over the rectangle \(R = [1,2] \times [1,2]\)
   (c) Compute the volume of the solid bounded below by the surface \(z = x^2 + \sin(y)\) and above by the surface \(z = 4-y^2\) over the rectangle \(R = [0,1] \times [-1,1]\)

   (a) \(\iint_D (x^3y + \cos(x)) \,dA = \int_0^{\frac{\pi}{2}} \int_y^{\frac{\pi}{2}} (x^3y + \cos(x)) \,dx\,dy = \int_0^{\frac{\pi}{2}} \int_0^x (x^3y + \cos(x)) \,dy\,dx\)
   ​\(\int_{0}^{\frac{\pi}{2}}\int_{y}^{\frac{\pi}{2}}(x^{3}y + \cos(x)) \,dx\,dy = \int _{0}^{\frac{\pi}{2}}\left(\frac{y}{4}x^{4} + \sin(x)\right)\bigg|_{y}^{\frac{\pi}{2}} \,dy = \int_{0}^{\frac{\pi}{2}}\left(\frac{y}{4}\frac{\pi^{4}}{2^{4}}+ 1 - \frac{y^{5}}{4} - \sin(y)\right)\,dy\)
   ​\(= \left[\frac{\pi^4}{2^7}y^2 + y - \frac{1}{4 \cdot 6}y^6 + \cos(y)\right]\bigg|_0^{\frac{\pi}{2}} = \frac{\pi^4}{2^7}\frac{\pi^2}{2^2} + \frac{\pi}{2} - \frac{1}{24}\frac{\pi^6}{2^6} - 1 = \frac{\pi^6}{2^9} + \frac{\pi}{2} - \frac{1}{3 \cdot 2^3}\frac{\pi^6}{2^6} - 1\)
   ​\(= \frac{2}{3}\frac{\pi^6}{2^9} + \frac{\pi}{2} - 1 = \frac{\pi^6}{3 \cdot 2^8} + \frac{\pi}{2} - 1 = \frac{\pi^6}{768} + \frac{\pi}{2} - 1\)

   \(\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{x}(x^{3}y + \cos x) dy dx = \int_{0} ^{\frac{\pi}{2}}\left( \frac{x^{3}}{2}y^{2}+ (\cos x)y \right) \Bigg|_{0}^{x}dx = \int_{0}^{\frac{\pi}{2}}\left( \frac{x^{5}}{2}+ x \cos x \right) dx = \int_{0}^{\frac{\pi}{2}} \frac{x^{5}}{2}dx + \int_{0}^{\frac{\pi}{2}}x \cos x dx \\\int_{0}^{\frac{\pi}{2}} \frac{x^{5}}{2}dx = \frac{1}{2}\times \frac{1}{6}x^{6}\Bigg|_{0}^{\frac{\pi}{2}}= \frac{1}{12}\cdot \frac{\pi^{6}}{2^{6}}= \frac{\pi^{6}}{3 \cdot 2^{8}}\\\int_{0}^{\frac{\pi}{2}} x \cos x dx = x \sin x \Bigg|_{0}^{\frac{\pi}{2}}- \int_{0}^{\frac{\pi}{2}}\sin x dx = \frac{\pi}{2}\sin \frac{\pi}{2}- 0 - [-\cos x]_{0}^{\frac{\pi}{2}}= \frac{\pi}{2} - (-\cos \frac{\pi}{2}- (-\cos 0)) = \frac{\pi}{2}- (0 - (-1)) = \frac{\pi}{2}- 1 \\\text{Thus, }\int_{0}^{\frac{\pi}{2}}\int_{0}^{x}(x^{3}y + \cos x) dy dx = \frac{\pi^{6}}{768} + \frac{\pi}{2}- 1\)

   \(\displaystyle\text{(b) }\int_1^2\int_1^2(\ln|xy|)dxdy=\int_1^2(x\ln|xy|\Big|_1^2-\int_1^21dx)dy=\int_1^2(2\ln(2y)-\ln y-1)dy \\=2\int_{1}^{2}\ln(2)dy+\int_{1}^{2}\ln ydy-\int_{1}^{2}1dy=4\ln(2)-2\)
   (c) \(\displaystyle\int_{-1}^{1} \int_{0}^{1} (x^{2} + \sin y) \, dx \, dy = \int_{-1}^{1} \left[ \frac{1}{3}x^{3} + (\sin y) x \right]_{0}^{1} \, dy = \int_{-1}^{1} \left( \frac{1}{3} + \sin y \right) \, dy = \left[ \frac{1}{3}y - \cos y \right]_{-1}^{1}\) \(\displaystyle= (\frac{1}{3} - \cos(1)) - (-\frac{1}{3} - \cos(-1)) = \frac{1}{3} - \cos(1) + \frac{1}{3} + \cos(1) = \frac{2}{3}\)
   ​\(\displaystyle\int_{-1}^{1} \int_{0}^{1} (4 - y^{2}) \, dx \, dy = \int_{-1}^{1} \left[ (4-y ^{2})x \right]_{0}^{1} \, dy = \int_{-1}^{1} (4 - y^{2}) \, dy = \left[ 4y - \frac{1}{3} y^{3} \right]_{-1}^{1} = \frac{22}{3}\)
   ​\(\text{Then } A = \frac{22}{3} - \frac{2}{3} = \frac{20}{3}\)

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