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Homework 9

  1. (20 points)
    Compute (if they exist) the first order partial derivatives of the following functions :
    (a) \(f(x, y) = e^{xy}\sin(x^2 + y)\).
    (b) \(g(x, y) = \begin{cases} \frac{x^3y}{x^4+y^2} & \text{if } (x, y) \neq (0, 0) \\ 0 & \text{if } (x, y) = (0, 0). \end{cases}\)

    (a) \(\frac{\partial f}{\partial x}= e^{xy}\cdot y \cdot \sin(x^{2}+y) + e^{xy}\cos(x^{2} +y) \cdot 2x\)

    \(\frac{\partial f}{\partial y} = e^{xy} \cdot x \cdot \sin(x^2+y) + e^{xy} \cos(x^2+y) \cdot 1\)

    (b) If \((x,y) \neq (0,0)\), \(\frac{\partial g}{\partial x} = \frac{3x^2y(x^4+y^2) - x^3y(4x^3)}{(x^4+y^2)^2} = \frac{3x^6y + 3x^2y^3 - 4x^6y}{(x^4+y^2)^2} = \frac{3x^2y^3 - x^6y}{(x^4+y^2)^2}\)

    \(\frac{\partial g}{\partial y} = \frac{x^3(x^4+y^2) - x^3y(2y)}{(x^4+y^2)^2} = \frac{x^7 + x^3y^2 - 2x^3y^2}{(x^4+y^2)^2} = \frac{x^7 - x^3y^2}{(x^4+y^2)^2}\)

    If \((x,y) = (0,0)\), \(\frac{\partial g}{\partial x}(0,0) = \lim_{x \to 0} \frac{g(x,0) - g(0,0)}{x-0} = \lim_{x \to 0} \frac{0-0}{x} = \lim_{x \to 0} \frac{0}{x} = 0\)

    \(\frac{\partial g}{\partial y}(0,0) = \lim_{y \to 0} \frac{g(0,y) - g(0,0)}{y-0} = \lim_{y \to 0} \frac{0-0}{y} = \lim_{y \to 0} \frac{0}{y} = 0\)

    Thus, \(\frac{\partial g}{\partial x} = \frac{3x^2y^3 - x^6y}{(x^4+y^2)^2}\) and \(\frac{\partial g}{\partial y} = \frac{x^7 - x^3y^2}{(x^4+y^2)^2}\) if \((x,y) \neq (0,0)\), and \(\frac{\partial g}{\partial x} = \frac{\partial g}{\partial y} = 0\) if \((x,y) = (0,0)\).

  2. (30 points)
    Prove or disprove that the following functions are differentiable at \((0,0)\):
    (a) \(f(x,y) = \begin{cases} \frac{x^3}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0). \end{cases}\)
    (b) \(g(x,y) = \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0). \end{cases}\)
    (c) \(h(x,y) = \begin{cases} (x^2+y^2)\sin(\frac{1}{\sqrt{x^2+y^2}}) & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0). \end{cases}\)

    (a) \(\frac{\partial f}{\partial x}(0,0) = \lim_{x\to 0} \frac{f(x,0)-f(0,0)}{x-0} = \lim_{x\to 0} \frac{\frac{x^3}{x^2}-0}{x} = \lim_{x\to 0} \frac{x}{x} = 1\)
    ​\(\frac{\partial f}{\partial y}(0,0) = \lim_{y\to 0} \frac{f(0,y)-f(0,0)}{y-0} = \lim_{y\to 0} \frac{0-0}{y} = 0\)

    Then \(\lim_{(x,y)\to(0,0)}\frac{f(x,y)-\left(f(0,0\right)+\nabla f(0,0)\cdot\left(x,y)\right)}{\|(x,y)\|} =\lim_{(x,y)\to(0,0)}\frac{\frac{x^{3}}{x^{2}+y^{2}}-x}{\sqrt{x^{2}+y^{2}}}=\lim_{(x,y)\to(0,0)} \frac{\frac{x^{3}-x(x^{2}+y^{2})}{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}}=\lim_{(x,y)\to(0,0)} \frac{-xy^{2}}{(x^{2}+y^{2})^{3/2}}\)
    Let \(y=x\). Then the limit is \(\lim_{x\to0}\frac{-x^{3}}{(2x^{2})^{3/2}}=\lim_{x\to0}\frac{-x^{3}}{|x|^{3}(2)^{3/2}}\)
    If \(x>0\), the limit is \(\frac{-1}{(2)^{3/2}}\). If \(x<0\), the limit is \(\frac{1}{(2)^{3/2}}\).
    Then limit does not exist. Thus, \(f\) is not differentiable at \((0,0)\).

    (b) \(\frac{\partial g}{\partial x} = \lim_{x \to 0} \frac{g(x, 0) - g(0,0)}{x-0} = \lim_{x \to 0} \frac{0}{x} = 0\)
    ​\(\frac{\partial g}{\partial y} = \lim_{y \to 0} \frac{g(0,y) - g(0,0)}{y-0} = \lim_{y \to 0} \frac{0}{y} = 0\)
    ​\(\nabla g(0,0) = (0,0), \text{ then } \lim_{(x,y) \to (0,0)} \frac{g(x,y) - (g(0,0) + \nabla g(0,0) (x,y))}{||(x,y)||} = \lim_{(x,y) \to (0,0)} \frac{xy \frac{x^2-y^2}{x^2+y^2}}{\sqrt{x^2+y^2}}\)
    ​\(= \lim_{(x,y) \to (0,0)}xy \frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{3/2}}\)

    Let \(x = r \cos \theta\), \(y = r \sin \theta\), then \(\lim_{r\to 0} \frac{r^2 \cos \theta \sin \theta \cdot r^2 (\cos^2 \theta - \sin^2 \theta)}{|r|} = 0\)

    Thus \(g\) is differentiable at \((0,0)\)

    (c) \(\frac{dh}{dx} = \lim_{x \to 0} \frac{h(x,0)-h(0,0)}{x-0} = \lim_{x \to 0} \frac{x^2 \sin(\frac{1}{\sqrt{x^2}})}{x} = \lim_{x \to 0} x \sin(\frac{1}{|x|}) = 0\)
    ​\(\frac{dh}{dy} = \lim_{y \to 0} \frac{h(0,y)-h(0,0)}{y-0} = \lim_{y \to 0} \frac{y^2 \sin(\frac{1}{\sqrt{y^2}})}{y} = \lim_{y \to 0} y \sin(\frac{1}{|y|}) = 0\)
    Then \(\nabla h(0,0) = (0,0)\). Then \(\lim_{(x,y)\to(0,0)}\frac{h(x,y)-h(0,0)-\nabla h(0,0)\cdot(x,y)}{\sqrt{x^{2}+y^{2}}} =\lim_{(x,y)\to(0,0)}\frac{(x^{2}+y^{2})\sin(\frac{1}{\sqrt{x^2+y^2}})}{\sqrt{x^{2}+y^{2}}}\)
    ​\(= \lim_{(x,y) \to (0,0)} \sqrt{x^2+y^2} \sin(\frac{1}{\sqrt{x^2+y^2}})\)
    Since \(|\sqrt{x^2+y^2} \sin(\frac{1}{\sqrt{x^2+y^2}})| \le |\sqrt{x^2+y^2}|\) and \(\sqrt{x^2+y^2} < \delta\), take \(\delta=\epsilon\).
    we get \(|\sqrt{x^2+y^2} \sin(\frac{1}{\sqrt{x^2+y^2}})| \le \delta = \epsilon\). Thus \(\lim_{(x,y) \to (0,0)} \sqrt{x^2+y^2} \sin(\frac{1}{\sqrt{x^2+y^2}}) = 0\).

  3. (30 points)
    (a) Find the directional derivative of the function \(f(x, y) = x^2 + xy + y^2\) at the point \((-1, 1)\) in the direction of the gradient of \(f\) at \((-1, 1)\).
    (b) Compute the directional derivative in the direction of the unit vector \(v = (v_1, v_2)\) of the function \(f(x, y) = \begin{cases} \frac{x^3y}{x^4+y^2} & \text{if } (x, y) \neq (0, 0) \\ 0 & \text{if } (x, y) = (0, 0). \end{cases}\) at \((0, 0)\). Is the function differentiable at \((0, 0)\)?

    Solution:
    (a) \(\frac{\partial f}{\partial x} = 2x+y\), \(\frac{\partial f}{\partial y} = x+2y\), then \(\nabla f(x, y) = (2x+y, x+2y)\)
    ​\(\nabla f(-1,1)=(-2+1,-1+2)=(-1,1)\)

    Then \(v=(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\), then \(\frac{\partial f}{\partial v}(-1,1) = \lim_{t \rightarrow 0}\frac{f(-1-\frac{t}{\sqrt{2}}, 1+\frac{t}{\sqrt{2}}) - f(-1,1)}{t}= \lim_{t \rightarrow 0}\frac{(1+\frac{t}{\sqrt2})^2-1}{t}\)
    ​\(\lim_{t\rightarrow0}\frac{\frac{t^{2}}{2}+\sqrt{2}t}{t}=\lim_{t\rightarrow0}\frac{t}{2}+\sqrt{2}=\sqrt{2}\)

    (b) \(\frac{\partial f}{\partial v}(0,0)=\lim_{t\to0}\frac{f(tv_{1},tv_{2})-f(0,0)}{t} =\lim_{t\to0}\frac{tv_{1}^{3}v_{2}}{t^{2}v_{1}^{4}+v_{2}^{2}}\).

    If \(v_2 \neq 0\), this limit is \(\frac{0}{0+v_2^2} = 0\). If \(v_2 = 0\), then \(v_1 = \pm 1\). \(f(tv_1, 0) = 0\)

    Thus \(\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0\).
    Then \(\lim_{(x,y) \to (0,0)} \frac{f(x,y) - (f(0,0) + \nabla f(0,0) (x,y))}{||(x,y)||} = \lim_{(x,y) \to (0,0)} \frac{\frac{x^3 y}{x^4 + y^2}}{\sqrt{x^2 + y^2}} = \lim_{(x,y) \to (0,0)} \frac{x^3 y}{(x^4 + y^2) \sqrt{x^2 + y^2}}\).
    Let \(y=x^2\), \(=\lim_{x\to0}\frac{x}{2|x|\sqrt{1+x^2}}\). If \(x \to 0^{+}\), \(\lim=\frac{1}{2}\). If \(x \to 0^{-}\), \(\lim=-\frac{1}{2}\).
    Thus \(f\) is not differentiable at \((0,0)\).

  4. (20 points)
    Find the tangent plane at \(P\) on the given level surfaces:
    (a) \(x+y+z=1\) at \(P=(0,1,0)\).
    (b) \((x^2+y^2)e^{-(x^2+y^2)} - z = 0\) at \(P=(0,0,0)\).

    (a) \(f(x, y) = 1-x-y\), \(\frac{\partial f}{\partial x} = -1\), \(\frac{\partial f}{\partial y} = -1\).
    ​\(z = f(0,1) + \frac{\partial f}{\partial x} (x-0) + \frac{\partial f}{\partial y} (y-1)\).
    ​\(z = -x-y+1\).

    (b) \(z = f(x,y) = (x^2+y^2) e^{-(x^2+y^2)}\). \(\frac{\partial f}{\partial x} = (2x) e^{-(x^2+y^2)} + (x^2+y^2) e^{-(x^2+y^2)} (-2x)\).
    ​\(\frac{\partial f}{\partial y} = (2y) e^{-(x^2+y^2)} + (x^2+y^2) e^{-(x^2+y^2)} (-2y)\).
    Then \(z=f(0,0)+\frac{\partial f}{\partial x}x+\frac{\partial f}{\partial y}y=2x_{0}^{2} e^{-(x_0^2+y_0^2)}-2x_{0}^{2}(x_{0}^{2}+y_{0}^{2})e^{-(x_0^2+y_0^2)}+2y_{0}^{2}e^{-(x_0^2+y_0^2)} -2y_{0}^{2}(x_{0}^{2}+y_{0}^{2})e^{-(x_0^2+y_0^2)}=0\).

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