Gradient & Directional Derivatives - Calculus 3

1. Given $u(x,y) = \frac{x^2 - y^2}{x^2 + y^2}$ for $(x,y) \neq (0,0)$, at the point $(2, 2)$, what is the directional derivative in the direction perpendicular to the gradient vector and pointing into the second quadrant?
   Answer: $0$ First, find $\nabla u$ using the quotient rule. Let $f = x^2 - y^2$ and $g = x^2 + y^2$. Then $u = \frac{f}{g}$ and $\nabla u = \frac{g\nabla f - f\nabla g}{g^2}$. We have $\nabla f = \langle 2x, -2y \rangle$ and $\nabla g = \langle 2x, 2y \rangle$. So: $\nabla u = \frac{(x^2 + y^2)\langle 2x, -2y \rangle - (x^2 - y^2)\langle 2x, 2y \rangle}{(x^2 + y^2)^2}$. At $(2, 2)$: $f = 4 - 4 = 0$, $g = 8$, $\nabla f = \langle 4, -4 \rangle$, $\nabla g = \langle 4, 4 \rangle$. Thus: $\nabla u(2, 2) = \frac{8\langle 4, -4 \rangle - 0\langle 4, 4 \rangle}{64} = \frac{\langle 32, -32 \rangle}{64} = \left\langle \frac{1}{2}, -\frac{1}{2} \right\rangle$. A vector perpendicular to $\left\langle \frac{1}{2}, -\frac{1}{2} \right\rangle$ is $\left\langle \frac{1}{2}, \frac{1}{2} \right\rangle$ or $\left\langle -\frac{1}{2}, -\frac{1}{2} \right\rangle$. The one pointing into the second quadrant (negative $x$, positive $y$) is $\left\langle -\frac{1}{2}, \frac{1}{2} \right\rangle$. The unit vector is $\frac{1}{\sqrt{1/2}}\left\langle -\frac{1}{2}, \frac{1}{2} \right\rangle = \frac{1}{\sqrt{2}}\langle -1, 1 \rangle$. The directional derivative in any direction perpendicular to the gradient is always $0$.
2. Consider the function $f(x,y) = e^{x^2+y^2}\sin(xy)$. At which of the following points does the gradient vector $\nabla f$ have the smallest magnitude?
   Answer: $(0, 0)$ We need $\nabla f = \langle f_x, f_y \rangle$ where $f_x = e^{x^2+y^2}[2x\sin(xy) + y\cos(xy)]$ and $f_y = e^{x^2+y^2}[2y\sin(xy) + x\cos(xy)]$. At $(0,0)$: $\nabla f(0,0) = e^0[\langle 0 + 0, 0 + 0 \rangle] = \langle 0, 0 \rangle$, so $||\nabla f(0,0)|| = 0$. At $(\pi, 0)$: $\nabla f(\pi,0) = e^{\pi^2}[\langle 2\pi \cdot 0 + 0 \cdot 1, 0 + \pi \cdot 1 \rangle] = e^{\pi^2}\langle 0, \pi \rangle$, so $||\nabla f|| = \pi e^{\pi^2}$. Similarly at $(0, \pi)$: $||\nabla f|| = \pi e^{\pi^2}$. At $(\sqrt{\pi}, \sqrt{\pi})$: The exponential factor is $e^{2\pi}$ and $\sin(\pi) = 0, \cos(\pi) = -1$, giving a non-zero magnitude. The smallest magnitude is $0$ at the origin.
3. The temperature distribution in a metal plate is given by $T(x,y) = 100 - x^2 - 4y^2$. A bug located at point $(2, 1)$ wants to move in such a direction that the temperature decreases most rapidly. If the bug moves $0.1$ units in this optimal direction, what is the approximate change in temperature?
   Answer: $-0.894$ To decrease temperature most rapidly, the bug should move in the direction of $-\nabla T$. We have $\nabla T = \langle -2x, -8y \rangle$, so at $(2,1)$: $\nabla T(2,1) = \langle -4, -8 \rangle$. The direction of steepest decrease is $-\nabla T = \langle 4, 8 \rangle$. The unit vector in this direction is $\hat{u} = \frac{1}{||\langle 4, 8 \rangle||}\langle 4, 8 \rangle = \frac{1}{\sqrt{16+64}}\langle 4, 8 \rangle = \frac{1}{\sqrt{80}}\langle 4, 8 \rangle = \frac{1}{4\sqrt{5}}\langle 4, 8 \rangle$. The directional derivative in this direction is $D_{\hat{u}}T = \nabla T \cdot \hat{u} = \langle -4, -8 \rangle \cdot \frac{1}{4\sqrt{5}}\langle 4, 8 \rangle = \frac{1}{4\sqrt{5}}(-16 - 64) = \frac{-80}{4\sqrt{5}} = \frac{-20}{\sqrt{5}} = -4\sqrt{5}$. Moving $0.1$ units gives approximate change: $\Delta T \approx 0.1 \cdot (-4\sqrt{5}) = -0.4\sqrt{5} \approx -0.4(2.236) = -0.894$.

Start this quiz