4.6 Directional derivatives and the gradient  (Page 6/8)

$f\left(x,y\right)=5-2x^2-\frac{1}{2}{y}^2$ at point $P\left(3,4\right)$ in the direction of $\text{u}=\left(\text{cos}\frac{\pi }{4}\right)i+\left(\text{sin}\frac{\pi }{4}\right)j$

$f\left(x,y\right)=y^2\text{cos}\left(2x\right)$ at point $P\left(\frac{\pi }{3},2\right)$ in the direction of $\text{u}=\left(\text{cos}\frac{\pi }{4}\right)i+\left(\text{sin}\frac{\pi }{4}\right)j$

Find the directional derivative of $f\left(x,y\right)=y^2\text{sin}\left(2x\right)$ at point $P\left(\frac{\pi }{4},2\right)$ in the direction of $u=5i+12j.$

$f\left(x,y\right)=xy,$ $P\left(0,\mathrm{-2}\right),$ $\text{v}=\frac{1}{2}i+\frac{\sqrt{3}}{2}j$

$h\left(x,y\right)=e^x\text{sin}y,P\left(1,\frac{\pi }{2}\right),v=\text{−}i$

$h\left(x,y,z\right)=xyz,P\left(2,1,1\right),\text{v}=2i+j-k$

$f(x,y)=xy,P(1,1),u=⟨\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}⟩$

$f(x,y)=x^2-y^2,\begin{array}{cc}u=⟨\frac{\sqrt{3}}{2},\frac{1}{2}⟩,\hfill & P(1,0)\hfill \end{array}$

$f(x,y)=3x+4y+7,\begin{array}{cc}u=⟨\frac{3}{5},\frac{4}{5}⟩,\hfill & P\left(0,\frac{\pi }{2}\right)\hfill \end{array}$

$\begin{array}{ccc}f(x,y)=e^x\text{cos}y,\hfill & u=⟨0,1⟩,\hfill & P=\left(0,\frac{\pi }{2}\right)\hfill \end{array}$

$\begin{array}{ccc}f(x,y)=y^{10},\hfill & u=⟨0,\mathrm{-1}⟩,\hfill & P=(1,\mathrm{-1})\hfill \end{array}$

$f(x,y)=\text{ln}(x^2+y^2),\begin{array}{cc}u=⟨\frac{3}{5},\frac{4}{5}⟩,\hfill & P\left(1,2\right)\hfill \end{array}$

$f(x,y)=x^{2}y,\begin{array}{cc}P(\mathrm{-5},5),\hfill & v=3i-4j\hfill \end{array}$

$f(x,y)=y^2+xz,\begin{array}{cc}P(1,2,2),\hfill & v=⟨2,\mathrm{-1},2⟩\hfill \end{array}$

$f\left(x,y\right)=x^2+2y^2,\theta =\frac{\pi }{6}$

$f\left(x,y\right)=\frac{y}{x+2y},\theta =-\frac{\pi }{4}$

$f\left(x,y\right)=\text{cos}\left(3x+y\right),\theta =\frac{\pi }{4}$

$w\left(x,y\right)=y{e}^x,\theta =\frac{\pi }{3}$

$\begin{array}{cc}f\left(x,y\right)=x\text{arctan}(y),\hfill & \theta =\frac{\pi }{2}\hfill \end{array}$

$\begin{array}{cc}f\left(x,y\right)=\text{ln}(x+2y),\hfill & \theta =\frac{\pi }{3}\hfill \end{array}$

Find the gradient of $f(x,y)=\frac{14-x^2-y^2}{3}.$ Then, find the gradient at point $P\left(1,2\right).$

Find the gradient of $f(x,y,z)=xy+yz+xz$ at point $P\left(1,2,3\right).$

Find the gradient of $f(x,y,z)$ at $P$ and in the direction of $u\text{:}$ $f(x,y,z)=\text{ln}(x^2+2y^2+3z^2),\begin{array}{cc}P(2,1,4),\hfill & u=\frac{\mathrm{-3}}{13}i-\frac{4}{13}j-\frac{12}{13}k\hfill \end{array}.$

$f(x,y,z)=4x^5{y}^2{z}^3,\begin{array}{cc}P(2,\mathrm{-1},1),\hfill & u=\frac{1}{3}i+\frac{2}{3}j-\frac{2}{3}k\hfill \end{array}$

$f(x,y)=x^2+3y^2,\begin{array}{cc}P(1,1),\hfill & Q(4,5)\hfill \end{array}$

$f(x,y,z)=\frac{y}{x+z},\begin{array}{cc}P(2,1,\mathrm{-1}),\hfill & Q(\mathrm{-1},2,0)\hfill \end{array}$

$f(x,y)=\mathrm{-7}x+2y,\begin{array}{cc}P(2,\mathrm{-4}),\hfill & u=4i-3j\hfill \end{array}$

$f(x,y)=\text{ln}(5x+4y),\begin{array}{cc}P(3,9),\hfill & u=6i+8j\hfill \end{array}$

[T] Use technology to sketch the level curve of $f(x,y)=4x-2y+3$ that passes through $P(1,2)$ and draw the gradient vector at $P.$

[T] Use technology to sketch the level curve of $f(x,y)=x^2+4y^2$ that passes through $P(\mathrm{-2},0)$ and draw the gradient vector at $P.$

$f(x,y)=x{y}^2-y{x}^2,P(\mathrm{-1},1)$

$f(x,y)=x{e}^y-\text{ln}(x),P\left(\mathrm{-3},0\right)$

$f(x,y,z)=xy-\text{ln}(z),P(2,\mathrm{-2},2)$

$f(x,y,z)=x\sqrt{y^2+z^2},P(\mathrm{-2},\mathrm{-1},\mathrm{-1})$

$f(x,y)=x^2+xy+y^2$ at point $\left(\mathrm{-5},\mathrm{-4}\right)$ in the direction the function increases most rapidly

$f(x,y)=e^{xy}$ at point $\left(6,7\right)$ in the direction the function increases most rapidly

$f(x,y)=\text{arctan}\left(\frac{y}{x}\right)$ at point $\left(\mathrm{-9},9\right)$ in the direction the function increases most rapidly

$f(x,y,z)=\text{ln}(xy+yz+zx)$ at point $\left(\mathrm{-9},\mathrm{-18},\mathrm{-27}\right)$ in the direction the function increases most rapidly

$f(x,y,z)=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ at point $\left(5,\mathrm{-5},5\right)$ in the direction the function increases most rapidly

$f(x,y)=x{e}^{\text{−}y},$ $\left(1,0\right)$

$f(x,y)=\sqrt{x^2+2y},$ $\left(4,10\right)$

$f(x,y)=\text{cos}(3x+2y),\left(\frac{\pi }{6},-\frac{\pi }{8}\right)$

The level curve $f(x,y,z)=12$ for $f(x,y,z)=4x^2-2y^2+z^2$ at point $\left(2,2,2\right).$

$f(x,y,z)=xy+yz+xz=3$ at point $\left(1,1,1\right)$

$f(x,y,z)=xyz=6$ at point $\left(1,2,3\right)$

$f(x,y,z)=x{e}^y\text{cos}z-z=1$ at point $\left(1,0,0\right)$

The temperature $T$ in a metal sphere is inversely proportional to the distance from the center of the sphere (the origin: $\left(0,0,0\right)).$ The temperature at point $\left(1,2,2\right)$ is $120\text{°}\text{C}.$

1. Find the rate of change of the temperature at point $\left(1,2,2\right)$ in the direction toward point $\left(2,1,3\right).$
2. Show that, at any point in the sphere, the direction of greatest increase in temperature is given by a vector that points toward the origin.

The electrical potential (voltage) in a certain region of space is given by the function $V(x,y,z)=5x^2-3xy+xyz.$

1. Find the rate of change of the voltage at point $\left(3,4,5\right)$ in the direction of the vector $⟨1,1,\mathrm{-1}⟩.$
2. In which direction does the voltage change most rapidly at point $\left(3,4,5\right)?$
3. What is the maximum rate of change of the voltage at point $\left(3,4,5\right)?$

If the electric potential at a point $\left(x,y\right)$ in the xy -plane is $V(x,y)=e^{\mathrm{-2}x}\text{cos}(2y),$ then the electric intensity vector at $\left(x,y\right)$ is $E=\text{−}\nabla V(x,y).$

1. Find the electric intensity vector at $\left(\frac{\pi }{4},0\right).$
2. Show that, at each point in the plane, the electric potential decreases most rapidly in the direction of the vector $E.$

In two dimensions, the motion of an ideal fluid is governed by a velocity potential $\phi .$ The velocity components of the fluid $u$ in the x- direction and $v$ in the y -direction, are given by $⟨u,v⟩=\nabla \phi .$ Find the velocity components associated with the velocity potential $\phi (x,y)=\text{sin}\pi x\text{sin}2\pi y.$