2.7E: Directional Derivatives and the Gradient (Exercises)

13.6: Directional Derivatives and the Gradient

In exercise 1, find the directional derivative using the limit definition only.

1) a. $f(x,y)=5−2x^2−\frac{1}{2}y^2$ at point $P(3,4)$ in the direction of $\vecs u=(\cos\frac{π}{4})\,\hat{\mathbf i}+(\sin\frac{π}{4})\,\hat{\mathbf j}$

Answer:

a. $D_{\vecs u}f(3,4) =−8\sqrt{2}$

1) b. $f(x,y)=y^2\cos(2x)$ at point $P(\frac{π}{3},2)$ in the direction of $\vecs u=(\cos\frac{π}{4})\,\hat{\mathbf i}+(\sin\frac{π}{4})\,\hat{\mathbf j}$

Answer:

b. $D_{\vecs u}f(\frac{π}{3},2) =−2\sqrt{6}-\sqrt{2}$

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

 

In exercises 3 - 13, find the directional derivative of the function in the direction of $\vecs v$ as a function of $x$ and $y$. Remember that you first need to find a unit vector in the direction of the direction vector. Then find the value of the directional derivative at point $P$.

3) $f(x,y)=xy, \quad P(-2,0), \quad \vecs v=\frac{1}{2}\,\hat{\mathbf i}+\frac{\sqrt{3}}{2}\,\hat{\mathbf j}$

Answer:

$D_{\vecs v}f(x, y) = \frac{1}{2}y + \frac{\sqrt{3}}{2}x$
$D_{\vecs v}f(-2, 0) = −\sqrt{3}$

4) $h(x,y)=e^x\sin y,\quad P(1,\frac{π}{2}),\quad \vecs v=−\,\hat{\mathbf i}$

5) $f(x,y)=x^2y,\quad P(−5,5),\quad \vecs v=3\,\hat{\mathbf i}−4\,\hat{\mathbf j}$

Answer:

$D_{\vecs v}f(x, y) = \frac{6}{5}xy - \frac{4}{5}x^2$
$D_{\vecs v}f(-5,5)= -50$

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

7) $f(x,y)=x^2−y^2, \quad P(1,0), \quad \vecs u=⟨\frac{\sqrt{3}}{2},\frac{1}{2}⟩$

Answer:

$D_{\vecs u}f(x, y) = x\sqrt{3} - y$
$D_{\vecs u}f(1,0)= \sqrt{3}$

8) $f(x,y)=3x+4y+7,\quad P(0,\frac{π}{2}), \quad \vecs u=⟨\frac{3}{5},\frac{4}{5}⟩$

9) $f(x,y)=e^x\cos y,\quad P=(0,\frac{π}{2}), \quad \vecs u=⟨0,5⟩$

Answer:

$D_{\vecs u}f(x, y) = -e^x\sin y$
$D_{\vecs u}f(0, \frac{π}{2})= −1$

10) $f(x,y)=y^{10},\quad \vecs u=⟨0,−3⟩,\quad P=(1,−1)$

11) $f(x,y)=\ln(x^2+y^2),\quad \vecs u=⟨2,-5⟩,\quad P(1,2)$

Answer:

$D_{\vecs u}f(x, y) = \frac{\sqrt{29}}{29}\left( \dfrac{4x-10y}{x^2 +y^2}\right)$
$D_{\vecs u}f(1,2)= -\frac{16\sqrt{29}}{145}$

12) $h(x,y,z)=xyz, \quad P(2,1,1),\quad \vecs v=2\,\hat{\mathbf i}+\,\hat{\mathbf j}−\,\hat{\mathbf k}$

Answer:

$D_{\vecs v}h(x, y, z) = \frac{\sqrt{6}}{6}(2yz + xz - xy)$
$D_{\vecs v}h(2,1,1) = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$

13) $f(x,y,z)=y^2+xz,\quad P(1,2,2),\quad \vecs v=⟨2,−1,2⟩$

Answer:

$D_{\vecs v}f(x, y, z) = \frac{2}{3}(z - y + x)$
$D_{\vecs v}f(1,2,2)= \frac{2}{3}$

 

In exercises 14 - 19, find the directional derivative of the function in the direction of the unit vector $\vecs u=\cos θ \,\hat{\mathbf i}+\sin θ \,\hat{\mathbf j}$.

14) $f(x,y)=x^2+2y^2,\quad θ=\frac{π}{6}$

15) $f(x,y)=\dfrac{y}{x+2y},\quad θ=−\frac{π}{4}$

Answer:

$D_{\vecs u}f(x,y) = \dfrac{−\sqrt{2}(x+y)}{2(x+2y)^2}$

16) $f(x,y)=\cos(3x+y),\quad θ=\frac{π}{4}$

17) $w(x,y)=ye^x,\quad θ=\frac{π}{3}$

Answer:

$D_{\vecs u}f(x,y) = \dfrac{e^x(y+\sqrt{3})}{2}$

18) $f(x,y)=x\arctan(y),\quad θ=\frac{π}{2}$

19) $f(x,y)=\ln(x+2y),\quad θ=\frac{π}{3}$

Answer:

$D_{\vecs u}f(x,y) = \dfrac{1+2\sqrt{3}}{2(x+2y)}$

 

In exercises 20 - 23, find the gradient $\vecs \nabla f$.

20) Find the gradient of $f(x, y) = 3x^2 + y^3 - 3x + y$. Then find it's value at the point $P(2,3)$.

21) Find the gradient of $f(x,y)=\dfrac{14−x^2−y^2}{3}$. Then, find the gradient at point $P(1,2)$.

Answer:

$\vecs \nabla f(x, y) = -\frac{2}{3}x\,\hat{\mathbf i} -\frac{2}{3}y\,\hat{\mathbf j}$
$\vecs \nabla f(1,2) = -\frac{2}{3}\,\hat{\mathbf i} -\frac{4}{3}\,\hat{\mathbf j}$

22) Find the gradient of $f(x,y)=\ln(4x^3 - 3y)$. Then, find the gradient at point $P(1,1)$.

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

Answer:

$\vecs \nabla f(x, y, z) = ⟨y+z, x+z, y + x⟩$
$\vecs \nabla f(1,2,3) = ⟨5,4,3⟩$

 

In exercises 24 - 25, find the directional derivative of the function at point $P$ in the direction of $Q$.

24) $f(x,y)=x^2+3y^2,\quad P(1,1),\quad Q(4,5)$

25) $f(x,y,z)=\dfrac{y}{x+z},\quad P(2,1,−1),\quad Q(−1,2,0)$

Answer:

$D_{\vecd{PQ}}f(x,y) = \frac{3}{\sqrt{11}}$

 

26) Find the directional derivative of $f(x,y,z))$ at $P$ and in the direction of $\vecs u: \quad f(x,y,z)=\ln(x^2+2y^2+3z^2),\quad P(2,1,4),\quad \vecs u=\frac{−3}{13}\,\hat{\mathbf i}−\frac{4}{13}\,\hat{\mathbf j}−\frac{12}{13}\,\hat{\mathbf k}$.

 

In exercises 27 - 29, find the directional derivative of the function at $P$ in the direction of $\vecs u$.

27) $f(x,y)=\ln(5x+4y),\quad P(3,9),\quad \vecs u=6\,\hat{\mathbf i}+8\,\hat{\mathbf j}$

Answer:

$D_{\vecs u}f(3,9) = \frac{31}{255}$

28) $f(x,y)=−7x+2y,\quad P(2,−4),\quad \vecs u=4\,\hat{\mathbf i}−3\,\hat{\mathbf j}$

29) $f(x,y,z)=4x^5y^2z^3,\quad P(2,−1,1),\quad \vecs u=\frac{1}{3}\,\hat{\mathbf i}+\frac{2}{3}\,\hat{\mathbf j}−\frac{2}{3}\,\hat{\mathbf k}$

Answer:

$D_{\vecs u}f(2,-1,1) = -320$

 

30) [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 $P$.

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

Answer:

 

 

In exercises 32 - 35, find the gradient vector at the indicated point.

32) $f(x,y)=xy^2−yx^2,\quad P(−1,1)$

33) $f(x,y)=xe^y−\ln(x),\quad P(−3,0)$

Answer:

$\vecs \nabla f(-3,0) = \frac{4}{3}\,\hat{\mathbf i}−3\,\hat{\mathbf j}$

34) $f(x,y,z)=xy−\ln(z),\quad P(2,−2,2)$

35) $f(x,y,z)=x\sqrt{y^2+z^2}, \quad P(−2,−1,−1)$

Answer:

$\vecs \nabla f(-2,-1,-1) = \sqrt{2}\,\hat{\mathbf i}+\sqrt{2}\,\hat{\mathbf j}+\sqrt{2}\,\hat{\mathbf k}$

 

In exercises 36 - 40, find the indicated directional derivative of the function.

36) $f(x,y)=x^2+xy+y^2$ at point $(−5,−4)$ in the direction the function increases most rapidly.

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

Answer:

$1.6(10^{19})$

38) $f(x,y)=\arctan\left(\dfrac{y}{x}\right)$ at point $(−9,9)$ in the direction the function increases most rapidly.

39) $f(x,y,z)=\ln(xy+yz+zx)$ at point $(−9,−18,−27)$ in the direction the function increases most rapidly.

Answer:

$\frac{5\sqrt{2}}{99}$

40) $f(x,y,z)=\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$ at point $(5,−5,5)$ in the direction the function increases most rapidly.

 

In exercises 41 - 43, find the maximum rate of change of $f$ at the given point and the direction in which it occurs.

41) $f(x,y)=xe^{−y}, \quad (-2,0)$

Answer:

$\text{max}\big\{D_{\vecs u} f(-2,0)\big\} = \sqrt{5}, \quad ⟨1,2⟩$

Solution:

$\vecs \nabla f(x, y) = e^{-y} \mathbf{\hat i} - xe^{-y} \mathbf{\hat j}.$   So $\text{max}\big\{D_{\vecs u} f(-2,0)\big\} = \|\vecs \nabla f(-2, 0)\| = \sqrt{5}.$
The direction in which it occurs will just be the direction of the gradient vector: $\vecs \nabla f(-2, 0) =  ⟨1,2⟩.$

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

43) $f(x,y)=\cos(3x+2y),\quad (\frac{π}{6},−\frac{π}{8})$

Answer:

$\text{max}\big\{D_{\vecs u} f\left(\frac{π}{6},−\frac{π}{8}\right)\big\} = \sqrt{\frac{13}{2}} = \frac{\sqrt{26}}{2},\quad ⟨−3,−2⟩$

 

In exercises 44 - 47, find equations of

a. the tangent plane and

b. the normal line to the given surface at the given point.

44) The level curve $f(x,y,z)=12$ for $f(x,y,z)=4x^2−2y^2+z^2$ at point $(2,2,2).$

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

Answer:

a. tangent plane equation: $x+y+z=3$,
b. normal line equations: $x−1=y−1=z−1$

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

47) $f(x,y,z)=xe^y\cos z−z=1$ at point $(1,0,0)$

Answer:

a. tangent plane equation: $x+y−z=1$,
b. normal line equations: $x−1=y=−z$

 

In exercises 48 - 51, solve the stated problem.

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

a. Find the rate of change of the temperature at point $(1,2,2)$ in the direction toward point $(2,1,3).$

b. 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.

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

a. Find the rate of change of the voltage at point $(3,4,5)$ in the direction of the vector $⟨1,1,−1⟩.$

b. In which direction does the voltage change most rapidly at point $(3,4,5)$?

c. What is the maximum rate of change of the voltage at point $(3,4,5)$?

Answer:

a. $\frac{32}{\sqrt{3}}$,
b. $⟨38,6,12⟩$,
c. $2\sqrt{406}$

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

a. Find the electric intensity vector at $(\frac{π}{4},0).$

b. Show that, at each point in the plane, the electric potential decreases most rapidly in the direction of the vector $E.$

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

Answer:

$⟨u,v⟩=⟨π\cos(πx)\sin(2πy),2π\sin(πx)\cos(2πy)⟩$

 

Contributors

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

• Paul Seeburger (Monroe Community College) created problems 20 and 22 and added a solution to problem 41.