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# 3.13: Directional Derivatives and the Gradient (Exercises)

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## 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 $$\vecs u=(\cos\frac{π}{4})\,\hat{\mathbf i}+(\sin\frac{π}{4})\,\hat{\mathbf j}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$D_{\vecs u}f(x, y) = \frac{\sqrt{29}}{29}\left( \frac{4x+10y}{x^2 +y^2}\right)$$
$$D_{\vecs u}f(1,2)= \frac{24\sqrt{29}}{145}$$

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

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

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

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

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

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

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

17) $$w(x,y)=ye^x,\quad θ=\frac{π}{3}$$

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

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

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

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

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

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

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 $$P(1,2,3).$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

$$1.6(10^{19})$$

38) $$f(x,y)=\arctan(\frac{y}{x})$$ 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.

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

40) $$f(x,y,z)=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$ 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) $$f(x,y)=xe^{−y}, \quad (1,0)$$

$$\sqrt{5}, \quad ⟨1,2⟩$$

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

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

$$\sqrt{\frac{13}{2}},\quad ⟨−3,−2⟩$$

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 $$(1,1,1)$$

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 $$(1,0,0)$$

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 $$(3,4,5)$$?

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 $$E=−\vecs ∇V(x,y).$$

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 $$⟨u,v⟩=\vecs ∇φ$$. Find the velocity components associated with the velocity potential $$φ(x,y)=\sin πx\sin 2πy.$$

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