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2.7E: 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)=52x212y2 at point P(3,4) in the direction of u=(cosπ4)ˆi+(sinπ4)ˆj

Answer:
a. Duf(3,4)=82

1) b. f(x,y)=y2cos(2x) at point P(π3,2) in the direction of u=(cosπ4)ˆi+(sinπ4)ˆj

Answer:
b. Duf(π3,2)=262

2) Find the directional derivative of f(x,y)=y2sin(2x) at point P(π4,2) in the direction of u=5ˆi+12ˆj.

 

In exercises 3 - 13, find the directional derivative of the function in the direction of 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,P(2,0),v=12ˆi+32ˆj

Answer:
Dvf(x,y)=12y+32x
Dvf(2,0)=3

4) h(x,y)=exsiny,P(1,π2),v=ˆi

5) f(x,y)=x2y,P(5,5),v=3ˆi4ˆj

Answer:
Dvf(x,y)=65xy45x2
Dvf(5,5)=50

6) f(x,y)=xy,P(1,1),u=22,22

7) f(x,y)=x2y2,P(1,0),u=32,12

Answer:
Duf(x,y)=x3y
Duf(1,0)=3

8) f(x,y)=3x+4y+7,P(0,π2),u=35,45

9) f(x,y)=excosy,P=(0,π2),u=0,5

Answer:
Duf(x,y)=exsiny
Duf(0,π2)=1

10) f(x,y)=y10,u=0,3,P=(1,1)

11) f(x,y)=ln(x2+y2),u=2,5,P(1,2)

Answer:
Duf(x,y)=2929(4x10yx2+y2)
Duf(1,2)=1629145

12) h(x,y,z)=xyz,P(2,1,1),v=2ˆi+ˆjˆk

Answer:
Dvh(x,y,z)=66(2yz+xzxy)
Dvh(2,1,1)=26=63

13) f(x,y,z)=y2+xz,P(1,2,2),v=2,1,2

Answer:
Dvf(x,y,z)=23(zy+x)
Dvf(1,2,2)=23

 

In exercises 14 - 19, find the directional derivative of the function in the direction of the unit vector u=cosθˆi+sinθˆj.

14) f(x,y)=x2+2y2,θ=π6

15) f(x,y)=yx+2y,θ=π4

Answer:
Duf(x,y)=2(x+y)2(x+2y)2

16) f(x,y)=cos(3x+y),θ=π4

17) w(x,y)=yex,θ=π3

Answer:
Duf(x,y)=ex(y+3)2

18) f(x,y)=xarctan(y),θ=π2

19) f(x,y)=ln(x+2y),θ=π3

Answer:
Duf(x,y)=1+232(x+2y)

 


In exercises 20 - 23, find the gradient f.

20) Find the gradient of f(x,y)=3x2+y33x+y. Then find it's value at the point P(2,3).

21) Find the gradient of f(x,y)=14x2y23. Then, find the gradient at point P(1,2).

Answer:
f(x,y)=23xˆi23yˆj
f(1,2)=23ˆi43ˆj

22) Find the gradient of f(x,y)=ln(4x33y). 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:
f(x,y,z)=y+z,x+z,y+x
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)=x2+3y2,P(1,1),Q(4,5)

25) f(x,y,z)=yx+z,P(2,1,1),Q(1,2,0)

Answer:
DaPQf(x,y)=311

 

26) Find the directional derivative of f(x,y,z)) at P and in the direction of u:f(x,y,z)=ln(x2+2y2+3z2),P(2,1,4),u=313ˆi413ˆj1213ˆk.

 

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

27) f(x,y)=ln(5x+4y),P(3,9),u=6ˆi+8ˆj

Answer:
Duf(3,9)=31255

28) f(x,y)=7x+2y,P(2,4),u=4ˆi3ˆj

29) f(x,y,z)=4x5y2z3,P(2,1,1),u=13ˆi+23ˆj23ˆk

Answer:
Duf(2,1,1)=320

 

30) [T] Use technology to sketch the level curve of f(x,y)=4x2y+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)=x2+4y2 that passes through P(2,0) and draw the gradient vector at P.

Answer:
Sketch of the level curve of f(x,y)=x^2+4y^2 that passes through P(−2,0) and showing the gradient vector at P.

 

 

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

32) f(x,y)=xy2yx2,P(1,1)

33) f(x,y)=xeyln(x),P(3,0)

Answer:
f(3,0)=43ˆi3ˆj

34) f(x,y,z)=xyln(z),P(2,2,2)

35) f(x,y,z)=xy2+z2,P(2,1,1)

Answer:
f(2,1,1)=2ˆi+2ˆj+2ˆk

 

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

36) f(x,y)=x2+xy+y2 at point (5,4) in the direction the function increases most rapidly.

37) f(x,y)=exy at point (6,7) in the direction the function increases most rapidly.

Answer:
1.6(1019)

38) f(x,y)=arctan(yx) 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:
5299

40) f(x,y,z)=xy+yz+zx 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)=xey,(2,0)

Answer:
max{Duf(2,0)}=5,1,2
Solution:
f(x,y)=eyˆixeyˆj.   So max{Duf(2,0)}=
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.

This page titled 2.7E: Directional Derivatives and the Gradient (Exercises) is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.

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