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. \( f(x,y)=5−2x^2−\frac{1}{2}y^2\) 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:
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\(D_{\vecs v}f(x, y) = \frac{1}{2}y + \frac{\sqrt{3}}{2}x \)
\(D_{\vecs v}f(-2, 0) = −\sqrt{3}\)
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:
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\(D_{\vecs v}f(x, y) = \frac{6}{5}xy - \frac{4}{5}x^2 \)
\(D_{\vecs v}f(-5,5)= -50\)
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:
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\(D_{\vecs u}f(x, y) = x\sqrt{3} - y \)
\(D_{\vecs u}f(1,0)= \sqrt{3}\)
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:
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\(D_{\vecs u}f(x, y) = -e^x\sin y \)
\(D_{\vecs u}f(0, \frac{π}{2})= −1\)
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:
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\(D_{\vecs u}f(x, y) = \frac{\sqrt{29}}{29}\left( \dfrac{4x-10y}{x^2 +y^2}\right) \)
\(D_{\vecs u}f(1,2)= -\frac{16\sqrt{29}}{145}\)
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:
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\( D_{\vecs v}h(x, y, z) = \frac{\sqrt{6}}{6}(2yz + xz - xy)\)
\( D_{\vecs v}h(2,1,1) = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}\)
13) \( f(x,y,z)=y^2+xz,\quad P(1,2,2),\quad \vecs v=⟨2,−1,2⟩\)
- Answer:
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\( D_{\vecs v}f(x, y, z) = \frac{2}{3}(z - y + x)\)
\(D_{\vecs v}f(1,2,2)= \frac{2}{3}\)
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:
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\( \vecs \nabla f(x, y) = -\frac{2}{3}x\,\hat{\mathbf i} -\frac{2}{3}y\,\hat{\mathbf j}\)
\( \vecs \nabla f(1,2) = -\frac{2}{3}\,\hat{\mathbf i} -\frac{4}{3}\,\hat{\mathbf j}\)
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:
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\( \vecs \nabla f(x, y, z) = ⟨y+z, x+z, y + x⟩\)
\( \vecs \nabla f(1,2,3) = ⟨5,4,3⟩\)
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
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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.