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Calculate D u f ( x , y , z ) and D u f ( 0 , −2 , 5 ) in the direction of v = −3 i + 12 j 4 k for the function f ( x , y , z ) = 3 x 2 + x y 2 y 2 + 4 y z z 2 + 2 x z .

D u f ( x , y , z ) = 3 13 ( 6 x + y + 2 z ) + 12 13 ( x 4 y + 4 z ) 4 13 ( 2 x + 4 y 2 z ) D u f ( 0 , −2 , 5 ) = 384 13

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Key concepts

  • A directional derivative represents a rate of change of a function in any given direction.
  • The gradient can be used in a formula to calculate the directional derivative.
  • The gradient indicates the direction of greatest change of a function of more than one variable.

Key equations

  • directional derivative (two dimensions)
    D u f ( a , b ) = lim h 0 f ( a + h cos θ , b + h sin θ ) f ( a , b ) h
    or
    D u f ( x , y ) = f x ( x , y ) cos θ + f y ( x , y ) sin θ
  • gradient (two dimensions)
    f ( x , y ) = f x ( x , y ) i + f y ( x , y ) j
  • gradient (three dimensions)
    f ( x , y , z ) = f x ( x , y , z ) i + f y ( x , y , z ) j + f z ( x , y , z ) k
  • directional derivative (three dimensions)
    D u f ( x , y , z ) = f ( x , y , z ) · u = f x ( x , y , z ) cos α + f y ( x , y , z ) cos β + f x ( x , y , z ) cos γ

For the following exercises, find the directional derivative using the limit definition only.

f ( x , y ) = 5 2 x 2 1 2 y 2 at point P ( 3 , 4 ) in the direction of u = ( cos π 4 ) i + ( sin π 4 ) j

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f ( x , y ) = y 2 cos ( 2 x ) at point P ( π 3 , 2 ) in the direction of u = ( cos π 4 ) i + ( sin π 4 ) j

−3 3

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Find the directional derivative of f ( x , y ) = y 2 sin ( 2 x ) at point P ( π 4 , 2 ) in the direction of u = 5 i + 12 j .

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For the following exercises, find the directional derivative of the function at point P in the direction of v .

f ( x , y ) = x y , P ( 0 , −2 ) , v = 1 2 i + 3 2 j

−1

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h ( x , y ) = e x sin y , P ( 1 , π 2 ) , v = i

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h ( x , y , z ) = x y z , P ( 2 , 1 , 1 ) , v = 2 i + j k

2 6

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f ( x , y ) = x y , P ( 1 , 1 ) , u = 2 2 , 2 2

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f ( x , y ) = x 2 y 2 , u = 3 2 , 1 2 , P ( 1 , 0 )

3

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f ( x , y ) = 3 x + 4 y + 7 , u = 3 5 , 4 5 , P ( 0 , π 2 )

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f ( x , y ) = e x cos y , u = 0 , 1 , P = ( 0 , π 2 )

−1.0

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

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f ( x , y ) = ln ( x 2 + y 2 ) , u = 3 5 , 4 5 , P ( 1 , 2 )

22 25

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f ( x , y ) = x 2 y , P ( −5 , 5 ) , v = 3 i 4 j

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f ( x , y ) = y 2 + x z , P ( 1 , 2 , 2 ) , v = 2 , −1 , 2

2 3

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For the following exercises, find the directional derivative of the function in the direction of the unit vector u = cos θ i + sin θ j .

f ( x , y ) = x 2 + 2 y 2 , θ = π 6

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f ( x , y ) = y x + 2 y , θ = π 4

2 ( x + y ) 2 ( x + 2 y ) 2

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f ( x , y ) = cos ( 3 x + y ) , θ = π 4

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w ( x , y ) = y e x , θ = π 3

e x ( y + 3 ) 2

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f ( x , y ) = x arctan ( y ) , θ = π 2

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f ( x , y ) = ln ( x + 2 y ) , θ = π 3

1 + 2 3 2 ( x + 2 y )

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For the following exercises, find the gradient.

Find the gradient of f ( x , y ) = 14 x 2 y 2 3 . Then, find the gradient at point P ( 1 , 2 ) .

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Find the gradient of f ( x , y , z ) = x y + y z + x z at point P ( 1 , 2 , 3 ) .

5 , 4 , 3

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Find the gradient of f ( x , y , z ) at P and in the direction of u : f ( x , y , z ) = ln ( x 2 + 2 y 2 + 3 z 2 ) , P ( 2 , 1 , 4 ) , u = −3 13 i 4 13 j 12 13 k .

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f ( x , y , z ) = 4 x 5 y 2 z 3 , P ( 2 , −1 , 1 ) , u = 1 3 i + 2 3 j 2 3 k

−320

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For the following exercises, find the directional derivative of the function at point P in the direction of Q .

f ( x , y ) = x 2 + 3 y 2 , P ( 1 , 1 ) , Q ( 4 , 5 )

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

3 11

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For the following exercises, find the derivative of the function at P in the direction of u .

f ( x , y ) = −7 x + 2 y , P ( 2 , −4 ) , u = 4 i 3 j

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f ( x , y ) = ln ( 5 x + 4 y ) , P ( 3 , 9 ) , u = 6 i + 8 j

31 255

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

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

The top of half of an ellipse centered at the origin with major axis horizontal and of length 4 and minor axis 2. The point (–2, 0) is marked, and there is an arrow pointing out from it to the left marked –4i.

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For the following exercises, find the gradient vector at the indicated point.

f ( x , y ) = x y 2 y x 2 , P ( −1 , 1 )

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f ( x , y ) = x e y ln ( x ) , P ( −3 , 0 )

4 3 i 3 j

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f ( x , y , z ) = x y ln ( z ) , P ( 2 , −2 , 2 )

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f ( x , y , z ) = x y 2 + z 2 , P ( −2 , −1 , −1 )

2 i + 2 j + 2 k

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For the following exercises, find the derivative of the function.

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

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

1.6 ( 10 19 )

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f ( x , y ) = arctan ( y x ) at point ( −9 , 9 ) in the direction the function increases most rapidly

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f ( x , y , z ) = ln ( x y + y z + z x ) at point ( −9 , −18 , −27 ) in the direction the function increases most rapidly

5 2 99

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f ( x , y , z ) = x y + y z + z x at point ( 5 , −5 , 5 ) in the direction the function increases most rapidly

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For the following exercises, find the maximum rate of change of f at the given point and the direction in which it occurs.

f ( x , y ) = x e y , ( 1 , 0 )

5 , 1 , 2

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f ( x , y ) = x 2 + 2 y , ( 4 , 10 )

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f ( x , y ) = cos ( 3 x + 2 y ) , ( π 6 , π 8 )

13 2 , −3 , −2

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For the following exercises, find equations of

  1. the tangent plane and
  2. the normal line to the given surface at the given point.

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

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f ( x , y , z ) = x y + y z + x z = 3 at point ( 1 , 1 , 1 )

a. x + y + z = 3 , b. x 1 = y 1 = z 1

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

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f ( x , y , z ) = x e y cos z z = 1 at point ( 1 , 0 , 0 )

a. x + y z = 1 , b. x 1 = y = z

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For the following exercises, solve the problem.

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 .

  1. Find the rate of change of the temperature at point ( 1 , 2 , 2 ) in the direction toward point ( 2 , 1 , 3 ) .
  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.
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The electrical potential (voltage) in a certain region of space is given by the function V ( x , y , z ) = 5 x 2 3 x y + x y z .

  1. Find the rate of change of the voltage at point ( 3 , 4 , 5 ) in the direction of the vector 1 , 1 , −1 .
  2. In which direction does the voltage change most rapidly at point ( 3 , 4 , 5 ) ?
  3. What is the maximum rate of change of the voltage at point ( 3 , 4 , 5 ) ?

a. 32 3 , b. 38 , 6 , 12 , c. 2 406

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If the electric potential at a point ( x , y ) in the xy -plane is V ( x , y ) = e −2 x cos ( 2 y ) , then the electric intensity vector at ( x , y ) is E = V ( x , y ) .

  1. Find the electric intensity vector at ( π 4 , 0 ) .
  2. Show that, at each point in the plane, the electric potential decreases most rapidly in the direction of the vector E .
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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 = φ . Find the velocity components associated with the velocity potential φ ( x , y ) = sin π x sin 2 π y .

u , v = π cos ( π x ) sin ( 2 π y ) , 2 π sin ( π x ) cos ( 2 π y )

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Practice Key Terms 2

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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