Calculus III
Practice Problems 5: Answers
1. Let f x y z x ln z 2yz. a) What is ∇ f ? b) Show that
∂2 f
∂ z∂ x
∂2 f
∂ x∂ z
Answer. Differentiate with respect to the variables:
∇f
ln zI 2zJ x
2y K
z
Now differentiate the coefficient of I with respect to z, and the coefficient of K with respect to x and compare;
∂f
∂ z∂ x
1
z
∂f
∂ x∂ z
1
z
2. Find the direction U of maximal change for the function w
DU w at this point?
x 3 y2 z xyz2 at the point (2,-1,2). What is
Answer. The direction of maximal change is that of the gradient. We calculate the gradient:
∇w 3x2 y2 z yz2 I 2x3 yz xz2 J x3 y2 2xyz K
Evaluating at the given point, we have ∇w 20I 24J. U is the unit vector in this direction, so U
5I 6J 61. At this point D U w ∇w 4 61.
3. Suppose that the function w f x y is differentiable at the point P in the plane. Let V I 2J W I J,
and that DV w 2 DW w 3 at P. What is ∇w P ?
Answer. Let ∇w P aI bJ. Then ∇w V D V w 2 ∇w W DW w 3, giving the equations
a 2b 2 a b 3
The solutions are a 8 3 b 1 3, so
∇w P 4. Suppose z 8
1
I
J
3
3
f x y is differentiable at (1,1), and suppose that
d
f 1 t 1 t 2 t dt
0
d
f 1 1 t t dt
3
0
2
What is ∇ f 1 1 ?
Answer. Let ∇ f 1 1 aI bJ. We are told the derivatives of f on the curves
γ1 : X1 t We have
1 t I dX1
dt
1 t2 J
I 2tJ γ2 : X2 t I dX2
dt
J
1 t J
so, at t0, dX1 dt
I dX2 dt J and
on γ1 :
3
df
dt
∇f dX1
dt
aI bJ I a
on γ2 :
2
df
dt
∇f dX2
dt
aI bJ I b
so ∇ f 1 1 3I 2J.
5. A particle moves in space according to the equation
X t t 2 I 1 t2 J 1 t K
For w xy yz2 xz2 , find dw dt along the trajectory. What is dw dt when t
Answer. We know that
dw
dt
∇w dX
dt
2?
Now,
dX
dt
∇w 2tI 2tJ K y z2 I x z2 J 2z x y K
Calculating the dot product:
dw
dt
At t
∇w dX
dt
y z2 2t x z2 2t 2z x y 2, we have x 4 y 3 z 1, so
dw
dt
3 1 4 4 1 4 2 1 4 3 22
6. Let w xyz and let γ be the helix given by
X t costI sintJ tk
Find dw dt at t
2π 3.
Answer. ∇w yzI xzJ xyK, and for the helix,
dX
dt
At t
sintI costJ K
2π 3 x cos 2π 3 1 2 y sin 2π 3 3 2 z 2π 3, so
∇w 3 2π
2 3
I
1 2π
2π
K
J
2 3
3
This gives
π
π
2π
I
J
K
3
3 3
dw
dt
∇w dX
dt
π
2
3
π
3
6
2π
3
dX
dt
and
2π
3
1
I
2
3
2
J K
7. Find the equation of the tangent plane to the surface
x1 y1 2 z1 2 0
2
at the point (4,9,25).
Answer. Take the differential of the defining equation:
1
x
2
dx 1 y 1 2dy 1 z 1 2 dy
1 2
2
2
Multiply by 2 and evaluate at (4,9,25):
dx
2
and replace the differentials by the increments:
dy
3
x 4 y 9
2
3
dz
5
0
z 25
5
0
When simplified, this comes down to 15x 10y 6z 0.
8. Find the equation of the tangent plane to the surface
x ln z 2yz 0
at the point
e 2 1 e2 .
Answer. Taking the differential, we have
ln zdx x
dz 2ydz 2zdy 0
z
Evaluate at the given point to get 0 dx e 2 e differentials by the increments we obtain
2
dz 2dz 2e2 dy dz 2e2 dy 0. Now replacing the
z e2 2e2 y 1 0 or 2e 2 y z 3e2