SOLUTIONS MATH 200 MIDTERM 3.
(10)I. The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out.
The resistance R is slowly increasing as the resistor heats up. Use Ohm’s Law, V = IR, to find
how the current I is changing at the moment when R = 1Ω, I = 0.08 A, dV /dt = −0.01 V /s, and
dR/dt = 0.03 Ω/s. Give a correct numerical expression for dI/dt but there is no need to simplify
completely.
Use the chain rule:
dV
dI
dR
dI
dR
= VI
+ VR
=R +I
dt
dt
dt
dt
dt
Substitute the given values to get
dI
+ 0.08(0.03)
−0.01 =
dt
dI
= −0.0124 A/s
dt
(5)II. Given x2 y 3 + xz + exyz = 10, use implicit differentiation to find ∂z/∂x and ∂x/∂z.
Use the formulas
∂z
Fx
Fz
∂x
=−
=−
and
∂x
Fz
∂z
Fx
So
and
2xy 3 + z + yz exyz
∂z
=−
∂x
x + xy exyz
∂x
x + xy exyz
=−
∂z
2xy 3 + z + yz exyz
(10)III. Use any method to find the point or points on the surface z = x2 /4 + y 2 that are closest to
the point (0, 0, 1). This one needs to be simplified.
The square of the distance from the general point (x, y, z) to (0, 0, 1) is f (x, y, z) = x2 +y 2 +(z −1)2 .
Substituting the value y 2 = z − x2 /4 gives g(x, z) = x2 + z − x2 /4 + (z − 1)2 . Simplify to get
g(x, z) = 3x2 /4 + z 2 − z + 1. The minimum occurs at a critical point. The partial derivatives are
gx = 3x/2 and gz = 2z − 1.
Setting the partial derivatives to zero gives 3x/2 = 0 and 2z − 1 = 0. So the critical point for g(x, z)
2
is (0, 1/2). Substituting these values of x and z into
√ the equation of the√surface gives 1/2 = y and
so there are two critical points on √
the surface, (0, 1/ 2, 1/2) and (0, −1/ 2, 1/2). The distance from
each of these points to (0, 0, 1) is 3/2. So both points are at the minumum distance from (0, 0, 1)
(15)IV. Let g(x, y, z) = xy + yz 2 and answer these questions about g. Simplify these answers.
a) What is the directional derivative of g at the point (3, 1, 2) in the direction of the vector
< 6, 3, 2 >,
b) As ~u varies what is the largest value of the directional derivative D~u g(3, 1, 2),
c) Which unit vector gives the smallest value for the directional derivative D~u g(3, 1, 2),
d) Give an equation for the plane that contains the origin and is parallel to all unit vectors with
D~u g(3, 1, 2) = 0,
e) Find the equation of the tangent plane to the level surface xy + yz 2 = 7 at the point (3, 1, 2).
a) ∇g = y~ı + (x + z 2 )~ + 2yz~k so ∇g(3, 1, 2) = ~ı + 7~ + 4~k The unit vector in the direction of h6, 3, 2i
is 1/7h6, 3, 2i. The directional derivative is the dot product ∇g(3, 1, 2) · ~u = 5
√
√
b)The largest directional derivative is k∇g(6, 3, 2)k = 1 + 72 + 42 = 66
c)The smallest value for the directional
derivative occurs when the unit vector points in the direction
√
opposite to the gradient, ~u = −1/ 66h1, 7, 4i
d) x + 7y + 4z = 0
e) (x − 3) + 7(y − 1) + 4(z − 2) = 0
(10)V. Find the critical points of f (x, y) = 2x3 + xy 2 + 5x2 + y 2 (there are four). Classify each critical
point as a local maximum, local minimum or saddle point.
First find the critical points. They are the solutions of ∇f = ~0. Now ∇f = (6x2 + y 2 + 10x)~ı +
(2xy + 2y)~.
First, 2xy + 2y = 2y(x + 1) = 0 gives y = 0 or x = −1.
If y = 0 then 6x2 + 0 + 10x = 2x(3x + 5) = 0 so x = 0 or x = −5/3, giving two critical points (0, 0)
and (−5/3, 0).
If x = −1 then 6(−1)2 + y 2 − 10 = y 2 − 4 = 0 so y = −2 or y = 2, giving two more critical points
(−1, 2) and (−1, −2).
The second order partial derivatives are fxx = 12x + 10, fxy = 2y and fyy = 2x + 2. The values of the
second order partials and Hessian at the critical points (0, 0), (−5/3, 0), (−1, 2) and (−1, −2) are
(0, 0) (−5/3, 0) (−1, 2) (−1, −2)
12x + 10
10
−10
−2
−2
2y
0
0
4
−4
2x + 2
2
−2
0
0
H(x, y) = 20
20
−16
−16
So, (0, 0) is a local minimum, (−5/3, 0) is a local maximmum, (−1, 2) and (−1, −2) are saddle points