Name:
MATH 208
Solutions
Exam # 2
This test is worth 100 points. Show all of your work to receive full and partial credit, making
sure that all work and answers are legible. Books, notes, formula sheets, etc. are not allowed.
(6 points)
1. Find a vector normal to the surface −x3 + 2y 2 − z = 4 at the point (1, 0, −1).
Let f (x, y, z) = −x3 + 2y 2 − z, so that the surface is the level surface f (x, y, z) = 4.
grad f = −3x2~i + 4y~j − ~k, so gradf (1, 0, −1) = −3~i − ~k is normal to the surface at the point (1,0,-1).
(8 points)
2. Below is a contour diagram for f (x, y).
(a) Which of ∇f (0, 3) and ∇f (1, 2) is bigger in magnitude?
∇f (1, 2) is bigger in magnitude because the steepest slope at (1, 2) is steeper than the steepest
slope at (0,3). You can see this because the contour lines near (1, 2) are closer together than those
at (0, 3).
(b) Draw ∇f (0, 3) and ∇f (1, 2) on the contour diagram.
See above. Note that both gradients are perpendicular to the contour lines at their respective
points. The length of the vectors is of course approximate, but having observed in part a) that
||∇f (1, 2)|| is larger, this vector should be drawn longer.
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MATH 208
Solutions
3. Consider the function f (x, y) =
√
2x − y.
(a) (10 points) Find all four second order partial derivatives of f (x, y) at the point (3, 5).
fx = 12 (2x − y)−1/2 (2) = (2x − y)−1/2 and fy = 12 (2x − y)−1/2 (−1) = − 21 (2x − y)−1/2 ,
so,
1
fxx = − (2x − y)−3/2 (2) = −(2x − y)−3/2
2
1
1 1
fyy = − (− )(2x − y)−3/2 (−1) = − (2x − y)−3/2
2 2
4
1
1
fxy = − (2x − y)−3/2 (−1) = (2x − y)−3/2
2
2
1
fyx = fxy = (2x − y)−3/2
2
Thus,
1
fxx (3, 5) = −1, fyy (3, 5) = − ,
4
1
1
fxy (3, 5) = , fyx (3, 5) = .
2
2
(b) (7 points) Find the closest quadratic approximation of f (x, y) near the point (3, 5). (You do NOT
need to expand and collect like terms.)
Q(x, y) = f (3, 5)+fx (3, 5)(x−3)+fy (3, 5)(y−5)+
fxx (3, 5)
fyy (3, 5)
(x−3)2 +fxy (3, 5)(x−3)(y−5)+
(y−5)2 ,
2
2
so,
1
1
1
1
Q(x, y) = 1 + (x − 3) − (y − 5) − (x − 3)2 + (x − 3)(y − 5) − (y − 5)2 .
2
2
2
8
(12 points)
4. Given z = 2xey , x = u2 + v 2 , y = u2 − v 2 , use the chain rule to find ∂z/∂v.
∂z
∂z ∂x ∂z ∂y
=
+
∂v
∂x ∂v
∂y ∂v
2
= 2ey (2v) + 2xey (−2v) = 4veu
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−v 2
− 4v(u2 + v 2 )eu
2
−v 2
MATH 208
Solutions
5. Let f (x, y) = 4 + x3 + y 3 − 3xy, with contour diagram shown below.
(a) (5 points) Use the diagram to predict the location of the critical points and whether each is a local
maximum, local minimum, or saddle point.
(0,0) looks like a saddle point.
(1,1) looks like a local minimum.
(b) (12 points) Use the formula given for f (x, y) and the Second Derivative Test to confirm your
predictions.
fx = 3x2 − 3y and fy = 3y 2 − 3x, so fxx = 6x, fxy = −3, and fyy = 6y.
Thus D = fxx fyy − (fxy )2 = 36xy − 9.
At (0, 0), D = −9 < 0, so (0, 0) is a saddle point.
At (1, 1), D = 36 − 9 = 27 > 0 and fxx (1, 1) = 6 > 0, so (1, 1) is a local minimum.
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MATH 208
Solutions
6. (a) (10 points) Use Lagrange multipliers to find the points (x0 , y0 ) at which f (x, y) = 3x − 4y has a
maximum or minimum subject to the constraint x2 + y 2 = 1.
Let g(x, y) = x2 + y 2 , so the constraint is g(x, y) = 1. Then ∇f = 3~i − 4~j and ∇g = 2x~i + 2y~j,
So we have 3~i − 4~j = λ(2x~i + 2y~j), giving 3 = λ2x and −4 = λ2y.
Solving for lambda gives:
2
3
=− ,
λ=
2x
y
and by cross multiplying we have:
4
3y = −4x, so y = − x.
3
Then
4
4
16
16
25
1 = g(x, − x) = x2 + (− x)2 = x2 + x2 = x2 (1 + ) = x2 ( ).
3
3
9
9
9
so
x2 =
and thus
9
25
3
x=± .
5
But then y = − 34 (± 35 ), so the points are ( 35 , − 45 ) and (− 53 , 45 ).
(b) (3 points) The maximum of f (x, y) subject to this constraint is:
3 4
3
4
25
f ( , − ) = 3( ) − 4(− ) =
= 5.
5 5
5
5
5
(c) (3 points) The minimum of f (x, y) subject to this constraint is:
3 4
3
4
25
f (− , ) = 3(− ) − 4( ) = −
= −5.
5 5
5
5
5
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MATH 208
Solutions
1
Z
Z
7. Consider the iterated integral
0
1
sin(x2 )dxdy.
y
(a) (8 points) Sketch the region of integration.
(b) (8 points) Reverse the order of integration.
Z
1
Z
0
x
sin(x2 )dydx
0
(c) (8 points) Evaluate the integral you found in part b). Write your answer as a decimal rounded to
two decimal places. (You may use the approximation cos(1) ≈ 0.54.)
Z
0
1
Z
0
x
sin(x2 )dydx =
Z
1
y sin(x2 )|x0 dx =
0
Z
1
x sin(x2 )dx
0
Using u-substitution with u = x2 we have:
Z
1
1 x=1
sin(u)du = − cos(x2 )|10
2 x=0
2
1
1
1
= − (cos(1) − cos(0)) ≈ − (0.54 − 1) = − (−0.46) = 0.23
2
2
2
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