Unformatted text preview: Answer Key 4
MATH 20550: Calculus III
Exam II Name: October 26, 2010 Instructor: As a member of the Notre Dame community, I will not participate in or tolerate academic dishonesty.
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Record your answers to the multiple choice problems by placing an × through one letter for each problem
on this page. There are 10 multiple choice questions worth 6 points each and 3 partial credits problems
worth 14 points each. On the partial credit problems try to simplify your answer and indicate your ﬁnal
answer clearly. You must show your work and all important steps to receive credit.
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DO NOT WRITE IN THIS COLUMN Multichoice
11
12
13
Total 1. a • c d e 6. a b c • e 2. • b c d e 7. a b • d e 3. a • c d e 8. a b • d e 4. a b c • e 9. a b c d • 5. • b c d e 10. a • c d e 1 1 Practice Exam 1
Problem (#11). Evaluate D x dA where D is the region bounded by the
curves y = x and y = x2 − x.
Solution. The region D looks like this: and the intersections of the to curves occur at x = 0 and x = 2. (See this
by solving the equation x = x2 − x.) Thus the integral we need to compute
is
2 x x dy dx 0 So, let’s do it:
2x
0 x2 −x x dy dx =
=
=
=
=
= x2 − x
2 x
0
x2 −x x dy dx
2 x
x x2 −x dy dx
02
x[x − (x2 − x)]dx
02 2
3
0 x − x dx
2 x3
x4 2
3 −4 0 because x is a const. with resp. to dy 4
3 Problem (#12). Find the absolute maximum value of the function f (x, y ) =
x2 y 3 on the region D = {(x, y )  x2 + y 3 ≤ 1}
Solution. 1. We ﬁrst ﬁnd the critical points of f (x, y ). We compute
∇f = 2xy 3 , 3x2 y 2
This is deﬁned for all x, y , so the only critical points are where ∇f (x, y ) =
0.
Now, ∇f = when at least one of x or y is 0. Noting that f (x, 0) = f (0, y ) =
0
0 for all x, y , we know that none of these points (x, 0) or (0, y ) yields the
1 absolute maximum we seek; f (x, y ) does attain positive values on the region
D.
2. We ﬁnd the the maxima of f (x, y ) on the boundary of D,
∂ D = (x, y ) : x2 + y 3 = 1 This means we need to solve the following constrained optimization problem:
Maximize f (x, y ),
Subject to g (x, y ) = x2 + y 3 = 1.
To solve this, we will use Lagrange multipliers. Setting ∇f = λ∇g , ﬁnd
ourselves with the following system of equations:
2xy 3 = 2λx
3x2 y 2 = 3λy 2
x2 + y 3 = 1
We know that points (x, y ) with x = 0 or y = 0 are irrelevant (they cannot
yield the maximum), so we can assume from now on that x = 0 and y = 0; in
particular, we can divide by x or y whenever we want. With this assumption,
the system yields that,
2y 3 = 2λ, 3x2 = 3λ
and then that
x2 = λ = y 3
Substituting these into the equation x2 + y 3 = 1, we get
2x2 = 1, 2y 3 = 1
2+ 3
So, the extrema of f (x, y ) on boundary {(x, y )  x y= 1} occur at
the
the points ( 1 , 3 1 ) and (− 1 , 3 1 ). Moreover, f ( 1 , 3 1 ) = 1/4 and
2
2
2
2
2
2
f ( − 1 , 3 1 ) = 1/4, so both points are absolute maxima of f on D, and
2
2
the maximum value is 1/4. Problem (#13). Find the tangent plane to the surface x2 + y 3 − xy = z 2 + x
at the point (1, 1, 0). 2 Solution. Let f (x, y, z ) = x2 + y 3 − xy − z 2 − x. The surface we are interested
in is really the level surface f (x, y, z ) = 0, and f (1, 1, 0) = 0 – that is, the
point (1, 1, 0) is on this surface. It follows that the gradient vector ∇f (1, 1, 0)
is orthogonal to the tangent plane of the surface f (x, y, z ) = 0 at (1, 1, 0).
Moreover, this tangent plane is determined by the base point r0 = 1, 1, 0
and normal vector n = ∇f (1, 1, 0).
We compute:
∇f = 2x − y − 1, 3y 2 − x, −2z
∇f (1, 1, 0) = 0, 2, 0
A vector equation of the tangent plane, then, is
0, 2, 0 · x − 1, y − 1, z = 0
which can be seen to be equivalent to the equation y = 1 once you compute
the dot product. 3 ...
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