Calculus Final Exam June 25, 2012
There are 9 questions with total 120 points in this exam.
1. (12 points) Find the interval of convergence of the series X∞
k=1
1 kxk.
Solution: Since 1 > lim
k→∞
¯¯
¯¯
¯¯
¯ 1 k + 1xk+1
1 kxk
¯¯
¯¯
¯¯
¯
= lim
k→∞
k
k + 1|x| = |x|, the radius of convergence is 1.
At x = 1, the series X∞
k=1
1
k diverges since it is a p−series with p ≤ 1.
At x = −1, the series X∞ k=1
1
k(−1)k = − X∞ k=1
1
(2k − 1)(2k) converges by comparing with the convergent series P∞
k=1
1 k2.
Therefore, the interval of convergence is [−1, 1).
2. Let f (x, y) =
2xy
x2+ y2, if (x, y) 6= (0, 0) 0 , if (x, y) = (0, 0).
(a) (10 points) Does the limit lim
(x,y)→(0,0) f (x, y) exist? Justify your answer.
Solution: Since lim
(x,x)→(0,0) f (x, x) = lim
(x,x)→(0,0)2 = 2 6= 0 = lim
(x,0)→(0,0)0 = lim
(x,0)→(0,0) f (x, 0), the limit does not exist.
(b) (10 points) Find ∂f
∂x(0, 0) if it exists.
Solution: Since lim
h→0
f (0 + h, 0) − f (0, 0)
h = lim
h→0
0
h = 0, we get ∂f
∂x(0, 0) = 0.
3. (10 points) Find the directional derivative of the function f (x, y) = x2 + 3y2 at the point (1, 1) in the direction of the vector (1, −1).
Solution: Since ∇f (x, y) = (2x, 6y), ∇f (1, 1) = (2, 6), and the directional derivative of f at the point (1, 1) in the direction of the vector (1, −1) is
f(1,−1)0 (1, 1) = ∇f (1, 1) · (1, −1)
√2 = (2, 6) ·(1, −1)
√2 = −2√ 2.
Calculus Final Exam (Continued) June 25, 2012
4. (12 points) Find the stationary points and the local extreme values of the function f (x, y) = 2x2 + y2− xy − 7y.
Solution: Solving ∇f (x, y) = (4x − y, 2y − x − 7) = (0, 0), we get the stationary point is (1, 4).
The second partials are
∂2f
∂x2 = 4, ∂2f
∂y∂x = −1 = ∂2f
∂x∂y, ∂2f
∂y2 = 2.
Thus, A = 4, B = −1, C = 2, and D = AC − B2 = 7 > 0. Since A > 0, it follows from the second-partials test that f (1, 4) = 2 + 16 − 4 − 28 = −14 is a local minimum.
5. (14 points) Use the method of Lagrange to find the maximum value of the function f (x, y) = xy on the circle x2 + y2 = 1.
Solution: By setting g(x, y) = x2 + y2− 1, we want to maximize f (x, y) = xy subject to the side condition g(x, y) = 0.
The gradients are
∇f (x, y) = (y, x), ∇g(x, y) = (2x, 2y).
Setting ∇f (x, y) = λ∇g(x, y), we have
y = 2λx, x = 2λy.
Multiplying the first equation by y and the second equation by x, we find that y2 = 2λxy, x2 = 2λxy
and therefore
y2 = x2.
The side condition x2+ y2 = 1 implies that 2x2 = 1 and therefore x = ±
√2
2 . The only points that can give rise to an extreme value are
(
√2 2 ,
√2
2 ), (−√ 2 2 ,
√2
2 ), (−
√2 2 , −
√2 2 ), (
√2 2 , −
√2 2 ), and f achieves the maximum value 1
2 at the first and third points.
6. (12 points) Determine whether or not the vector function (P (x, y), Q(x, y)) = (xy2, x2y) is a gradient. If so, find all the functions f with its gradient ∇f = (P, Q).
Solution: Since Py = 2xy = Qx, the vector function
(P (x, y), Q(x, y)) = (xy2, x2y) is a gradient of some function f, and f =
Z
y=constant
xy2dx = x2y2
2 + g(y), where g is a function of y to be determined.
Since x2y = Q = fy = x2y + g0(y), we get g0(y) = 0, and f = x2y2
2 + C, where C is an arbitrary constant.
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Calculus Final Exam (Continued) June 25, 2012
7. (a) (6 points) Let Ω be the triangular region bounded by y = x, y = 0, x = 1.
Sketch Ω and fill in the blanks of the following double integral Z ¤
¤
Z ¤
¤
ex2dy dx.
Solution: R1
0
Rx
0 ex2dy dx.
(b) (6 points) Evaluate the double integral Z 1
0
Z 1
y
ex2dx dy.
Solution: By part (a), we have Z 1
0
Z 1
y
ex2dx dy = Z 1
0
Z x
0
ex2dy dx = Z 1
0
xex2dx = ex2
2 |10 = e − 1 2 .
8. (a) (6 points) Find the Jacobian of the transformation x = r cos θ, and y = r sin θ.
Recall that the Jacobian is the determinant of the square matrix J(r, θ) =
¯¯
¯¯
¯¯
¯
∂x
∂r
∂y
∂x ∂r
∂θ
∂y
∂θ
¯¯
¯¯
¯¯
¯ .
Solution: The Jacobian is J(r, θ) =
¯¯
¯¯
¯¯
¯¯
¯
∂x
∂r
∂y
∂r
∂x
∂θ
∂y
∂θ
¯¯
¯¯
¯¯
¯¯
¯
=
¯¯
¯¯
¯¯
cos θ sin θ
−r sin θ r cos θ
¯¯
¯¯
¯¯= r
(b) (8 points) Evaluate Z ∞
0
e−x2dx.
Solution: Set I = Z ∞
0
e−x2dx.
Then I2 =¡Z ∞
0
e−x2dx¢ ¡Z ∞
0
e−y2dy¢
= Z ∞
0
Z ∞
0
e−x2−y2dx dy
= Z π/2
0
Z ∞
0
e−r2r dr dθ = −¡e−r2 2 |∞0 ¢ ¡
θ|π/20 ¢
= π 4. Thus,
Z ∞
0
e−x2dx =
√π 2 .
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Calculus Final Exam (Continued) June 25, 2012 9. (a) (6 points) Find the Jacobian of the transformation x = u + v
2 , and y = u − v 2 .
Solution: The Jacobian is J(u, v) =
¯¯
¯¯
¯¯
¯¯
¯
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
¯¯
¯¯
¯¯
¯¯
¯
=
¯¯
¯¯
¯¯
¯¯
¯ 1 2
1 2 1 2 −1
2
¯¯
¯¯
¯¯
¯¯
¯
= −1 2
(b) (8 points) Evaluate Z Z
Ω
¡x + y¢ ¡ x − y¢
dxdy, where Ω is the parallelogram bounded by x + y = 0, x + y = 1, x − y = 0, x − y = 2.
Solution: Setting u = x + y, and v = x − y, we get x = u + v
2 , and y = u − v 2 . Then the Jacobian is J = −1
2 and Z Z
Ω
¡x+y¢ ¡ x−y¢
dxdy = Z 2
0
Z 1
0
u v |J|du dv = 1 2
Z 2
0
Z 1
0
u v du dv = 1 8
¡u2|10)¡ v2|20¢
= 1 2.
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