Critical Point - Multivariable Calculus - Exam, Exams of Calculus

This is the Exam of Multivariable Calculus and its key important points are: Critical Point, Saddle Point, Local Minimum, Local Maximum, Total Di Erential, Linearization, Standard Linear Approximation, Tangent Plane, Surface, Function

Typology: Exams

2012/2013

Uploaded on 02/14/2013

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1. Let z=x2ey, where x=u2v1 and y=uv 2. The partial derivative z
∂u at
(u, v) = (1,2) is
(1) 2 (2) 8 (3) 10 (4) 16
2. For the function f(x, y) = x312x+y2,
(1) there are no critical points.
(2) fhas a saddle point at the critical point (2,0).
(3) fhas a local minimum at the critical point (0,0).
(4) fhas a local maximum at the critical point (2,0).
3. Let f(x, y) = (1 + sin(2x))y3. Using the total differential (or the linearization or
standard linear approximation) of f(x, y) at the point P(0,1), one can approximate the
value of (1 + sin(0.02))(1.01)3by
(1) 1.02 (2) 1.03 (3) 1.05 (4) 1.07
4. The tangent plane to the surface z=xy2at the point P(1,1,1) contains the point
P(0,0, k) when kis
(1) 3 (2) 0 (3) 2 (4) 5
5. At the point P(1,1,1) the function
f(x, y, z) = 2x2y+z4
y2
increases fastest in the direction of the vector
(1) i+k(2) i+j+k(3) 4i(4) jk
6. The value of
Z6
0Z3
y/2
ex2dx dy
is
(1) e91 (2) e36 1 (3) 5e91 (4) e18 1
A-1
pf3
pf4

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  1. Let z = x^2 ey^ , where x = u^2 v − 1 and y = uv − 2. The partial derivative (^) ∂u∂z at (u, v) = (1, 2) is (1) 2 (2) 8 (3) 10 (4) 16
  2. For the function f (x, y) = x^3 − 12 x + y^2 ,

(1) there are no critical points. (2) f has a saddle point at the critical point (− 2 , 0). (3) f has a local minimum at the critical point (0, 0). (4) f has a local maximum at the critical point (− 2 , 0).

  1. Let f (x, y) = (1 + sin(2x))y^3. Using the total differential (or the linearization or standard linear approximation) of f (x, y) at the point P (0, 1), one can approximate the value of (1 + sin(0.02))(1.01)^3 by

(1) 1. 02 (2) 1. 03 (3) 1. 05 (4) 1. 07

  1. The tangent plane to the surface z = xy^2 at the point P (1, 1 , 1) contains the point P (0, 0 , k) when k is (1) − 3 (2) 0 (3) − 2 (4) 5
  2. At the point P (1, 1 , 1) the function

f (x, y, z) = 2x^2 y + z^4 y^2

increases fastest in the direction of the vector

(1) i + k (2) i + j + k (3) 4i (4) j − k

  1. The value of (^) ∫ (^6)

0

y/ 2

ex 2 dx dy

is (1) e^9 − 1 (2) e^36 − 1 (3) 5e^9 − 1 (4) e^18 − 1

  1. The area of the region that lies outside the cardioid r = 1 + sin θ and inside the circle r = 1 is

(1)

∫ (^) π

0

∫ (^) 1+sin θ

1

r dr dθ

∫ (^) π

0

1+sin θ

r dr dθ

−π

1+sin θ

r dr dθ

∫ (^2) π

π

∫ (^) 1+sin θ

1

r dr dθ

  1. The integral of f (x, y, z) = z over the solid in the first octant bounded by the planes y = 0, x = 0, z = 1, and z = x + y equals

0

0

0

z dx dy dz

0

0

x+y

z dz dx dy

0

∫ (^1) −y

0

x+y

z dz dx dy

0

∫ (^1) −y

0

∫ (^1) −x−y

0

z dz dx dy

  1. A solid has its volume given in cylindrical coordinates by

∫ (^2) π

0

0

∫ √ 16 −r 2

r/√ 3

r dz dr dθ.

When re-expressed in spherical coordinates, this volume equals

∫ (^2) π

0

∫ (^) π/ 3

0

0

ρ^2 sin φ dρ dφ dθ

∫ (^2) π

0

∫ (^) π/ 3

0

∫ (^4) / cos φ

0

ρ^2 sin φ dρ dφ dθ

∫ (^2) π

0

∫ (^) π/ 6

0

0

ρ^2 sin φ dρ dφ dθ

∫ (^2) π

0

∫ (^) π/ 6

0

√3 cos φ^ ρ

(^2) sin φ dρ dφ dθ

  1. Suppose that the numbers an are positive numbers satisfying lim n→∞

an+ an

= 3. Then

the open interval of convergence for the power series

∑^ ∞

n=

an 2 n^

xn

is (1) (− 6 , 6) (2) (−

  1. The first three nonzero terms of the Taylor series generated by f (x) = sin(x) with center at the point a = π 2 are

(1) (x − π 2

(x − π 2

)^3 +

(x − π 2

)^5

(x − π 2

)^2 +

(x − π 2

)^4

π 2

π 2

)^3 +

π 2

)^5

x^2 +

x^4

  1. Suppose that the series

∑^ ∞

k=

ak has partial sums sn =

∑^ n

k=

ak satisfying sn = 3 +

2 n^

Then

(1) the series

∑^ ∞

k=

ak diverges because (^) nlim→∞ sn = 3.

(2) the series

∑^ ∞

k=

ak converges to 4 because

∑^ ∞

n=

2 n^

(3) the series

∑^ ∞

k=

ak diverges because lim k→∞ ak 6 = 0.

(4) the series

∑^ ∞

k=

ak converges to 3 because (^) nlim→∞ sn = 3.