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Practice Final Exam SP 20
Multivariate Calculus (Drexel University)
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The Final Exam is comprehensive and covers all topics of the course. Total time of the
exam is 2 hours +1 hour to make a copy/picture of the writing paper and upload it on the
Black Board or send by email)
No calculators as well as other computing devices are allowed.
In order to obtain full credit for the Free Response problems all your work (step by step)
must be shown. Answers without work will be graded by score of zero.
For Multiple Choice Questions: show some steps of your solution. Partial credit will be
awarded for significant progress towards the correct answer. Only 50% of the credit will
be given for unsupported correct answers.
I. Free Response Problems
1.
Given vectors a = i + 4 j + 8 k and b = i + 2 j − 2 k .
1.1. Express the vector a as the sum of a vector parallel to b and a vector perpendicular
to b .
1.2. Write an equation of the plane that is parallel both vectors a and b and passes
through the point A (0, 1, 2) .
2
2
2
2 . Let f ( x, y, z ) = 2 ( x − 3) + ( y − 3) + ( z − 7 ) , M (4, 5, 9).
.
2.1. Find the rate of change of the function f ( x, y , z ) at the point M in the direction of
the vector l = 3 i + 2 j − 6 k .
2.2. Find a unite vector in the direction in which f ( x, y , z ) increases most rapidly at M
and find the rate of change of f ( x, y , z ) at M in that direction.
2.3. Find the equation for the level surface S of f ( x, y , z ) that passes through the point M..
Describe the surface.
2.4. Find the equation of the tangent plane to the surface S at the point M.
1
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3. Identify all critical points of the function f ( x, y ) = 3 x 2 − x 3 + 3 y 2 + 4 y and classify
each critical point as a relative maximum, relative minimum, or saddle point.
4. Use the double integral in polar coordinates to find the volume of the solid which is
inside of the cylinder x 2 + y 2 = 9 above the plane z = 0 and under the plane z = 3 − x .
5. Use the transformation
u = x − 2y
v = 3x− y
to compute the area of the region D on xy-plane
that is bounded by lines
x − 2 y = 0, x − 2 y = 3, 3 x − y = 1, 3 x − y = 7 .
1
2
u + v
5
5 .
Hint: the inverse transformation is
3
1
y=− u + v
5
5
x= −
II. Multiple Choice Section
6. Which of the following points belongs to the tangent line to the
curve γ 1 : r1 (t) = (3 + t) i + (1 − t 2 ) j + (4 − 5t + 4 t 4 ) k at the point P (3, 1, 4) ?.
a) (3, 1, 0)
b) (2, 0, 3)
c) (1, 0, -5)
d) (4, 1, -1)
e) none of the above
2
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7. Which of the following functions is the linear approximation of
f(x, y) = ln ( x − 3 y ) at the point (7, 2)?
a) L ( x, y ) = x − 3 y − 1
b) L ( x, y ) = 3 x − y
c) L ( x, y ) = 1 − x + 3 y
d) L ( x, y ) = x − 3 y
e) L ( x, y ) =
1
3x − y
8. If x 2 y z 2 − x − y − z = 0 , then
2x − 1
2z − 1
2 xyz 2 − 1
a) −
b) −
c) −
∂z
equals
∂x
2 x 2 yz − 1
2 x 2 yz − 1
2 xyz 2 − 1
2z − x − y − 1
d) −
2x − 1 − y − z
2x − 1 − y − z
e) −
2z − x − y − 1
1 3
9. Evaluate the integral
0 1
y
e2 z
dx dy dz .
y
0
2
a) e − 1
b) 2 e 2 − 1
2
c) 3 e − 1
d) 2 e − 2
e)
2e −1
3
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1 e
y
f ( x, y ) dx dy
10. Reverse the order of integration:
0 0
e ln x
f ( x, y ) dy dx
a)
1 0
e 1
f ( x, y ) dy dx
b)
0 ln x
e 1
f ( x, y ) dy dx +
c)
d)
f ( x, y ) dy dx
0 0
0 ln x
1 1
e
1
f ( x, y ) dy dx +
0 0
1 1
e)
1
e
f ( x, y ) dy dx
1 ln x
e 1
f ( x, y ) dy dx +
0 ln x
f ( x, y ) dy dx
1 ln x
4
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11. Let D be the solid that is inside the sphere x 2 + y 2 + z 2 = 4 and above the cone
z = 3(x 2 + y 2 ) .
Which of the following iterated integrals represents the volume of this solid in spherical
coordinates?
π
2π 3 2
d ρ dφ dθ
a)
0
0 0
π
2π 3 2
ρ d ρ dφ dθ
b)
0
0 0
π
2π 4 2
c)
0
ρ 2 sin ϕ d ρ dφ dθ
0 0
π
2π 6 2
ρ d ρ dφ dθ
d)
0
0 0
π
2π 6 2
e)
0
ρ 2 sin ϕ d ρ dφ dθ
0 0
5
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ANSWERS
1.
2.
1.2
16 50 58
− 7 − 14 14
,
,
,
,
+
9
9
9
9 9
9
12 x − 5 y + z + 3 = 0
2.1
−
1.1 a =
4
7
1
1
,
,
3
3
2.2
1
3
,
4 3
2 ( x − 3)2 + ( y − 3)2 + ( z − 7 )2 = 10 . Ellipsoid, centered at C (3, 3, 7) ,
2.3
Semi-axis: a = 5 , b = c = 10
x + y + z − 18 = 0
2.4
2
3
3. relative minimum at M 1 (0, − ) ;
4.
27π
5,
18
5
6.
d) (4, 1, -1)
7.
a) L ( x, y ) = x − 3 y − 1
8.
b)
9. a) e
2
2
M 2 (2, − ) is the saddle point
3
−1
1 1
e
1
f ( x, y ) dy dx +
10. d)
0 0
f ( x, y ) dy dx
1 ln x
π
2π 6 2
11. e)
0
ρ 2 sin ϕ d ρ dφ dθ
0 0
6
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