MATH 2210 - SAMPLE FINAL EXAM
1) Consider the surface x2 + z = 2 and the parametrized curve
σ(t) = (1, t3 , t).
a) Find the point P where the curve intersects the surface.
b) Write a parametric equation of the line tangent to σ(t) at the point P .
c) Write an equation of the tangent plane to the surface at the point P .
Answers: a) (1, 1, 1) (for t = 1); b) (1, 1, 1) + t(0, 3, 1); c) 2x + z = 3.
2)
a) Sketch the region of integration for the integral
Z 1Z 1
√
y cos(x4 )dxdy.
√
0
y
b) Compute the integral (Hint: reverse the order of integration).
Answer: sin6 1 .
3) Find the minimum and the maximum value of the function f (x, y) = 3x + 4y along the
circle x2 + y 2 = 1.
Answer: Maximum is f (3/5, 4/5) = 5 and minimum f (−3/5, −4/5) = −5.
4) Find and analyze the critical points of the function
f (x, y) = 2x4 − x2 + 3y 2 .
Solution: fx = 8x3 − 2x = 2x(4x2 − 1), fy = 6y 2 . Thus fx = 0 on vertical lines x = 0, 1/2
and −1/2, and fy = 0 on the horizontal line y = 0. Thus fx = 0 AND fy = 0 for three points
(0, 0), (1/2, 0) and (−1/2, 0). These are the critical points. Since
fxx fxy
24x2 − 2 0
=
fyx fyy
0
6
second derivative test says that (0, 0) is a saddle point, while (±1/2, 0) are local minima.
5) Evaluate the line integral by finding a potential function.
Z (2,2)
(3x2 + 2xy 2 )dx + (3y 2 + 2x2 y)dy.
(1,1)
Solution: f (x, y) = x3 + x2 y 2 + y 3 , so the integral is f (2, 2) − f (1, 1) = 32 − 3 = 29.
6) Consider the vector field V = xi + yj. Calculate the flux of V accross the boundary of
the triangle with vertices (0, 0), (1, 0) and (0, 1) in two ways:
a) Directly.
1
2
MATH 2210 - SAMPLE FINAL EXAM
b) Using the Divergence theorem.
Solution: Part a). The flux across C is given by
Z
V · n ds
C
where n is the vector of length one and perpendicular to the curve C. The curve C is a
triangle, so the vector n is constant along each edge. Consider first the edge C1 connecting
(0, 1) and (1, 0). This edge lies on the line x + y = 1 i.e. on the graph of y = 1 − x, with
√ . Thus
0 ≤ x ≤ 1. Along this edge n = i+j
2
x+y
1
V·n= √ = √ .
2
2
p
√
Since ds = 1 + (f 0 (x))2 dx and f (x) = 1 − x for the edge in question, ds = 2dx. Putting
everything together
Z 1
Z
1 √
√ 2 dx = 1.
Vϕ · n ds =
2
0
C1
The flux along the other two edges is 0, since V · n = 0 there, so the total flux is 1. Part b)
is easy, since divV(x, y) = 2, so the double integral of the triangle is 1.
7) Consider the vector field V(x, y) = −yi+xj. Let Vϕ be the vector field obtained by rotating
V counter-clockwise for the angle ϕ. Find the flux of Vϕ accros the circle C x2 + y 2 = 1 two
ways:
a) Directly.
b) Using the divergence theorem.
Solution:
Vϕ (x, y) = (−y cos ϕ − x sin ϕ)i + (−y sin ϕ + x cos ϕ)j.
The flux across C is given by
Z
Vϕ · n ds
C
where n is the vector of length one and perpendicular to the curve C. (Of course n depends
on the point on C.) For this particular curve, unit circle, n = xi + yj. Since x2 + y 2 = 1 on
C,
Vϕ · n = −x2 sin ϕ − y 2 sin ϕ = sin ϕ.
Thus
Z
Z
Vϕ · n ds =
C
Z
− sin ϕ ds = − sin ϕ
C
ds = −2π sin ϕ
C
since the length of C is 2π. This is “direct” calculation of the flux. Using the divergence
theorem, let S be the interior of the unit circle,
Z Z
Z Z
Z Z
divVϕ dxdy =
−2 sin ϕ dxdy = −2 sin ϕ
dxdy = −2π sin ϕ
S
since the area of S is π.
S
S
MATH 2210 - SAMPLE FINAL EXAM
3
7’) Consider the vector field V(x, y) = −yi + xj. Let Vϕ be the vector field obtained by
rotating V counter-clockwise for the angle ϕ. Find the work done of Vϕ along the circle
x2 + y 2 = 1 in the counterclockwise direction in two ways:
a) Directly.
b) Using Green’s theorem.
The work done along C is given by
Z
(−y cos ϕ − x sin ϕ) dx + (−y sin ϕ + x cos ϕ) dy.
C
In order to compute this line integral, we need to parameterize the curve C. This is done by
x = cos t and y = sin t where 0 ≤ t ≤ 2π. Since dx = − sin t dt and dy = cos t dt, the integral
is
Z 2π
(− sin t cos ϕ − cos t sin ϕ)(− sin t) dt + (− sin t sin ϕ + cos t cos ϕ)(cos t) dt =
0
Z
=
2π
2
Z
2
(sin t cos ϕ + cos t cos ϕ) dt =
0
2π
cos ϕ dt = 2π cos ϕ.
0
Using Green’s theorem,
Z Z
Z Z
(−y sin ϕ + x cos ϕ)x − (−y cos ϕ − x sin ϕ)y dxdy =
2 cos ϕ dxdy = 2π cos ϕ.
S
S
8) Let C be the closed curve obtained by going from (0, 0) to (1, −1) along y = −x2 , then
going to (1, 1) along x = 1, and then coming back to (0, 0) along y = x2 . Calculate
Z
xdy
C
two ways:
a) Directly.
b) Using Green’s theorem.