Practice Exam 1
1.
(0 points) Consider the curve x = sin t, y = cos2 t for 0 ≤ t ≤ π2
a) Graph the curve, indicating initial and terminal points, and the direction of the curve.
cos2 t = 1 − sin2 t, so y = 1 − x2 . Initial point = (0, 1), terminal point = (1, 0).
b) Find the equation of the tangent line at the point with t = π6 .
At t = π/6, dy/dx = −2 sin(π/6) = −1. Also, at t = π/6, x = 1/2 and y = 3/4. So, the equation of the
tangent line is
y − 3/4 = −1(x − 1/2).
c) Set up the equation (do not evaluate) indicating the length of the given curve.
R π/2 p
(− cos t)2 + (−2 cos t sin t)2 dt
L= 0
2.
(0 points)
a) Identify the conic section 25y 2 + 4x2 − 100 = 0
The equation we get is
x2
y2
+
=1
52
22
So the conic section is an ellipse.
b) Identify the vertices and foci.
vertices
√ = (±5, 0)
√
√
√
c = a2 − b2 = 25 − 4 = 21. foci = (± 21, 0)
c) Graph the curve.
Use the fact that we know that the following points are on the ellipse: (±5, 0), (0, ±2).
3. (0 points) Consider the curve r = 1 + 2 cos(θ).
a) Graph the curve.
The trick to graphing a polar curve that you can’t immediately identify, is to first treat r as y, and θ
as x. Graph this in the xy−plane. Then, translate this to the polar plane by seeing how y changes as
x increases. When θ = 0, r = 3, and r = 0 when θ − 4π/3, 5π/3. So, the curve is like a cradiod, but
it curves in inside, and touches the polar axis when θ = π, since at that point, r = −1. Then it curves
back upwards and then crosses 0 when θ = 5π/3, and then finally reaches back to the inital point when
θ = 2π. Verify with graphing calculator.
b) Evaluate the slope of the tangent line at θ = π4 .
Use the polar equation form of dy/dx here. Note, on an exam, you do not need to simplify unless it has
been explicitly
stated to do so.
√
−1+√2
dy/dx = −3−
2
c) Setup, but do not evaluate the integral for the arc length of of the curve.
L=
R 2π p
0
(1 + 2 cos θ)2 + (−2 sin θ)2 dθ.
4.
(0 points) Find the area of the region that lies inside the first curve, and outside the second curve
r = 3 sin θ, r = 2 − sin θ
The first step is to graph both of these functions. Do not think about this problem in the context of
finding areas using traditional calculus. Think about the area as a sector of a circle being swepted out
as θ increases.
After graphing both, we see that there are three regions we need to subtract from the area of r = 3 sin θ.
Note that both curves intersect when θ = π/6 (solve for θ: 3 sin θ = 2 − sin θ).
So, the first sliver of the area we want to subtract from the area of r = 3 sin θ corresponds to:
1) 3 sin θ for θ from 0 to π/6.
Then, the next part we want to subtract comes from r = 2 − sin θ. Note that both curves also intersect
when θ = 5π/6. So, we want to also subtract the area:
2) 2 − sin θ for θ from π/6 to 5π/6. The final part of the puzzle comes from r = 3 sin θ, and corresponds
to subtracting the area: 3) 3 sin θ from 5π/6 to π.
Now what remains is setting up the formula for the area of the region we are interested in. It will be in
the form of:
(Area of the entire region of r = 3 sin θ) - (Area of 1) + Area of 2) + Area of 3)).
So we have:
!
Z π
Z π/6
Z 5π/6
Z π
2
2
2
2
A = (1/2)
(3 sin θ) dθ− (1/2)
(3 sin θ) dθ + (1/2)
(2 − sin θ) dθ + (1/2)
(3 sin θ) dθ .
0
0
5.
π/6
(0 points) Determine whether the vectors are parallel, orthogonal, or neither.
a) a =< 8, 2, −1 >, b =< 1, −3, 5 >
Neither.
b) a = −2i + 2j − k, 6i − 6j + 3k
Parallel.
6. (0 points) Let a =< −7, 5, 2 >, and b =< 3, 2, 3 >
a) Find proja b
Standard computation.
b) Find projb a
Standard computation.
5π/6
7.
~ + DE
~ − BE
~ + BC
~ − DC
~
(0 points) Use the figure to write AD
Note, this problem has been changed, since there was no quick and simple way to solve the original
practice problem. To solve, we start at A, move to D, move to E, then B, then C, and end at D. So, the
~
answer is AD.
8.
(0 points) Find the volume of the parallelepiped determined by
a =< −1, −3, −2 >, b =< 2, −2, −3 >, and c =< 1, 1, 1 >.
V = |a · (b × c)| = 24.
Question: Let a =< 1, −1, 1 > . If a · b = 0 and a × b = 0, what can we say about b?