53 sections 202/204 Quiz 1 Solutions
Problem 1 (10 pts). (a) Show that the parametric equations
x = x1 + (x2 − x1 )t and y = y1 + (y2 − y1 )t
where 0 ≤ t ≤ 1, describe the line segment that joints the points (x1 , y1 ) and (x2 , y2 ).
Solution: Assume x1 6= x2 . To show that the parametric curve lies on the line segment, show that
they satisfy the equation
y2 − y1
y − y1 =
(x − x1 )
x2 − x1
(the equation for y in terms of x for the line through the two points) by plugging in the given
expressions for x and y in terms of t.
To show that the parametric curve attains all values on the segment between (x1 , y1 ) and (x2 , y2 ), use
the intermediate value theorem on either x(t) or y(t), evaluating at t = 0 and t = 1.
To show that the parametric curve does not go beyond the endpoints (x1 , y1 ) and (x2 , y2 ), there are a
number of methods: you can argue geometrically, can solve for the t for which (x(t), y(t)) would be
beyond the segment, or you can do the following: show that the distance between (x(t), y(t)) and
(x
p2 , y2 ) is a decreasing function of t (remember that the distance between two points (a, b) and (c, d) is
(c − a)2 + (d − b)2 ) by differentiating it. Then evaluate (x(0), y(0)); this should be the farthest from
(x2 , y2 ) you can get. Do a similar evaluation with the increasing distance from the point (x1 , y1 ).
Assume x1 = x2 . Then the line segment is (x1 , y) for y1 ≤ y ≤ y2 . The parametric curves are
x = x1 and y = y1 + (y2 − y1 )t
which is certainly contained on the line (x1 , y). A similar analysis to the case for x1 6= x2 shows that
you get precisely those points with y1 ≤ y ≤ y2 .
(b) Find parametric equations to represent the line segment from (−2, 7) to (3, −1).
Solution: Using part (a), the equations are
x = −2 + (3 + 2)t = 5t − 2 and y = 7 + (−1 − 7)t = −8t + 7
1
Problem 2 (10 pts). Find the area of the region enclosed by one loop of the curve r = 4 cos 3θ.
Solution: One loop of r = 4 cos 3θ is enclosed by taking − π6 ≤ θ ≤ π6 ; certainly r = 4 cos 3 ± π6 = 0,
and since cosine is nonzero for − π2 < θ < π2 there are no values of θ strictly between ± π6 for which r is
Rθ
zero. We can then use the area formula, A = 12 θ12 r2 dθ:
Z π
1 6
A=
(4 cos 3θ)2 dθ
2 − π6
Z π
6
=8
cos2 3θ dθ
− π6
Z
π
6
=4
1 + cos 6θ dθ
− π6
1
π
= 4 θ + sin 6θ −6 π
6
6
π
=4
+0
3
4π
=
3
2