Exam 3 Solutions
1.) Included with your exam is a page of 10 plots of r vs θ. The plots on the
left are cartesian plots and the ones on the right are polar plots. Draw a line
from each cartesian plot to its corresponding polar plot.
1 → 3, 2 → 4, 3 → 2, 4 → 5, 5 → 1.
How do we do this? Well, the first distinction we might notice is that plot
#3 is strictly decreasing. Thus the corresponding polar graph will have a radius which decreases with increasing θ. In general, when we have r smoothly
increasing or decreasing with respect to θ, the graph will be a spiral. Thus,
3 → 2. 5 → 1 perhaps most simply because they are the only remaining
pair whose radius is always properly less than 1. Similarly, 4 → 5 because
this pair is the only one remaining whose radius values exceed 1. Finally,
1 → 3 because it is the only remaining one for which r(0) = 1. 2 → 4 by
process of elimination. Obviously, there are many equally valid ways to do
this problem.
2.) Sketch the plots of curves given (in cartesian coordinates) by the following parametric equations.
i.) x = sin(t), y = csc(t), 0 < t < π/2
y = csc(t)
1
=
sin(t)
1
= ,
x
so just plot y = 1/x from x = sin(0) = 0 to x = sin(π/2) = 1
1
ii.) x = 4 cos(t), y = 5 sin(t), 0 < t < 2π
1 = cos2 (t) + sin2 (t)
x 2 y 2
=
+
4
5
so the curve is an ellipse with axes 4 and 5.
2
iii.) x = t2 + 2t + 1, y = 1 + et+1 , 0 < t < 10
y = 1 + et+1
√
2
= 1 + e (t+1) since t > 0
√
=1+e
√
=1+e
t2 +2t+1
x
.
3.) For this problem, let c be a real number, and let f be a continuously
differentiable real-valued function with f 0 (c) 6= 0. Also, let g be a real valued
function, and let x = f (t), y = g(t) for t in some open interval containing c.
i.) Show that there is an open interval (a, b) containing c for which f 0 (c) 6= 0
f 0 (t) 6= 0. Hint: Since f 0 is continuous at c we know that for every positive
number , there is an open interval (a, b) containing c such that for all t in
(a, b), f 0 (t) − < f 0 (c) < f 0 (t) + .
Since f 0 (c) 6= 0, we know that there is a positive number such that
< |f 0 (c)|. That way, any real number r such that r − < f 0 (c) < r + will
not be equal to 0 either. But we know from the − δ definition of the limit
that we can choose an open interval (a, b) containing c such that for all t in
(a, b), f 0 (t) − < f 0 (c) < f 0 (t) + . Thus, f 0 (t) 6= 0 for t in (a, b).
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ii.) Use (i) to show that f is strictly increasing or strictly decreasing on the
interval (a, b).
Since f 0 6= 0 on (a, b) and is continuous, it must always either be strictly
greater than or strictly less than 0 on this interval. Thus, f is either increasing or decreasing
iii.) Show that there is a function F such that y = F (x) for f (a) < x < f (b).
Hint: Show that f −1 exists.
Since f is strictly increasing or decreasing on (a, b), it is 1-1. In other words,
for any t1 6= t2 in (a, b), f (t1 ) 6= f (t2 ). To show this, suppose we have two
distinct numbers in (a, b). Call the lesser one t1 and the greater one t2 . Then
if f is increasing, f (t1 ) < f (t2 ) and if it is decreasing, f (t1 ) > f (t2 ). Either way, f (t1 ) 6= f (t2 ). The range of f on the interval (a, b) is (f (a), f (b))
because f is increasing or decreasing continuously from f (a) to f (b). Thus
f −1 exists on the domain (f (a), f (b)). But since x = f (t), t = f −1 (x), so
y = g (f −1 (x)).
4
4.) Consider the curve given in polar coordinates by r = sin(θ)e−θ , 0 < θ < π.
Find the point along the curve which is farthest from the origin.
dr
= 0
We want to maximize r with respect to θ, so we will find where dθ
dr
for 0 < θ < π. Now, dθ
= e−θ (cos(θ) − sin(θ)), so we must find θ such that
cos(θ) = sin(θ), and the only θ between 0 and π for which this is true is
θ = π/4. Thus, the point we are seeking is (θ, r) = (π/4, sin(π/4)e−π/4 ).
5.) Consider the curve given in cartesian coordinates by the parametric
equations x = t − sin(t), y = 1 − cos(t), 0 < t < 2π.
See example 10.2.3 in your book.
i.) Sketch the curve.
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ii.)Find the area between the curve and the x-axis.
6
2
r
1
1.0
0.5
Θ
1
2
3
4
5
0.5
-1.0 -0.5
6
1.0
-0.5
-1
-1.0
-2
1.0
r
1.4
1.2
0.5
1.0
0.8
0.6
-0.6-0.4-0.2
0.2 0.4 0.6
0.4
0.2
0.0
-0.5
Θ
0
1
2
3
4
5
6
-1.0
r
100
1.0
80
0.8
60
0.6
0.4
40
0.2
20
-1.0
0.5
-0.5
1.0
Θ
2
4
6
8
10
1.0
r
0.5
1.5
1.0
0.5
Θ
1
2
3
4
5
-1.0
0.5
-0.5
6
1.0
-0.5
-1.0
-0.5
-1.5
-1.0
60
r
40
1.4
1.2
20
1.0
7
0.8
0.6
-80
-60
-40
20
-20
0.4
-20
0.2
0.0
Θ
0
20
40
60
80
100
-40
40