Assignment 2
Math 110-003
Due Thurs Jan 28
Hand in full solutions to the questions below. Make sure you justify all your work and include
complete arguments and explanations. Your answers must be clear and neatly written, as well
as legible (no tiny drawings or micro-handwriting please!). Your answers must be stapled,
with your name and student number at the top of each page.
1. The height of a right-angled triangle is increasing at a rate of 5cm/min while its area
stays constant. How fast must the base be decreasing at the moment when the height
is five times the base?
Hint: use the product rule. Also, note that the fact that the triangle is right-angled is
immaterial.
Solution.
Let h be the height and b the base. Then h0 = 5 and the area A is given by A =
By the product rule,
1
1
A0 = (b0 h + bh0 ) = (b0 h + 5b).
2
2
0
Since the area is constant, A = 0. Thus, when the height is five times the
(h = 5b),
1
5
0 = A0 = (5bb0 + 5b) = b(b0 + 1).
2
2
0
Now solve for b (note that b 6= 0):
5
0 = b(b0 + 1) ⇒ 0 = b0 + 1
2
⇒ −1 = b0 .
1
bh.
2
(1)
base
(2)
(3)
(4)
Thus, the base is decreasing at 1cm/min.
2. By Pythagoras’s theorem,
sin2 x + cos2 x = 1
(5)
y = − cos4 x + sin4 x.
(6)
for any x.
Suppose that
1
1
d2 x
d2 y
dx
= .
=
0,
find
when
2
2
dt
dt
dt
2
Solution.
If
Note that this question had a typo in its original formulation (the minus sign was
missing in fron tof cos4 x). The problem can still be solved with this typo but is
significantly harder. The solution below is for the correct formulation of the problem.
By Pythagoras’s theorem,
y 0 = −4 cos3 (x)(− sin(x))x0 + 4 sin3 (x) cos(x)x0
= 4x0 sin x cos x(sin2 x + cos2 x)
= 4x0 sin x cos x.
Differentiating once more and plugging in x00 = 0, we get
y 00 = 4x00 sin x cos x + 4x0 (sin x cos x)0
= 4x0 (cos2 x − sin2 x).
(7)
(8)
y 00 = 2(cos2 x − sin2 x).
(9)
Thus, when x0 = 1/2,
3. A man starts walking north at 4km/h from a point P . Thirty minutes later, a woman
starts jogging in the same direction at 8km/h from a point 100m due east of P . Both
people maintain a constant speed throughout their (endless) respective journeys.
(a) How far have the man and the woman each gone 15 minutes after the woman
started jogging?
(b) What is the distance between the two people 15 minutes after the woman started
jogging?
(c) What is the total amount of time the woman must jog in order to overtake the
man?
(d) At what speed is the distance between the two people changing 15 minutes after
the woman started jogging?
Hint: use Pythagoras’s theorem and implicit differentiation.
Note: mind the units!
Solution.
(a) The man has been walking for 30 + 15 = 45 minutes, which is 3/4 hours, at 4km/h
so has gone (3/4)(4) = 3km. The woman has jogged for 15 minutes (1/4 hours) at
8km/h, so has gone (1/4)(8) = 2km.
(b) By (a), the north-south distance between them is 3 − 2 = 1km. The east-west distance between them is constantly 100m
by Pythagoras’s theorem,
√
√ = 0.1km. Thus,
2
2
the (total) distance between them is 0.1 + 1 = 1.01km.
2
(c) By (a), after 45 minutes, the man and woman have gone 3km and 2km, respectively.
After t more minutes have passed, the man and woman have gone an additional
4tkm and 8tkm. Thus, after 45 minutes and t hours, the man and woman have
gone 3 + 4tkm and 2 + 8tkm, respectively. The woman has overtaken the man once
these two distances between equal, i.e. when
3 + 4t = 2 + 8t.
(10)
We find t by solving this equation:
3 + 4t = 2 + 8t ⇒ 3 − 2 = 8t − 4t
⇒ 1 = 4t
⇒ t = 1/4.
(11)
(12)
(13)
Thus, the woman has overtaken the man after 45 minutes and 1/4 hours, which
is 45 + 60/4 = 60 minutes. Since the woman didn’t start jogging until 30 minutes
in, she overtakes the man 60 − 30 = 30 minutes after having started jogging.
(d) Let y be the north-south distance between the two people and let z be the (total)
distance between them (both in km). Since the east-west distance between them
is 0.1km, z 2 = 0.12 + y 2 = 0.01 + y 2 . Differentiating with respect to time t, we get
2zz 0 = 2yy 0 .
(14)
Fifteen minutes
after the woman has started jogging, we know by (b) that y = 1
√
and z = 1.01. Thus, the equation above becomes
√
2 1.01z 0 = 2(1)y 0 .
(15)
Now y 0 is the rate of change of the north-south distance between the two people.
Thus, y 0 is their relative (north-south) velocity, which is the difference between
their respective north-south velocities. In other words, y 0 = 4 − 8 = −4km/h.
Plugging this in to the equation above, we get
√
(16)
2 1.01z 0 = 2(−4) = −8
and solving for z 0 yields
−8
4
z0 = √
= −√
.
2 1.01
1.01
(17)
√
Thus, the distance between the two people at this time is decreasing at 4/ 1.01km/h.
3