MATH 100 V1A
November 19th – Practice problems
Hints and Solutions
1. In mechanics, a force exerted on a steel rod is called stress, and denoted σ. The
deformation of the rod due to stress is called strain, and denoted . Stress and strain
are related by the Ramberg-Osgood equation
= aσ + bσ n ,
where a, b and n are constants.
Suppose the stress applied to a steel rod increases at a constant rate of R. Determine
the rate at which the strain is increasing when the stress is equal to σ0 .
, when
Solution: We want to find the rate of change of strain with respect to time, d
dt
dσ
d
= R and σ = σ0 . To find dt , we differentiate the Ramberg-Osgood equation
dt
implicitly. In the case that n 6= 0, 1, we get that
dσ
dσ
d
=a
+ bnσ n−1 ,
dt
dt
dt
so when
dσ
dt
= R and σ = σ0 , we get that
d
dt
= aR + bnσ0n−1 · R.
If n = 1, then we get that
so when
dσ
dt
dσ
d
= (a + b) ,
dt
dt
d
= R and σ = σ0 , we get that dt = (a + b)R.
Similarly, when n = 0,
dσ
dt
= R and σ = σ0 , we get that
d
dt
= aR.
2. Oceanographers on research ships use a device called a CTD probe to make underwater
measurements at extreme depths. The probe drops vertically through the water at 4
m/s. Simultaneously, the research ship moves away from the drop site at 5 m/s. What
is the rate of change of the distance between the probe and the ship 10 seconds after
the probe is dropped?
Hint: There are multiple ways to do this problem. One way is to let x denote the
distance the ship has travelled in the horizontal direction after the probe was dropped
and to let y denote the distance the probe has travelled in the vertical direction
p after
the probe was dropped. The the distance between
the
ship
and
the
probe
is
x2 + y 2 .
√
Use this to finish the problem (the answer is 41).
3. The Space Shuttle launches vertically. In the initial part of its flight, it’s altitude in
kilometres after t seconds is given by
h(t) =
t2
.
300
The launch is supervised from the Launch Control Centre (LCC), 5 kilometres away
from the launch pad. Determine the rate at which the distance between the LCC and
the Space Shuttle is increasing 30 seconds after liftoff.
Hint: The distance between the LCC and the Space Shuttle at time t is
p
D(t) = h(t)2 + 25.
Use this to finish the problem (the answer is
2
√3 ).
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