WORKSHOP 1·11
Solutions
Warm-up Problem
Consider two functions f and g such that
• f 0 (x) < 0 for x > 0 and f 0 (x) > 0 for x < 0,
• both g(x) and g 0 (x) are negative for all values of x.
Find at what rate the composite function f (g(x)) is changing when x = 2. Can the rate of
change of f (g(x)) be positive for some value of x?
Solution
The rate of change of the composite function f (g(x)) is given by the chain rule:
[f (g(x))]0 = f 0 (g(x)) · g 0 (x).
We can then use x = 2:
[f (g(2))]0 = f 0 (g(2)) · g 0 (2).
Since we know that x = 2 > 0, then we know that both g(2) < 0 and g 0 (2) < 0, therefore f 0 (g(2)) > 0. The
rate of change is then a product of a positive and a negative number and is thus negative.
Note that no matter what value of x we use both g(x) and g 0 (x) will be negative and therefore the derivative
of f (g(x)) will always be negative.
Part B
A small group of mathematician just started a company that produces soup with noodles
shaped like mathematical symbols. Being mathematicians, they were quite confident of their
success and devised an employment and production model using functions. As the company
grows, they could hire more and more employees with time and the more employees they hire
the more soup they can produce. The model uses the following functions for the employment
and production functions E(t) and P(t):
√
E(t) = at2 + bt + 5, P (x) = c x,
where t is measured in time and x is the number of employees. a, b and c are positive constants
that can be adjusted. Find an expression of a function that express the rate of change of
production with time of this model.
Solution: The function giving the instantaneous rate of change would be the time derivative of the production
function. However, That function’s input are employees, so we need to use a function composition. The
function P (E(t)) gives the production value with time. To find its derivative we need to use the chain rule.
We can compute the derivative of each function:
E 0 (t) = 2at + b,
c
P 0 (x) = √ ,
2 x
The derivative then becomes
(P (E(t)))0 = P 0 (E(t))E 0 (t)
c
= p
(2at + b)
2 E(t)
c
√
(2at + b).
2 at2 + bt + 5
1
We can then use specific values for a, b and c to get the production rate at any specific time, but being
mathematicians, they were not interested in that and moved along.
5. Part C [50 minutes]
While designing their future production plant, one of the entrepreneurs imagines a large tank in which the
various ingredients would be mixed. As is often the case with mathematician’s designs the shape of the tank
is spherical, as it requires the least amount of material to hold the maximum amount of soup.
If a spherical tank of radius 4 metres is filled up to h metres with soup, then the volume of soup in the tank
is given by the following formula:
V =
π 2
h (12 − h)
3
Questions:
(a) At what rate is the volume of soup in the tank changing with respect to the height of the
soup when h = 1 m? What are the units on this quantity?
(b) Now suppose that the height of soup in the tank is being regulated by an inflow and
outflow (e.g., a faucet and a drain) so that the height of the soup at time t is given by
the rule h(t) = sin(πt) + 1, where t is measured in hours (and h is still measured in metres).
At what rate is the height of the soup changing with respect to time at the instant t = 2?
(c) Continuing under the assumptions in (b), at what rate is the volume of soup in the tank
changing with respect to time at the instant t = 2?
(d) What are the main differences between the rates found in (a) and (c)? Include a discussion
of the relevant units.
(e) Suppose now that the height of soup in the tank is given by the piecewise function
(
sin(πt) + 1, 0 ≤ t ≤ 2
h(t) = 1
1
2 cos(πt) + 2 , t > 2
Is the Volume function with time V (h(t)) continuous at t = 2? Is it differentiable at t = 2?
Solution:
(a) The rate of change of volume with respect to soup height is given by the derivative
height of 1m:
dV
=
dh
dh
(1) =
dh
2π
π
h(12 − h) − h2
3
3
2π
π
· 11 − = 7π m2 .
3
3
2
dV
dh
evaluated at a
(b) The rate of change of the soup’s height with respect with time is given by the derivative
at t = 2s. Note here that we use the chain rule:
dh
dt
evaluated
h(t) = f (g(t)),
df
= cos(x)
dx
f (x) = sin(x) + 1,
g(t) = πt,
dg
dt
= π.
dh
d
d
=
(sin(x)) · (πt)
dt
dx
dt
= cos(πt) · π
dh
(2) = cos(2π) · π = πm/hr.
dt
(c) The Volume of soup as a function of time is given by the composite function V (h(t)). Using the chain
rule and the results from the previous sections we obtain the following
d
d
V (h(t)) · h(t)
dh
dt
d
d
V (h(2)) · h(2)
dh
dt
d
=
V (sin(2π) + 1) · π
dh
d
=
V (1) · π = 7π 2 m3 /hr.
dh
d
V (h(t)) =
dt
d
V (h(2)) =
dt
(d) The first is with respect to soup height in the tank and the other is with respect to time. The units
are also different. The first result is in volume units per distance units and the second is in volume
units per time units.
(e) Each piece being continuous and differentiable we only need to check for continuity at the interface
(t = 2hr) only. Since we have
lim h(t) = sin(2π) + 1 = 1 =
t→2−
1
1
cos(2π) + = lim+ h(t),
2
2 t→2
we know that the function h(t) is continuous for all time. Using the limit laws we get
π
( lim h(t))2 (12 − lim h(t))
3 t→2−
t→2−
π
= h(2)2 (12 − h(2)), as h is continuous
3
= V (2),
lim− V (t) =
t→2
and similarly for the right limit. thus, The volume function is continuous.
To check for differentiability, we want to know if the function has a corner at t = 2hr. To do this we
verify if the one-sided limits of the derivative at that time match. We already computed the left limit
3
in part (c) and we obtained a value of 7π 2 m3 /h. To find the right limit we only need the derivative of
the second case of h(t):
d 1
1
dh
=
cos(πt) +
dt
dt 2
2
1
= − sin(πt) · π
2
dh
1
(2) = − sin(2π) · π = 0m/hr.
dt
2
We can then use the chain rule to find the derivative of the volume:
dV
dV
dh
(2) =
(2) ·
(2) = 0m3 /hr.
dt
dh
dt
We obtained that the slope of the tangent line goes suddenly from 7π 2 to zero. That means that there
is a corner at t = 2 and that the volume function is not differentiable at t = 2 hours.
4