Math 100 - Related Rates - *not to hand in*
Basic skills required to work through the problems:
• differentiate (explicitly or implicitly) a mathematical expression;
• apply Pythagoras’ theorem;
• write down proportionality relations between the sides of similar triangles;
• use volume formulas to find the volume of simple solids.
Learning Goals: After completing this problem set, you should be able to apply the
problem-solving strategy below and solve problems involving related rates.
Problem-Solving Strategy: Steps to approach a problem.
1. Understand the problem: Read the problem and make sure you understand
it clearly. Ask yourself:
• What do I know?
– What are the given quantities?
– What are the given conditions?
• What do I need to find?
–
–
–
–
What is the problem asking?
What is the unknown?
Do I need to introduce suitable notation?
Do I need to draw a sketch?
2. Think of a plan: Find the connection between the given information and the
unknown. Ask yourself:
• What calculus concept(s) do I need to apply?
– Do I need to differentiate? With respect to what variable do I differentiate?
– Do I need to solve an equation? Why? What variable do I solve for?
– Do I need to apply a certain theorem or definition? Why?
3. Carry out your plan: Do your computations!
4. Reflect on what you are doing: Does your answer make sense?
1. (WITH SOLUTION) Read the following problem and answer the questions below.
Remember to draw a diagram!
“Two roads, one in the east-west direction and the other north-south, intersect
at right angles. Car A is 100 km directly south of the intersection point of the
roads and is travelling north at 80 km/h. Car B is 60 km west of the intersection
and is travelling west at 50 km/hr. At what rate is the distance between the cars
changing? You do not need to simplify your answer in this question.”
(a) UNDERSTAND THE PROBLEM: What is the problem asking? Choose the
correct answer from the ones below.
(a.1) The problem asks to find the rate of change of the distance of Car A from
the intersection point at a certain moment in time.
(a.2) The problem asks to find the rate of change of the distance of Car B from
the intersection point at a certain moment in time.
(a.3) The problem asks to find the rate of change of the distance of Car A from
Car B at a certain moment in time.
(a.4) The problem asks to find the rate of change of the distance between the
cars at any time.
Solution: The correct answer is (a.3). The problem asks to find the rate
of change of the distance of Car A from Car B at a certain moment in
time. We have information about the cars at a specific moment of time
when Car A is 100 kilometres directly south of the intersection and driving
north at 80 km/hr and Car B is 60 kilometres west of the intersection and
driving west at 50 km/hr, so any calculations involving this information
will be referring to that specific moment of time.
(b) UNDERSTAND THE PROBLEM: Assign names to variables. For example,
suppose D is the distance between the cars, x is the distance of Car A from the
intersection, y is the distance of Car B from the intersection. Draw a picture
if necessary. As you determined above, you need to find the rate of change of
= ..., dy
= .... What are the
D at a specific time when x = ..., y = ..., dx
dt
dt
missing values in this sentence?
Solution: Placing the origin of the axes in the intersection point and
taking the positive x-axis to point east of the intersection and the positive
y-axis to point north of the intersection, then the problem asks for the rate
= +50, and
of change of D at a specific time when x = 60, y = 100, dx
dt
dy
= −80 (negative because the distance from car B to the intersection is
dt
decreasing).
(c) THINK OF A PLAN: Which of the following actions best describes a strategy
to solve the given problem? Choose the correct answer from the ones below.
(c.1) Finding the derivative of D with respect to x.
(c.2) Finding the derivative of D with respect to y.
(c.3) Finding the derivative of D with respect to time.
Solution: The correct answer is
(c.3) Finding the derivative of D with respect to time.
(d) THINK OF A PLAN: Find an expression relating D to x and y.
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Solution: If you draw a picture of the position of the cars relative to the
intersection, you find that D2 = x2 + y 2 . This relation is true at any
time (assuming the cars continue to drive in the same direction). Since
both cars are moving, all the distances x, y, and D are functions of time.
Thus the expression found in the previous part can be written as [D(t)]2 =
[x(t)]2 + [y(t)]2 .
(e) THINK OF A PLAN: Which of the distances defined in part (b) changes with
time?
(f) CARRY OUT YOUR PLAN: Use all the information collected in parts (a) (e) and solve the given problem.
(g) REFLECT ON WHAT YOU ARE DOING: What units is your answer expressed in? Make sure that your calculation is dimensionally consistent.
Solution: Based on the strategy determined above we need to find dD
.
dt
We do not have an explicit expression for D as a function of time so we
differentiate implicitly the equation found in part (c). So
dD
dx
dy
= 2x + 2y
dt
dt
dt
1
dx
dy
dD
=
2x + 2y
dt
2D
dt
dt
p
√
Using Pythagoras we find that D = x2 + y 2 = 10000 + 3600 so
2D
dD
1
√
=
(2 × 50 × (60) + 2 × 100 × (−80))
dt
2 × 13600
from which we conclude
dD
dt
=
−5000
√
13600
km/hr.
2. (Warm up) Suppose an ice sculpture has a conical shape. The sculpture was left in
the sun and is now slowly melting in such a way that the height of the ice cone is
always equal to its diameter. Let r be the radius of the cone at a certain time. Find
an expression for the volume of the cone V as a function of r. No other variables,
besides r, should appear in your answer.
Solution:
The volume of a cone is V = πr2 h/3, where r is the radius of the cone’s base
and h is its height. We are given that h equals the diameter, so h = 2r. So,
V = πr2 (2r)/3 = (2π/3)r3 .
3. A container whose shape is a right circular cone is turned upside-down (vertex is
at the bottom) and used to store sand; the height of cone is 16 m, the diameter of
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the top is 6 m. The container is filled half-way (by depth) with sand draining out
at a constant rate of 3 m3 per second. How fast is the depth of the sand changing
at that time? Hint: Draw a diagram.
Solution: For a cone of radius r and depth h, the volume is
1
V = π(r2 h)
3
and so
dV
1
dr
2 dh
= π (2r) h + r
.
dt
3
dt
dt
The cone is losing sand so dV
= −3. We need to find dh
when the cone is half
dt
dt
filled so when h = 8. We would still appear to have too many variables, however
3 dh
3
h so dr
= 16
. Hence,
by similar triangles (use a picture) we have that r = 16
dt
dt
we have
3 3 dh
3 2 dh
1
·8+( )
−3 = π 2 ·
,
3
2 16 dt
2 dt
4
and so dh
= − 3π
. Thus the depth
from which we solve −3 = 13 π 92 + 49 dh
dt
dt
4
is getting lower at 3π m per second. The problem can also be solved (more
easily) by substituting the expression for r into the expression for V before
differentiating.
4. An extension ladder can lengthen or shorten. An extension ladder is leaning against
a wall. The ladder has its top end fixed on a hook 15m above the ground on the
vertical wall. The bottom of the ladder is on the ground, 8m from the base of the
wall. The bottom of the ladder is slipping and moving at 2 m/min away from the
base of the wall. How fast is the ladder lengthening?
Solution:
Let a denote the distance of the top of the ladder from the base of the wall, let
b denote the distance of the base of the wall from the bottom of the ladder, and
let c denote the length of the ladder. Then
a2 + b 2 = c 2
Since a is a constant (the top of the ladder is fixed), namely 15, we have
225 + b2 = c2
and so
db
dc
= 2c .
dt
dt
We have been told that b = 8 and db
=
2. We compute using Pythagoras
dt
√
c = a2 + b2 = 17, and hence we have
2b
2 · 8 · 2 = 2 · 17 ·
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dc
,
dt
from which we conclude
dc
dt
=
16
17
m/min.
5. A circular ferris wheel with radius 10 metres is revolving at the rate of 10 radians
per minute. How fast is a passenger on the wheel rising when the passenger is 6
metres higher than the centre of the wheel and is rising? Include units with your
answer.
Solution:
Draw a right triangle with horizontal and vertical sides and whose hypotenuse
has ends the centre of the ferris wheel and the position of the passenger. This
hypotenuse has length 10 and the vertical side has length y where the angle θ
between the horizontal side and the hypotenuse satisfies sin θ = y/10. Differentiate with respect to θ to get (cos θ)(dθ/dt) = (1/10)(dy/dt). Now substitute
y = 6 and note the 3-4-5 right triangle to get cos θ = 4/5 and also substitute
dθ/dt = 10 to get a final answer dy/dt = 80 m/min.
6. At a certain instant an aircraft is flying due east at 400 km/hr and passes directly
overhead a car traveling due southeast at 100 km/hr on a straight, level road. If
the aircraft is traveling at an altitude of 1 km, how fast is the distance between
the aircraft and the car increasing 36 seconds after the airplane passes directly over
the car? Hint: Draw a very careful diagram! You shall also need the Cosine Law:
For a triangle with sides a, b, and c and the angle µ opposite the side c, one has
c2 = a2 + b2 − 2ab cos(µ).
Solution:
Let C be the position of the car when the aircraft is directly overhead and A be
the position of the aircraft at that time. Let s = s(t) be the separation between
the aircraft and the car at any given time t (in hours). Using the Pythagorean
Theorem and the Law of Cosines,
s 2 = 1 + x2 + y 2 −
√
2xy,
where x is the distance of the car from C at time t and y is the distance of the
aircraft from A at time t. Differentiating with respect to t gives
2s
√ dy
ds
dx
dy √ dx
= 2x + 2y − 2 y − 2x .
dt
dt
dt
dt
dt
We have dx/dt = 400 and dy/dt = 100. Also, when t =
seconds in
√ 1/100 (36 √
hours), we have x = 4, y = 1, and s2 = 1 + 16 + 1 − 4 2 = 18 − 4 2, which
gives s ≈ 3.5133. Solving for ds/dt above gives
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√
√
1
ds
= (400(8 − 2) + 100(2 − 4 2)) ≈ 323.
dt
2s
So, the separation is increasing at a rate of ≈ 323 km/h after 36 seconds.
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