Related Rate Problems
Problem 1: (Depth) A conical (cone-shaped) tank (with vertex down) is 10 feet across the top
and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the
rate of change of the depth of the water when the water is 8 feet deep.
12 ’
h
Figure 1: Conical Tank
Step 1: Identify what is given in the problem.
Top of conical tank has a diameter of 10 feet (radius = 5 feet).
Height of conical tank is 12 feet.
Rate at which the water is flowing into the tank :dV
dt= 10 ft3/min
Want to find the rate of change of the depth of the water when water is
8 feet deep:dh
dt= ?
Step 2: Develop a model or formula
In this problem we have a conical (cone) shaped tank.
V =13πr2h (Volume of a cone where, V = volume, h = height, r = radius)
Step 3: Use the model and the given information to solve the problem
V =13πr2h (1)
Substitute in r =512
h (see similar triangles section below) and simplifying we end up with
V =25432
πh3 (2)
Differentiate both sides of equation(3) with respect to t (time)
dV
dt=
75432
πh2 dh
dt(3)
Equation (3) is called a related rate problem, because there is a relation between the rate at which
the water is flowing into the tank and the rate of change of the depth of the water.
Solve fordh
dtwhen height of the water is 8 feet (h = 8). Substitute in
dV
dt= 10 and h = 8 into
equation (3),
10 =75432
π(8)2dh
dt(4)
Therefore,dh
dt=
910π
feet/minute.
Similar Triangles:
r
5
12
h
r
5=
h
12
Solving for r, r =512
h
Problem 2: (Air Traffic Control) An airplane is flying at an altitude of 5 miles and passes directly
over a radar antenna. When the plane is 10 miles away (s = 10), the radar detects that the distance
s is changing at a rate of 240 miles per hour. What is the speed of the plane ?
y
x
s
Figure 2: Airplane
Step 1: Identify what is given in the problem.
Plane is 10 (s = 10) miles away from the radar.
Plane is flying at an altitude of 5 miles. (Let y = 5 )
Rate at which the distance between the plane and radar is changing:ds
dt= 240 miles / hour.
Want to find the the speed of the plane when s = 10 ?
Note: speed = | velocity |.So we really want to find the velocity when s = 10 ?
dx
dt= ?
Step 2: Develop a model or formula
We must apply the Pythagorean Theorem: s2 = x2 + y2
Step 3: Use the model and the given information to solve the problem
Using implicit differentiation,
2sds
dt= 2x
dx
dt+ 2y
dy
dt(5)
Since the plane is flying in the ”x-direction” then there is no change in y, hencedy
dt= 0.
Equation (5) becomes
2sds
dt= 2x
dx
dt(6)
This further reduces to
sds
dt= x
dx
dt(7)
Looking at the above figure, when s = 10 and y = 5, implies that x =√
102 − 52 = 5√
3.
Substitutingds
dt= 240, s = 10, and x = 5
√3 into equation (7), we get that
dx
dt= 480/
√3 ≈ 277.13
miles / hour. Therefore, the speed is 277.13 miles / hour.
Problem 3: (Electricity) The combined electrical resistance R of R1 and R2, connected in parallel,
is given by1R
=1
R1+
1R2
(8)
where R, R1, and R2 are measured in ohms. R1 and R2 are increasing at rates of 1 and 1.5 ohms
per second, respectively. At what rate is R changing when R1 = 50 ohms and R2 = 75 ohms ?
Step 1: Identify what is given in the problem.
Increasing rate at resistor 1:dR1
dt= +1 ohms/sec
Increasing rate at resistor 2:dR2
dt= +1.5 ohms/sec
Want to find the the rate of R when R1 = 50 ohms and R2 = 75 ohms:dR
dt= ?
Step 2: Develop a model or formula
In this case we are provided a model to work with.
1R
=1
R1+
1R2
(9)
Step 3: Use the model and the given information to solve the problem
We can re-write equation (9) as
R−1 = R−11 + R−1
2 (10)
Using implicit differentiation on equation we get the following related rate equation,
1R2
dR
dt=
1R2
1
dR1
dt+
1R2
2
dR2
dt(11)
Solving fordR
dt
dR
dt= R2
(1
R21
dR1
dt+
1R2
2
dR2
dt
)(12)
We are givendR1
dt,
dR2
dt, R1, and R2 but not provided with R. Using equation
(9) we can solve for R:1R
=150
+175⇒ R = 30
Solving fordR
dt:
dR
dt= 302
(1
502(1) +
1752
(1.5))
(13)
Therefore,dR
dt= 0.6 ohms/sec.
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