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.