Related Rates Examples

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.

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. dV = 10 ft3 /min dt Want to find the rate of change of the depth of the water when water is dh 8 feet deep: =? dt

12 ’

Rate at which the water is flowing into the tank :

Step 2: Develop a model or formula

h

Figure 1: Conical Tank

In this problem we have a conical (cone) shaped tank. 1 V = πr2 h (Volume of a cone where, V = volume, h = height, r = radius) 3 Step 3: Use the model and the given information to solve the problem 1 V = πr2 h 3 Substitute in r =

(1)

5 h (see similar triangles section below) and simplifying we end up with 12 V =

25 πh3 432

(2)

Differentiate both sides of equation(3) with respect to t (time) dV 75 dh = πh2 dt 432 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. dh dV when height of the water is 8 feet (h = 8). Substitute in = 10 and h = 8 into Solve for dt dt equation (3),

10 = Therefore,

dh 9 feet/minute. = dt 10π

Similar Triangles:

5

r

12 h

r h = 5 12 Solving for r, r =

5 h 12

75 dh π(8)2 432 dt

(4)

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 ?

Step 1: Identify what is given in the problem. x

Plane is 10 (s = 10) miles away from the radar. Plane is flying at an altitude of 5 miles. (Let y = 5 )

y s

Rate at which the distance between the plane and radar is changing: ds = 240 miles / hour. dt 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

Figure 2: Airplane

Step 2: Develop a model or formula

We must apply the Pythagorean Theorem: s2 = x2 + y 2 Step 3: Use the model and the given information to solve the problem

Using implicit differentiation,

2s

ds dx dy = 2x + 2y dt dt dt

Since the plane is flying in the ”x-direction” then there is no change in y, hence

(5) dy = 0. dt

Equation (5) becomes

2s This further reduces to

dx ds = 2x dt dt

(6)

s

ds dx =x dt dt

(7)

√ √ Looking at the above figure, when s = 10 and y = 5, implies that x = 102 − 52 = 5 3. √ √ ds dx Substituting = 240, s = 10, and x = 5 3 into equation (7), we get that = 480/ 3 ≈ 277.13 dt dt 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 by 1 1 1 = + R R1 R2

(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 = +1 ohms/sec dt

Increasing rate at resistor 2:

dR2 = +1.5 ohms/sec dt

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. 1 1 1 = + R R1 R2

(9)

Step 3: Use the model and the given information to solve the problem

We can re-write equation (9) as

R−1 = R1−1 + R2−1

(10)

Using implicit differentiation on equation we get the following related rate equation, 1 dR 1 dR1 1 dR2 = 2 + 2 2 R dt R1 dt R2 dt

(11)

Solving for

dR dt dR = R2 dt

µ

1 dR1 1 dR2 + 2 2 R1 dt R2 dt

¶ (12)

dR1 dR2 , , R1 , and R2 but not provided with R. Using equation dt dt 1 1 1 (9) we can solve for R: = + ⇒ R = 30 R 50 75 dR Solving for : dt We are given

dR = 302 dt Therefore,

dR = 0.6 ohms/sec. dt

µ

1 1 (1) + 2 (1.5) 2 50 75

¶ (13)