Related Rates

Related Rates

Related Rates (Definition and Process)

Another synonym for the word derivative is rate or rate of change.  When you hear the word rate you should identify \( \frac{d}{dt} \), since rate always corresponds to the derivative with respect to time.  

To solve a related rate problem you should do to following:

1)  Draw the picture (if applicable).

2)  Identify what derivatives are known.

3)  Identify what derivative is asked for.

4)  Find an equation that relates only the variables for which the derivative are known and the variable for which the derivative is asked for.

5)  Implicitly take the derivative with respect to \( t \) (apply \( \frac{d}{dt} \) to both sides of the equation)

6)  Plug in all known constants

7)  Answer the question.

Examples

Example

The radius of a spherical balloon is increasing at a rate of 5 inches per minute.  Find the rate of change of the volume when \( r = 3 \) inches.

Solution:  

(1)        

(2)  \( \frac{dr}{dt} = 5 \)

(3)  \( \frac{dV}{dt} = ? \)

(4)  \( V = \frac{4}{3} \pi r^3 \)  Notice that \( r \) is a variable (not 3).

(5)  \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \)

(6)  \( \frac{dV}{dt} = 4 \pi (3)^2 (5) \)

(7)  The volume is expanding at a rate of 180 \( \pi \) inches per minute.

Example

Two cars start moving from the same point.  One travels south at 40 m/hr and the other travels west at 30 m/hr.  At what rate is the distance between the cars increasing two hours later?

Solution

We construct a triangle as shown below.

the rate of change of the distance between the two cars is given by

        \( \frac{ds}{dt} \)

We know

        \( \frac{dx}{dt}  =  40 \)        and        \( \frac{dy}{dt}  =  30 \)

The Pythagorean theorem gives

        \( x^2 + y^2 = s^2 \)

Now implicitly differentiate with respect to t to get

        2x dx/dt + 2y dy/dt  =  2s ds/dt  \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2s \frac{ds}{dt}  \)

Now it is time to plug in our numbers.  We know that after 2 hours, 

        \( x  =  40(2)  =  80 \)

and

        \( y  =  30(2)  =  60 \)

We find s by 

        \( s^2 = (80)^2 + (60)^2 = 10,000 \)

Taking the square root gives

        \( s = 100 \)

Now we can plug in to get

        \( 2(80)(40) + 2(60)(30)  =  2(100) \frac{ds}{dt} \)

Dividing by 200 gives

        \( 32 + 18  =  \frac{ds}{dt} \)

        ds/dt  \( \frac{ds}{dt} =  50 \)

The distance between the two cars is changing at a rate of 50 miles per hour.

Exercises

A.  At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at the rate of 20 cubic feet per minute.  The diameter of the base of the cone is approximately three times the altitude.  At what rate is the height of the pile changing when it is 10 feet high?

B.  A boat is pulled by means of  a winch on the dock 12 feet above the deck of the boat.  The winch pulls in rope at the rate of  3 feet per second.  Determine the speed of the boat when there is 13 feet of rope out.  

C) Biologists are attempting to restore a ruined coral reef.  Once the coral has been established, will will grow in all directions along a large rock.  The growth rate of the coral is a constant 40 square centimeters per year.  How fast will the radius of the coral reef be growing when the radius reaches 100 cm?

 D)  A light shines from the ground onto a wall 25 feet away.  A man that is 6 feet tall walks away from the spotlight at a rate of 2 feet per second.  How fast is the height of his shadow decreasing when he is 10 feet away from the light?

E)  Boyle's Law states that when an ideal gas is compressed, the product of the volume and the pressure stays constant.  A cylindrical piston with radius 2 inches and height 12 inches is compressed at a rate of 3 cm³/sec.  If the pressure is 80 kPa when the cylinder is at fully extended, how fast is the pressure increasing when the cylinder is compressed to a height of 4 inches? 

Back to the Differentiation page