Example 1

EXAMPLE 1: The side of a square increases at a rate of 3 cm/sec. Find the rate of change of the square’s perimeter and it’s area when the length of the side is 4 cm.

Solution: What rates are we trying to relate? What rates are given?

         Known:   ds/dt = 3 cm/sec    Unknown:     dP/dt = ?   dA/dt = ?

                                                                           when      s = 4 cm

Draw a sketch of the square and label the side, s. What equation relates s and P?


What equation relates s and A?

        P = 4s             A = s2

Now take the derivative of each of these equations with respect to time.

        dP/dt = 4

        dA/dt = 2s ds/dt         (Remember to use the Chain Rule.)

Plug in values to finish the problem.

       dP/dt = 4 cm/sec

        dA/dt = 2(4cm) (3 cm/sec) = 24 cm2/sec


Click here for EXAMPLE 2.