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?
NEED SKETCH
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.
