#### Question

# Calculus rate of change?

"The length L of a rectangle is decreasing at a rate of 2cm per second, and teh width W is increasing at the rate of 4cm per second. Find the rate of change of the area when L = 13cm and W = 10cm."

Can someone please explain the above and work through it, i am not understanding these types of questions.

#### Answer

Area = A = L W

dA/dt = L dW/dt + W dL/dt

when L = 13 cm and W = 10 cm

and dL/dt = -2 cm/s and dW/dt = 4 cm/s

dA/dt = 13(4) + 10(-2) = 52-20 = 32 cm/s

May 5 at 13:4

= ([13 - 2][10 + 4]) - 13(10)

= 11(14) - 130

= 154 - 130

= 24

Answer: increasing 24 sq cm per second

May 5 at 16:50

Area = LW

dArea/dt = L (dW/dt) + W (dL/dt)

dArea/dt = 13 (4) + 10 (-2)

dArea/dt = 32 cm2/s

May 5 at 20:59