Rates question (calc1)

Question

The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 2 square cm/minute. At what rate is the base of the triangle changing when the altitude is 9.5 centimeters and the area is 81 square centimeters?

The base is changing at ________ cm/min.

Answer 1

A = bh/2

Area = base times height divided by 2

A=81, h= 9.5

81=b(9.5)/2

162 = 9.5b

b = 162/9.5

take the derivative of A, using the product rule

A' = (1/2)(bh' + hb')

2 = (1/2)(162(2)/9.5 + 9.5b')

4 = 324/9.5 + 9.5b'

9.5b' = 4 - 324/9.5

b' = 4/9.5 - 324/90.25 = 38/90.25 - 324/90.25 = - 286/90.25

= about -3.17 cm/min

Answer 2

If A = 81 cm² and h = 9.5 cm then b = 2A/h = 162cm²/9.5 cm = 17.05 cm; A = 0.5hb;dA/dt = 0.5hdb/dt + 0.5bdh/dt; db/dt = (2dA/dt - bdh/dt)/h = (2·2cm²/min - 17.05 cm·2cm/min)/9.5 cm = - 3.17 cm/min