calculus solution

Question

The altitude of a triangle is increasing at a rate of  centimeters/minute while the area of the triangle is increasing at a rate of  square centimeters/minute. At what rate is the base of the triangle changing when the altitude is  centimeters and the area is  square centimeters?

Answer 1

Area of the triangle = A = (1/2)bh = one half the base times the height or altitude = 92

bh= 184

b=2A/h= 2(92)/8.5

Take the derivative of A(t) = A' = dA/dt = 5 = (1/2)[bh' + hb'] = (1/2)[h'184/h + hb']

The derivative of a product is the sum of first term times derivative of the 2nd, plus

2nd term times derivative of the first. Use the chain rule, given all variables are

function of time t

h'=dh/dt = 3

h=8.5

5 = (1/2)[3(184)/8.5 + 8.5(b')

10= 552/8.5 + 8.5b'

85 = 552 + 72.25 b'

72.25b' = 85-552 = -467

b' = db/dt = rate of change of the base with respect to time = -467/72.25 = about - 6.46 centimeters per minute. The base is shrinking