Homework Help Please

Question

The height 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 per minute. At what rate is the base of the triangle changing when the height is  centimeters and the area is  square centimeters?

Remember that you're using the formula for the area of a triangle: Area = 1/2 b⋅h

__________=cm/min

Answer 1

Calculus is the study of variations and related rates...

So we have the following area for the formula for the triangle --> A = 1/2 * b * h

The change in the area of the triangle is related to the change in the base as well as the height ( in general ), hence why they are related rates. b and h are "variables" in this case...

dA/dx = 1/2 * d/dx(b*h)

dA/dx = 1/2 * (b * dh/dx + h * db/dx)

Now the question is based on the problem, what rates are changing? Area and height are changing, which means the base is constant. Finding the rate of the base changing is a matter of solving for the db/dx...

2 * dA/dx - b * dh/dx = h * db/dx

db/dx = (2 * dA/dx - b * dh/dx) / h

The values of dA/dx and dh/dx are given and the value of b and h are related to the specific time of the computation of A = 1/2 * b *h