Calculus Question about rate of a changing base

Question

The altitude (i.e., height) of a triangle is increasing at a rate of 2.5 cm/minute while the area of the triangle is increasing at a rate of 5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 88 square centimeters?

Answer 1

Hi Stefania!

To answer this question, we want to relate the rates of change of altitude, base, and area.

To do so, let's first look at the formula for the area of a triangle: Area = .5 * base * altitude, which I'll write as A = .5*b*h (h = height = altitude).

We then differentiate - dA/dt = .5(b*dh/dt + h*db/dt), where I used the product rule to differentiate.

Now that we've related all our rates of change, we simply plug in our values. We're told the area is 88 cm and altitude is 8 cm, allowing us to solve for the base. I'll call that value b^*.

Plugging in the known rates of change, we then have 5 cm^2/min = .5(b^**2.5 cm/min + 8 cm * db/dt). Solving for db/dt is now simply algebra, which I will leave for you to complete.