Rate of Change Calculus

Question

A triangle has a base that is decreasing at a rate of 10 feet per hour with the height being held constant. What is the rate of change of the area of the triangle if the height is 6 feet?

Answer 1

A triangle has a base that is decreasing at a rate of 10 feet per hour with the height being held constant. What is the rate of change of the area of the triangle if the height is 6 feet?

OK, I will set this up for you and then you can either take it the rest of the way or schedule an appointment with me to do that with you.

The area, A, of any triangle is (1/2)BH where B is the base dimension and H is the height dimension. Note that, while B is the length of the leg opposite the top vertex, H is often not: it is the perpendicular distance from the top of the triangle to the base.

OK, so we have A=(1/2)BH so

2dA/dt=BdH/dt +HdB/dt

Since we know the height is constant, dH/dt=0 so

2dA/dt=HdB/dt

We know dB/dt=-10 ft/hr and we need to know dA/dt when H=6 ft. You can do the rest!

Answer 2

Let the base of the triangle = b, the height of the triangle = h and the area of the triangle = A

Area of a triangle, A = 1/2(bh)

Rate of change of the area, A is

dA/dt = d(1/2(bh))/dt = 1/2 (db/dt)h + 1/2(bdh/dt)

At constant height, dh/dt = 0 ===> dA/dt = 1/2(db/dt)h

We are told that the base is decreasing at a rate of 10 feet per hour, i.e db/dt = -10ft/h, and height, h = 6ft.

Therefore, dA/dt = 1/2(-10ft/h)6ft + 0 = -30ft²/h.

Hence, the area of the triangle is decreasing at a rate of 30ft² per hour.