The altitude of a triangle is increasing at a rate of 1.5 centmeters

Question

The altitude of a triangle is increasing at a rate of 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 3 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10 centimeters and the area is 84 square centimeters?

Answer 1

Relate the two rates:

Altitude is height (I'm assuming), Area = 1/2 Base * height. (1)

They give dh/dt = 1.5 cm/min and dArea/dt = 3 cm^2 / min.

note that since they give 2 rates, its likley you will get a formula with both of them in. Lets call dArea/dt : dA/dt, base b and height h.

Then using the product rule, dA/dt = 1/2 ( b*dh/dt + h*db/dt). (2)

Plug what you have into (1), you get b = 16.8

Now plug everything into (2).

Note that you have everything except db/dt. After some algebra, the solution is db/dt = -1.92. This just means that the base is shrinking as the height and area expand. If the base started out really small, it would end up expanding as the area increases.

Don't be thrown by the negative sign in the solution.

Answer 2

A = 1/2bh. dh/dt = 1.5 , dA/dt = 3. , h = 10 , A = 84. , b = 16.8. , db/dt = ?

dA/dt = 1/2(h·db/dt + b·dh/dt)

6 = (10·db/dt + 16.8·1.5)

db/dt = - 1.92 cm/min