Question

The altitude of a triangle is increasing at a rate of 1.5 cm/min while the area of the triangle is increasing at a rate of 5 square cm/min. At what rate is the base of the triangle changing when the altitude is 9 cm and the area is 81 square cm?

Answer 1

This is a related rates (of change) type problem.

The variables of interest are
`a` = altitude
`A` = area and, since the area of a triangle is `A=1/2ba`, we need
`b` = base.

The given rates of change are in units per minute, so the (invisible) independent variable is `t` = time in minutes.

We are given:
`(da)/dt = 3/2` cm/min

`(dA)/dt = 5` cm`""^2`/min

And we are asked to find `(db)/dt` when `a = 9` cm and `A = 81`cm`""^2`

`A=1/2ba`, differentiating with respect to `t`, we get:

`d/dt(A)=d/dt(1/2ba)`.

We'll need the product rule on the right.

`(dA)/dt = 1/2 (db)/dt a + 1/2b (da)/dt`

We were given every value except `(db)/dt` (which we are trying to find) and `b`. Using the formula for area and the given values of `a` and `A`, we can see that `b=18`cm.

Substituting:

`5= 1/2 (db)/dt (9)+1/2(18)3/2`

Solve for `(db)/dt = -17/9`cm/min.

The base is decreasing at `17/9` cm/min.