HElp question down below

Question

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min.

 At what rate is the base of the triangle changing when the altitude is 20 cm

 and the area is 150 cm²?

Answer 1

A = hb/2= half height times base = area of a triangle

take the derivative with respect to time

A' =dA/dt = (1/2)[hb' + bh']

h'=1

A'=2

A= 150

h=20

b= 2(150)/20=15 cm = base length

A= (1/2)20(15) = 150

plug in the values

A' = (1/2)[hb' +bh']

2= (1/2)20(b') + (1/2)(15)(1)

4 = 20b' +15

b'= (4-15)/20 = -11/20 = -.55

b' = -0.55 cm/second= rate of change of the base

Answer 2

The calc AP exams refer to this type of problem as "related rates". The idea is to write down a simple equation relating the two variables, get a differential equation, then solve for the desired quantity. In this case the equation will be the area formula for a triangle, A = 1/2*base*height. The variable that is changing, the independent variable, is time, so you think of the variables as functions of time and then differentiate both sides.

A= 1/2 * b(t) * h(t)

dA/dt = 1/2 * b * dh/dt + 1/2 * db/dt * h

Now substitute the first two numbers for dA/dt and dh/dt.

(2) = 1/2 * b * (1) + 1/2 * db/dt * h

2 = 1/2 * b + 1/2*20*db/dt

2 = b/2 + 10*db/dt

We have the area as 150cm^2, so that means the base at that time is b=2*150/20 = 15 cm. Plugging this in and solving

2 = 15/2+10*db/dt

-11/2 = 10*db/dt

db/dt = -11/20 ≈ -0.55 cm/min.

As with all such problems double check the units at the end make sense. areas are in cm^2, speeds are all in cm/min. The speed of the base is negative which means it is shrinking. Intuitively the height is larger, but not twice as large as the base, so with relative rates of +1cm/min for the height and -0.55 cm/min for the base the area should increase not decrease.