Question

i have done this problem multiple times and still cant get it. Help please





(1 pt)

The altitude of a triangle is increasing at a rate of 2.5 centimeters/minute while the area of the triangle is increasing at a rate of 1 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.5 centimeters and the area is 97?

Answer 1

A = 1/2 * b * h
dA/dt = 1/2 * (b*dh/dt + h * db/dt)
Plug in the numbers and find db/dt.

Answer 2

I have done that, and im not getting the anwer. I get something like 5.424, every time which is wrong

Answer 3

A = 1/2 * b * h
A = 97, h = 9.5
97 = 0.5 * b * 9.5
b = 20.42

dA/dt = 1/2 * (b*dh/dt + h * db/dt)
dA/dt = 1
dh/dt = 2.5
b = 20.42
h = 9.5
1 = 0.5 * (20.42 * 2.5 + 9.5 * db/dt)
2 = 20.42 * 2.5 + 9.5 * db/dt
2 = 51.05 + 9.5 * db/dt
db/dt = (2-51.05)/9.5 = -5.16 cm / minute.

Answer 4

ok, I see what I did there. It was the area, I never took over the 1/2. Thank u!

Answer 5

You are welcome.