Related Rates?

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 cen/min. At what rate is the base of the triangle changing when the altitude is 9 cen and the area is 81 square cen?

Answer 1

A: Area
h: height/altitude
b: base

A = 81cm²
h = 9cm
b = 2*(81/9) = 18cm
dA/dt = 5cm²/min
dh/dt = 1.5cm/min

A = (1/2)hb
dA/dt = (1/2)(h.db/dt + b.dh/dt)
5 = (1/2)(9*db/dt + 18*1.5)
10 = 9db/dt + 27
9db/dt = -17
db/dt = -17/9 cm/min
≈ -1.89 cm/min

The length of the base is decreasing at a rate of approximately 1.89 cm/min.

Answer 2

first lets find your related rates equation. (You want to find a way to relate base, height, and area.)
This is:
Area = .5 b * h
now, differentiate implicitly.
da/dt = .5b * (dh/dt) + h * .5 (db/dt).
plug in the information you are given.
da/dt = 5
dh/dt = 1.5
h = 9
a = 81.
So we get:
5 = 81/2 * 1.5 + 9 *.5(db/dt)
solving this equation for db/dt, we get:
-223/18.
Therefore the base is decreasing at a rate of
-223/18 centimeters/min

Answer 3

You need a formula, so.....
Area = 0.5 * base * altitude
A= 0.5 b * h in symbols (find b for A=81)
dA/dt = 0.5 [( b *dh/dt) + (h*db/dt)]
You have everything except db/dt, so solve for it.