The altitude of a triangle is increasing at a rate of 3.000 centimeters/minute while the area of the triangle is increasing at a rate of 2.500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.000 centimeters and the area is 93.000 square centimeters?
Note: The "altitude" is the "height" of the triangle in the formula "Area=(1/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shape the triangle really is.
To solve this problem, we can use the chain rule from calculus.
Let's denote the altitude of the triangle as h, the base as b, and the area as A.
We are given that the altitude is increasing at a rate of 3.000 cm/minute and the area is increasing at a rate of 2.500 cm^2/minute. We want to find the rate at which the base is changing, db/dt, when the altitude is 9.000 cm and the area is 93.000 cm^2.
We know that the area of a triangle is given by the formula A = (1/2) * b * h. Taking the derivative of both sides with respect to time t, we have:
dA/dt = (1/2) * (dh/dt) * b + (1/2) * h * (db/dt)
Now, let's substitute the given values into this equation:
dA/dt = 2.500 cm^2/minute
dh/dt = 3.000 cm/minute
h = 9.000 cm
A = 93.000 cm^2
Plugging these values into the equation above, we have:
2.500 = (1/2) * 3.000 * b + (1/2) * 9.000 * (db/dt)
Simplifying further:
2.500 = 1.500b + 4.500(db/dt)
Now, let's solve for db/dt:
2.500 - 1.500b = 4.500(db/dt)
(db/dt) = (2.500 - 1.500b) / 4.500
Substituting the given values of b = ? (unknown base length), we can find db/dt at the desired point.
This formula allows us to find the rate of change of the base of the triangle when the altitude is 9.000 cm and the area is 93.000 cm^2.