The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 4 square cm/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 97 square centimeters?

a = bh/2

da/dt = 1/2 (h db/dt + b dh/dt)

Now just plug and chug, solving for db/dt

To find the rate at which the base of the triangle is changing, we can use the relationship between the altitude, the base, and the area of a triangle.

The area of a triangle is given by the formula: A = (1/2) * base * height, where A is the area, base is the length of the base, and height is the altitude.

Differentiating both sides of the equation with respect to time (t), we get:

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

Where dA/dt is the rate of change of the area, dh/dt is the rate of change of the altitude, db/dt is the rate of change of the base, and h is the altitude.

In this case, we are given:

dh/dt = 2 cm/minute (rate of change of the altitude)
dA/dt = 4 square cm/minute (rate of change of the area)
h = 10.5 cm (altitude)
A = 97 square cm (area)

Substituting these values into the equation, we get:

4 = (1/2) * (db/dt * 10.5 + b * 2)

To solve for db/dt, we need to rearrange the equation and isolate db/dt:

8 = db/dt * 10.5 + 2b

db/dt = (8 - 2b) / 10.5

Now, we can substitute the given values of b (base) and A (area) into the equation to find the rate at which the base is changing:

db/dt = (8 - 2 * b) / 10.5
= (8 - 2 * 97) / 10.5
= (8 - 194) / 10.5
= -186 / 10.5
= -17.714 cm/minute

Therefore, the base of the triangle is changing at a rate of -17.714 cm/minute when the altitude is 10.5 centimeters and the area is 97 square centimeters. The negative sign indicates that the base is decreasing.