Question

The altitude of a triangle is increasing at a rate of 3 centimeters/minute while the area of the triangle is increasing at a rate of 5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11 centimeters and the area is 91 square centimeters

Answer 1

The base of the triangle is decreasing at a rate of 3.60 centimeters/minute.

According to the statement

we have to find that the rate at which the base of triangle changing.

So, For this purpose,

Area of triangle: A = (1/2)(b * h)

Related Rates - Derivative with respect to time:

(d/dt)[A = (1/2)(b * h)]

Then

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

Given: h = 11 centimeters, A = 91 square centimeters, b = ??

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

Then it becomes

b = 2A/h

substitute the value then

b = 2(91)/11

b = 16.54 centimeters

Then

h = 11 centimeters, b = 16.54 centimeters, dA/dt = 5 square centimeters/minute, dh/dt = 3 centimeters/minute, db/dt = ?? centimeters/minute

So,

(d/dt)[A = (1/2)(b * h)]

Then it become

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

5= (1/2)[16.54 * 3 + 11 * db/dt]

5= 49.62/2 + 11/2 * db/dt

11/2 * db/dt = 5- 49.62/2

5.5*db/dt = 5 - 24.81

db/dt = -19.81/5.5

db/dt = -3.60 centimeters/minute

So, The base of the triangle is decreasing at a rate of 3.60 centimeters/minute.

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