Related Rates: The altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute.

Question

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

Answer: ______cm/min

Answer 1

Related Rates: The altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11.5 centimeters and the area is 93 square centimeters?

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

Related Rates - Derivative with respect to time:

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

Given: h = 11.5 centimeters, A = 93 square centimeters, b = ??

A = (1/2)(b * h) ==> b = 2A/h ==> b = 2(93)/11.5 ==> b = 16.17391304 centimeters

h = 11.5 centimeters, b = 16.17391304 centimeters, dA/dt = 3.5 square centimeters/minute, dh/dt = 1 centimeters/minute, db/dt = ?? centimeters/minute

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

3.5 = (1/2)[16.17391304 * 1 + 11.5 * db/dt]

3.5 = 16.17391304/2 + 11.5/2 * db/dt

11.5/2 * db/dt = 3.5 - 16.17391304/2

5.75*db/dt = 3.5 - 8.086956522

db/dt = -4.586956522/5.75 = -0.7977 centimeters/minute

db/dt = -0.7977 centimeters/minute

The base of the triangle is decreasing at a rate of 0.7977 centimeters/minute

Answer 2

A = 1/2bh where "A" is the area, "b" is the base, "h" is the altitude (aka 'height')

Taking the derivative with respect to time when both b and h are changing (are variables), means you would use the product rule [(u•v)' = u'v + uv' and u = 1/2b, v = h, u' = 1/2(1)•db/dt, and v' = (1)•dh/dt]:

or, simplified:

dA/dt = 0.5(h•db/dt + b•dh/dt)

We know:

dA/dt = 3.5

h = 11.5

dh/dt = 1

We are looking for db/dt

Notice we still need "b". We can calculate "b" from A = 1/2bh using A = 93 and h = 11.5:

93 = 1/2b(11.5)

b = (93)(2)/11.5 = 16.1739

So plugging in the numbers we get:

3.5 = 0.5[(db/dt)•11.5 + (1)(16.1739)]

7 = 11.5•db/dt + 16.1739

-9.1739 = 11.5•db/dt

db/dt = -0.7977 cm/min