Calculus: Rates of change (problem below)

Question

The lower base of a trapezoid is 6 m. If its height is increasing at the rate of 1 m/sec while its area is increasing at the rate of 2 m^2/sec. At what rate is the upper base of the trapezoid changing when the height is 4 m and the area is 20 m^2?

Answer 1

Before we start, let's define our terms (which are all parts of the trapezoid and their derivatives):

b1 = lower base

b2 = upper base

h = height

A = area

Anything with a quote is its time derivative. For instance, A' equals the change in area with time.

Now, let's start with the fundamental equation relating the variables:

A = (b1 + b2)*h/2

We want to take a derivative of this equation, so let's first identify which parameters are constant: b1 only. Now let's take the derivative. This requires the product rule.

A' = (b1 + b2)*h'/2 +h/2*b2'

We want to solve this equation for b2'.Let's figure out which of these values we are given. We know, A', b1, h' and h. We still need b2. This is where we use the original equation for A before differentiating to find b2. Solving we get:

b2 = 2A/h - b1

Plugging in the values yields:

b2 = 2*20/4 - 6

b2 = 4 m

Now we need to solve the differentiated equation for b2'. This yields:

b2' = 2/h*[A' - (b1 + b2)*h'/2]

Plugging in the values we now have:

b2' = 2/4*[2 - (6+4)*1/2]

b2' = -1.5 m/s

This means that at the given point in time, the upper base is decreasing at a rate of 1.5 m/s.