Related Rates

Related rate problems are a type of problem where we have a rate, which is related to another, and we want to determine the rate using the one we already know. Solving these problems doesn’t require any new theories or skills, it is just a straight application of our current knowledge.

Example: Air is being pumped into a spherical balloon at a rate of 6cm^3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 15cm.

First, we identify the information we have been given. We know the rate at which the spherical balloon inflates, and the ending diameter of the balloon. We want to know the rate that the radius is increasing at the time where the radius is 15cm.

Let’s define two functions, v(t), which is the volume of the balloon at a time t, and r(t), which is the radius of the balloon at a time t.

We know that the balloon inflates at a rate of 6cm3/min, therefore, v’(t) = 6.

We also know that the volume of the sphere is v(t) = \frac{4}{3}\pi[r(t)]^3. This tells us how the volume relates to the radius; however, we want to know how the change in radius relates to the change in volume. To do this, we can take the derivative of both sides to get:

v'(t) = \frac{4}{3}\pi r^2r'(t), which we get by using the chain rule on the function r(t). From here, we can simply substitute our values and solve the function.

6 = \frac{4}{3} \pi (15^2)(r'(t))
\Longrightarrow \frac{18}{4\pi*15^2} = r'(t)
\Longrightarrow \frac{18}{900\pi} = r'(t)
\Longrightarrow \frac{1}{50\pi} = r'(t)

Example: A ladder 10 ft long rests along a vertical wall. If the bottom of the ladder sliders away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6ft from the wall?

As with the first question, we need to consider what rate we have, and which we need. For this question, a diagram will help us understand exactly what is being asked.

This diagram makes it clear that the ladder against the wall will make a right-angled triangle. We know that the ladder is falling at a rate of 1 ft/s, so this tells us that the rate, \frac{dx}{dt} = 1 ft/s. We are being asked to find \frac{dy}{dt} at x = 6ft.

Another piece of information that we know looking at the diagram is the relationship between x, y, and the ladder. By Pythagorean’s theorem, we can conclude that x^2+y^2=100

From here, we can proceed using implicit differentiation.

2x * \frac{dx}{dt} + 2y * \frac{dy}{dt} = 0

We want to know the rate \frac{dy}{dt}, so we isolate it to get:

\frac{dy}{dt} = - \frac{x}{y}\frac{dx}{dt}

From here, we just need to figure out the x and y values we want to solve. We want to determine \frac{dy}{dt} at x = 6ft, so we plug in x = 6 into the original equation to get:

x^2 + y^2 = 100
\Longrightarrow 36 + y^2 = 100
\Longrightarrow y^2 = 64
\Longrightarrow y = 8

Therefore, we have \frac{dy}{dx} = - \frac{6}{8}(1) = - \frac{3}{4}

Just to clear up a few possible questions, the value is negative since the ladder is falling. The change in y over t is a decreasing value, so we expect the rate of change to be negative. In addition, we could substitute \frac{dx}{dt} = 1 in the equation, since it was provided in the question.

The general idea of any related rates problem is to determine what rates we have, what rates we need, and solve them using an equation we can build.  

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