Related Rates
In this section we are going to look at an application of implicit differentiation. Most
of the applications of derivatives are in the next chapter however there are a couple of
reasons for placing it in this chapter as opposed to putting it into the next chapter with
the other applications. The first reason is that it’s an application of implicit
differentiation and so putting right after that section means that we won’t have
forgotten how to do implicit differentiation. The other reason is simply that after
doing all these derivatives we need to be reminded that there really are actual
applications to derivatives. Sometimes it is easy to forget there really is a reason that
we’re spending all this time on derivatives.
For these related rates problems it’s usually best to just jump right into some problems
and see how they work.
Example 1
Air is being pumped into a spherical balloon at a rate of 5 cm
3
/min. Determine the
rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.
Solution
The first thing that we’ll need to do here is to identify what information that we’ve been given
and what we want to find. Before we do that let’s notice that both the volume of the balloon and
the radius of the balloon will vary with time and so are really functions of time,
i.e.
and
.
We know that air is being pumped into the balloon at a rate of 5 cm
3
/min. This is the rate at
which the volume is increasing. Recall that rates of change are nothing more than derivatives and
so we know that,
We want to determine the rate at which the radius is changing. Again, rates are derivatives and so
it looks like we want to determine,
Note that we needed to convert the diameter to a radius.
Now that we’ve identified what we have been given and what we want to find we need to relate
these two quantities to each other. In this case we can relate the volume and the radius with the
formula for the volume of a sphere.
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View Full DocumentAs in the previous section when we looked at implicit differentiation, we will typically not use
the
part of things in the formulas, but since this is the first time through one of these we
will do that to remind ourselves that they are really functions of
t
.
Now we don’t really want a relationship between the volume and the radius. What we really
want is a relationship between their derivatives. We can do this by differentiating both sides with
respect to
t
. In other words, we will need to do implicit differentiation on the above formula.
Doing this gives,
Note that at this point we went ahead and dropped the
from each of the terms. Now all
that we need to do is plug in what we know and solve for what we want to find.
We can get the units of the derivative by recalling that,
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 Fall '08
 sc
 Derivative, Implicit Differentiation, Pythagorean Theorem, triangle

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