Related Rates

Related Rates - Related Rates In this section we are going...

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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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As 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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Related Rates - Related Rates In this section we are going...

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