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 Calculus I - Notes

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 it 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 cm3/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 cm3/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.                                                               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,

The units of the derivative will be the units of the numerator (cm in the previous example) divided by the units of the denominator (min in the previous example).

Let’s work some more examples.

 Example 3  Two people are 50 feet apart.  One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0.01 rad/min.  At what rate is distance between the two people changing when  radians? Solution This example is not as tricky as it might at first appear.  Let’s call the distance between them at any point in time x as noted above.  We can then relate all the known quantities by one of two trig formulas.                                                   We want to find  and we could find x if we wanted to at the point in question using cosine since we also know the angle at that point in time.  However, if we use the second formula we won’t need to know x as you’ll see.  So, let’s differentiate that formula.                                                               As noted, there are no x’s in this formula.  We want to determine  and we know that  and  (do you agree with it being positive?).  So, just plug in and solve.

So far we we’ve seen three related rates problems.  While each one was worked in a very different manner the process was essentially the same in each.  In each problem we identified what we were given and what we wanted to find.  We next wrote down a relationship between all the various quantities and used implicit differentiation to arrive at a relationship between the various derivatives in the problem.  Finally, we plugged into the equation to find the value we were after.

So, in a general sense each problem was worked in pretty much the same manner.  The only real difference between them was coming up with the relationship between the known and unknown quantities.  This is often the hardest part of the problem.  In many problems the best way to come up with the relationship is to sketch a diagram that shows the situation.  This often seems like a silly step, but can make all the difference in whether we can find the relationship or not.

Let’s work another problem that uses some different ideas and shows some of the different kinds of things that can show up in related rates problems.

In the second part of the previous problem we saw an important idea in dealing with related rates.  In order to find the asked for rate all we need is an equation that relates the rate we’re looking for to a rate that we already know.  Sometimes there are multiple equations that we can use and sometimes one will be easier than another.

Also, this problem showed us that we will often have an equation that contains more variables that we have information about and so, in these cases, we will need to eliminate one (or more) of the variables.  In this problem we eliminated the extra variable using the idea of similar triangles.  This will not always be how we do this, but many of these problems do use similar triangles so make sure you can use that idea.

Let’s work some more problems.

 Example 5  A trough of water is 8 meters deep and its ends are in the shape of isosceles triangles whose width is 5 meters and height is 2 meters.  If water is being pumped in at a constant rate of .  At what rate is the height of the water changing when the water has a height of 120 cm?   Solution Note that an isosceles triangle is just a triangle in which two of the sides are the same length.  In our case sides of the tank have the same length.   We definitely need a sketch of this situation to get us going here so here.  A sketch of the trough is shown below. Now, in this problem we know that  and we want to determine  when .  Note that because  is in terms of meters we need to convert h into meters as well.  So, we need an equation that will relate these two quantities and the volume of the tank will do it.   The volume of this kind of tank is simple to compute.  The volume is the area of the end times the depth.  For our case the volume of the water in the tank is,                                                       As with the previous example we’ve got an extra quantity here, w, that is also changing with time and so we need to eliminate it from the problem.  To do this we’ll again make use of the idea of similar triangles.  If we look at the end of the tank we’ll see that we again have two similar triangles.  One for the tank itself and one formed by the water in the tank.  Again, remember that with similar triangles ratios of sides must be equal.  In our case we’ll use,     Plugging this into the volume gives a formula for the volume (and only for this tank) that only involved the height of the water.                                                        We can now differentiate this to get,                                                                     Finally, all we need to do is plug in and solve for .                                   So, the height of the water is rising at a rate of 0.25 m/sec.

 Example 7  A spot light is on the ground 20 ft away from a wall and a 6 ft tall person is walking towards the wall at a rate of 2.5 ft/sec.  How fast is the height of the shadow changing when the person is 8 feet from the wall?  Is the shadow increasing or decreasing in height at this time?   Solution Let’s start off with a sketch of this situation and the sketch here will be similar to that of the previous problem.  The top of the shadow will be defined by the light rays going over the head of the person and so we again get yet another set of similar triangles.   In this case we want to determine  when the person is 8 ft from wall or .  Also, if the person is moving towards the wall at 2.5 ft/sec then the person must be moving away from the spotlight at 2.5 ft/sec and so we also know that .   In all the previous problems that used similar triangles we used the similar triangles to eliminate one of the variables from the equation we were working with.  In this case however, we can get the equation that relates x and y directly from the two similar triangles.  In this case the equation we’re going to work with is,     Now all that we need to do is differentiate and plug values into solve to get .                     The height of the shadow is then decreasing at a rate of 2.0833 ft/sec.

Okay, we’ve worked quite a few problems now that involved similar triangles in one form or another so make sure you can do these kinds of problems.

It’s now time to do a problem that while similar to some of the problems we’ve done to this point is also sufficiently different that it can cause problems until you’ve seen how to do it.

 Example 8  Two people on bikes are separated by 350 meters.  Person A starts riding north at a rate of 5 m/sec and 7 minutes later Person B starts riding south at 3 m/sec.  At what rate is the distance separating the two people changing 25 minutes after Person A  starts riding?   Solution There is a lot to digest here with this problem.  Let’s start off with a sketch of the situation.     Now we are after  and we know that  and .  We want to know  after Person A had been riding for 25 minutes and Person B has been riding for  minutes.  After converting these times to seconds (because our rates are all in m/sec) this means that at the time we’re interested in each of the bike riders has rode,                            Next, the Pythagorean theorem tells us that,                                                                                                              (2)   Therefore, 25 minutes after Person A starts riding the two bike riders are                        apart.   To determine the rate at which the two riders are moving apart all we need to do then is differentiate (2) and plug in all the quantities that we know to find .                                            So, the two riders are moving apart at a rate of 7.9958 m/sec.

Every problem that we’ve worked to this point has come down to needing a geometric formula and we should probably work a quick problem that is not geometric in nature.

 Example 9  Suppose that we have two resistors connected in parallel with resistances  and  measured in ohms (  ).  The total resistance, R, is then given by,                                                                   Suppose that  is increasing at a rate of 0.4  /min and  is decreasing at a rate of 0.7  /min.  At what rate is R changing when  and ?   Solution Okay, unlike the previous problems there really isn’t a whole lot to do here.  First, let’s note that we’re looking for  and that we know  and .  Be careful with the signs here.   Also, since we’ll eventually need it let’s determine R at the time we’re interested in.                         Next we need to differentiate the equation given in the problem statement.                                                 Finally, all we need to do is plug into this and do some quick computations.                               So, it looks like R is decreasing at a rate of 0.002045  /min.

We’ve seen quite a few related rates problems in this section that cover a wide variety of possible problems.  There are still many more different kinds of related rates problems out there in the world, but the ones that we’ve worked here should give you a pretty good idea on how to at least start most of the problems that you’re liable to run into.

 Calculus I - Notes