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.

Example 2  A 15 foot ladder is resting against the wall.  The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of  ft/sec.  How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?

Solution

The first thing to do in this case is to sketch picture that shows us what is going on.

We’ve defined the distance of the bottom of the latter from the wall to be x and the distance of the top of the ladder from the floor to be y.  Note as well that these are changing with time and so we really should write  and .  However, as is often the case with related rates/implicit differentiation problems we don’t write the  part just try to remember this in our heads as we proceed with the problem.

Next we need to identify what we know and what we want to find.  We know that the rate at which the bottom of the ladder is moving towards the wall.  This is,

Note as well that the rate is negative since the distance from the wall, x, is decreasing.  We always need to be careful with signs with these problems. 

We want to find the rate at which the top of the ladder is moving away from the floor.  This is .  Note as well that this quantity should be positive since y will be increasing.

As with the first example we first need a relationship between x and y.  We can get this using Pythagorean theorem.

All that we need to do at this point is to differentiate both sides with respect to t, remembering that x and y are really functions of t and so we’ll need to do implicit differentiation.  Doing this gives an equation that shows the relationship between the derivatives.

                                                                                                                     (1)

Next, let’s see which of the various parts of this equation that we know and what we need to find.  We know  and are being asked to determine  so it’s okay that we don’t know that.  However, we still need to determine x and y. 

Determining x and y is actually fairly simple.  We know that initially  and the end is being pushed in towards the wall at a rate of  ft/sec and that we are interested in what has happened after 12 seconds.  We know that,

So, the end of the ladder has been pushed in 3 feet and so after 12 seconds we must have .  Note that we could have computed this in one step as follows,

To find y (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the values of x that we just found above.

Now all that we need to do is plug into (1) and solve for .

Notice that we got the correct sign for .  If we’d gotten a negative then we’d have known that we had made a mistake and we could go back and look for it.

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.

Example 4  A tank of water in the shape of a cone is leaking water at a constant rate of .  The base radius of the tank is 5 ft and the height of the tank is 14 ft. 

(a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?

(b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

Solution

Okay, we should probably start off with a quick sketch (probably not to scale) of what is going on here.

As we can see, the water in the tank actually forms a smaller cone with the same central angle as the tank itself.  The radius of the “water” cone at any time is given by r and the height of the “water” cone at any time is given by h.  The volume of water in the tank at any time t is given by,

and we’ve been given that .

(a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?

For this part we need to determine  when  and now we have a problem.  The only formula that we’ve got that will relate the volume to the height also includes the radius and so if we were to differentiate this with respect to t we would get,

So, in this equation we know  and h and want to find , but we don’t know r and .  As we’ll see finding r isn’t too bad, but we just don’t have enough information, at this point, that will allow us to find  and  simultaneously. 

To fix this we’ll need to eliminate the r from the volume formula in some way.  This is actually easier than it might at first look.  If we go back to our sketch above and look at just the right half of the tank we see that we have two similar triangles and when we say similar we mean similar in the geometric sense.  Recall that two triangles are called similar if their angles are identical, which in the case here.  When we have two similar triangles then ratios of any two sides will be equal.  For our set this means that we have,

If we take this and plug into our volume formula we have,

This gives us a volume formula that only involved the volume and the height of the water.  Note however that this volume formula is only valid for our cone, so don’t be tempted to use it for other cones!  If we now differentiate this we have,

At this point all we need to do is plug in what we know and solve for .

So, it looks like the height is decreasing at a rate of 0.1386 ft/hr.

(b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

In this case we are asking for  and there is an easy way to do this part and a difficult (well, more difficult than the easy way anyway….) way to do it.  The “difficult” way is to redo the work part (a) above only this time use,

to get the volume in terms of V and r and then proceed as before.

That’s not terribly difficult, but it is more work that we need to so.  Recall from the first part that we have,

So, as we can see if we take the relationship that relates r and h that we used in the first part and differentiate it we get a relationship between  and .  At this point all we need to do here is use the result from the first part to get,

Much easier that redoing all of the first part.  Note however, that we were only able to do this the “easier” way because it was asking for  at exactly the same time that we asked for  in the first part.  If we hadn’t been using the same time then we would have had no choice but to do this the “difficult” way.