# Section 8: More Tricks with Derivatives

## 8.2 The Dance of the Derivatives: Related Rates

### A Shaggy Dog Story

That little doggie looked so cute when you first brought him home. How could any creature so adorable make such a mess of your house? Here is his latest (though by far not his messiest) caper. It seems he has discovered the bathroom. And in it is this wonderful toy hanging on the wall with a soft end dangling down that is just begging to be pulled. Which is what he is now in the habit of doing. He trots into the bathroom, grabs the end of the toilet paper, and then runs out the door with a streamer unreeling behind him.

With all this happening, amazingly you are thinking about calculus and asking yourself, "If the dog runs at 2 meters per second, how quickly is the radius of the toilet paper roll decreasing?" Of course you are.

Before we dive into the how to of this problem, we need a little more information. One is how thick is the toilet paper. Why do we need to know this? Well if the paper were very thick then we'd expect that each turn of the roll would reduce the radius by more than if the paper were thin. We'll say that this particular brand is 0.00005 meters thick (that's about 2 mils in the English system of measures). Also since the radius of the roll is changing at a rate which is perhaps not constant, we need the problem to state at exactly what moment we want to know this rate of change. So let's say it is at the moment when the radius of the roll is 0.05 meters (about 2 inches).

The first trick in this problem is to relate the radius of the roll to how much toilet paper the dog has already pulled off. Visualize the roll in crossection. It has a crossectional area of A = πr2, where A is area and r is radius. I could, at this point, substitute 0.05 meters for r, but usually it is good practice to wait until the end of the problem before you substitute values for variables. So that's what we'll do.

Now consider the thickness of the toilet paper, ξ (which the problem stated was equal to 0.00005 meters), and think about how much area a length, s, of toilet paper of that thickness contributes to the crossectional area of the roll. If you look at that piece of toilet paper edge on, it is a very long and thin rectangle. And the area of that rectangle is thickness times length, or ξs.

The rate at which the dog pulls toilet paper off the roll is given in the problem. And that rate is the time derivative (where t is the time variable) of the length of paper he has already pulled.

```   ds
=  2 meters/sec                                            eq. 8.2-1
dt
```
Do you see what that equation is saying? The rate at which length of paper, s, is being pulled off the role with respect to time, t, is 2 meters per second.

With each meter of paper the dog pulls off, the crossectional area of the roll is reduced by ξ times one meter. The rate at which he pulls the paper off is given in eq. 8.2-1 in meters per second. So the equation for how fast the crossectional area of the roll is reduced is

```   dA        ds
=  -ξ                                                      eq. 8.2-2
dt        dt
```
What this equation says is that the crossectional area of the roll, A, is reduced (that's where the minus sign comes from), and the rate with respect to time, t, that it is reduced is the paper's thickness, ξ, times the rate it is being pulled off, ds/dt, in meters per second. Note that dA/dt has units of square meters per second.

But how is all this crossectional area stuff related to the radius of the roll? The crossection of the roll is a circle. So we get this relationship from how the area of a circle is related to its radius.

```   A  =  πr2                                                      eq. 8.2-3a
```
But both A and r are functions of time, t. That is, it would be reasonable to express them as A(t) and r(t) respecively. So another more explicit way to write this is
```   A(t)  =  πr2(t)                                                eq. 8.2-3b
```
Now look at what happens when you take the time derivative of both sides of equation 8.2-3b. Clearly on the left we get dA/dt. On the right, though, we have a function of a function -- that is a composite. It is a composite of the radius function of time, r(t), and the function of squaring the radius. So we have to apply the chain rule.

The derivative of squaring the radius is 2r(t). And the derivative of the radius function itself, r(t), is dr/dt. So the derivative of the radius function squared is 2r(t) dr/dt. That gives us, for the entire equation of the derivative of equation 8.2-3b:

```   dA            dr
=  2πr(t)                                                  eq. 8.2-4
dt            dt
```
Observe what the problem asks for. It is asking for the rate at which the radius of the roll decreases. Isn't that dr/dt? So we need to solve for that:

 ``` dA 1 1 = dt 2π r(t) ``` ``` dr eq. 8.2-5a dt ```
We already have an expression for dA/dt from equation 8.2-2. So we substitute that in for dA/dt.

 ``` -ξ ``` ``` ds 1   dt 2π ``` ``` 1 = r(t) ``` ``` dr eq. 8.2-5b dt ```
And now it all works out because ds/dt, ξ, and r(t) are given in the problem. Remember

```   ds
=  2 meters/second
dt

r(t)  =  0.05 meters

ξ      =  0.00005 meters
```
So finishing the solution is just a matter of plugging in values. My calculator tells me that
```   dr
=  -0.0003183 meters/second                                eq. 8.2-5c
dt
```
which is about -0.012 inches per second. Observe that the result is negative. This indicates that the radius of the roll is decreasing with time, as we would expect.

Seeing that the sign of the result is as you would expect it is just one of the things you can do to check your solution for correctness. Here are some others:

1. If you increase the thickness, ξ, of the paper, do you expect that the roll will shrink faster or slower? It would shrink faster because each turn of the roll takes off more area from the crossection of the roll. Now look at equation 8.2-5b. Does it predict that increasing the thickness makes the roll shrink faster?
2. If the dog runs faster, that is if you increase ds/dt, do you expect the roll to shrink faster or slower? Faster of course. Does equation 8.2-5b predict that?
3. If the radius, r(t), is smaller, do we expect the roll to shrink faster or slower? Think about this one. With a smaller radius it's circumference is less, so with each meter of paper pulled you will get more turns of the roll. So the roll shrinks faster when it is smaller. Does the way r(t) appears in equation 8.2-5b predict this?

Perhaps you have not been in the habit of doing checks like the ones I have shown above. Now is a good time to start. You will save yourself a lot of wrong answers by performing reasonableness checks on all your solutions to problems, especially if you are headed into engineering or the sciences. Remember that if someday you are solving problems as part of your profession, there will be no answers in the back of the book for you to check against. You will have to know whether or not your answer is right on your own.

### The Seven Veils: The Questions to Ask Yourself When Doing Related Rates Problems

Question 1: What is the independent variable in this problem? The independent variable is usually (but not always) time, which you traditionally notate using the symbol, t. The key phrases that indicate time to be the independent variable are "how fast," "how often," "at what speed," "at what rate is this or that changing," and so on. In rare cases a related rate problem might have something other than time as its independent variable. If, for example, the problem asks "how much water boils for each gallon of fuel," you can be relatively sure that the independent variable is gallons of fuel. The word, "per," or the phrase, "for each," gives you a clue that the word or phrase that follows is likely to be an independent variable.

Question 2: What dependent variables does the problem reveal? In the case of the doggie problem above, radius of the roll is clearly a dependent variable, since the problem asks you about it. Anything that the problem mentions that is a function of the independent variable you arrived at by asking yourself the first question is probably going to be a dependent variable. Since the problem mentions the length of paper being pulled (or at least the rate of length of paper being pulled) that means that length of paper being pulled is also a dependent variable.

Question 3: What other dependent variables (if any) does the problem imply? For this you have to use your own knowledge of how the world works. Often this is not an easy question to answer. In the case of the doggie problem, the implied dependent variable was the crossectional area of the roll. We were only able to arrive at this by thinking carefully about what happens when you pull paper off a roll. In other problems you will have to use your knowledge of geometry or trig to answer this question. But answering it always starts with taking a moment to clearly visualize what is going on in the problem. And quite often, the problem will indeed state all the variables you need in order to solve it, in which case the answer to this question will be none.

Question 4: What are the relationships between the variables? Again this often involves tapping your world knowledge as well as what you learned in geometry and trig. In the doggie problem, we have real-world knowledge that toilet paper rolls usually have a circular crossection. And we know from junior high school that the area of a circle is πr2. That gives us the relationship between dependent variable, radius, and dependent variable, area.

Question 5: What do you get when you take the derivative with respect to the independent variable of both sides of the relationship equation(s)? This just means you have to take some derivatives. You will almost always have to apply the chain rule, because one of your dependent variables will very likely be squared or cubed or square rooted, or have some trig function taken of it. Here are some examples using t as the independent variable:

1. If you have dependent variable, y, and you see the expression, y3 - y, then the derivative of that expression is:

 ``` (3y2 - 1) ``` ``` dy   dt ```

2. If you have dependent variable, θ, and you see the expression, sin(θ), then the derivative of that expression is:

 ``` cos(θ) ``` ``` dθ   dt ```

3. If you have dependent variables, u and v, and you see the expression, uv, you have to apply the product rule, and you will get as the derivative of uv:
```     dv       du
u     +  v
dt       dt
```
Each of the above is an example of implicit differentiation.

Question 7: What variables and derivatives does the problem give actual values for? In the doggie problem it stated the thickness of the paper, the rate the paper was being pulled, and the radius at which you were to establish the rate (observe that the thickness is neither a dependent nor an independent variable -- it is a constant. That is because it does not change as the dog pulls the paper.) We did identify length of paper pulled as a dependent variable. We observe that the problem gives a rate for that, which means that it assigns a value to the derivative of length of paper pulled. It also assigns a value to the radius (and not its derivative). Once you have identified all the values given, go back to the equation where you solved for what the problem was asking for and plug all the values in. That will enable you to get an answer.

### Some Worked Examples

Example 1: Two taxicabs begin at the same time at the intersection of Park Ave and 59th St. One taxicab heads uptown (North) on Park Ave at 40 mph. The other heads west on 59th St at 30 mph. Both have the unbelievable good fortune not to be stopped by traffic lights or slowed by traffic. When the taxicab on Park Ave. has gone half a mile, how fast is the distance (as the New York crow flies) between them increasing?

Answering Question 1: The problem asked "how fast." So the independent variable is time, t.

Answering Question 2: The problem talks of distance, and indeed we have distance north of Park and 59th and we have distance west of Park and 59th. We'll call them y and x respectively. The problem asks for the distance between the taxicabs, so that is also a dependent variable, which we shall call, s. Observe that these are all functions of time, since both taxicabs are moving.

Answering Question 3: There are no hidden aspects of taxicabs that are relevant to this problem, so there are no inferred variables.

Answering Question 4: There are two relationships you need to know about in this problem. One is that the taxicabs are traveling at right angles to each other. This means that the Pythagorean distance formula applies. The distance, s between the taxicabs is the hypotenuse of the right triangle formed by the line connecting the northbound taxicab with Park and 59th and the line connecting the westbound taxicab with Park and 59th. The lengths of those sides are y and x respectively. So you have the relationship:

```   s2  =  x2 + y2
```
There is another relationship in this problem that we can infer from our experience doing motion problems in junior high school. The problem gives us the speeds of the taxicabs and that they start from the same point at the same time. Remember that the northbound taxicab goes 40 mph and the westbound one goes 30 mph. So we can conclude that at any time:

 ``` y = 40 ``` ``` x eq. 8.2-6 30 ```
or indeed that

```   3
y  =  x                                                      eq. 8.2-7
4
```

Answering Question 5: This just means taking the derivatives of the relationship equations we got when we answered question 4. For the Pythagorean relationship (eq. 8.2-5), that gives:

```      ds        dx        dy
2s     =  2x     +  2y                                         eq. 8.2-8
dt        dt        dt
```
Taking the derivative of the other relationship yields
```   3 dy     dx
=                                                        eq. 8.2-9
4 dt     dt
```
which is the same as saying that three fourths of the speed of the northbound taxicab is equal to the speed of the same as the speed of the westbound taxicab. And we already knew that.

Answering Question 6: The problem is asking for the rate at which the distance, s, between the two taxicabs is increasing. So it is asking for ds/dt. Observe that this parameter already appears in one of our equations. To solve for it, we simply have to divide the derivative of the Pythagorean equation (eq. 8.2-8) by 2s.

 ``` ds 1 =   dt s ``` ``` ``` ``` dx dy x + y   dt dt ``` ``` ``` ``` eq. 8.2-10 ```

Answering Question 7: Now all we have to do is plug in numbers for all the parameters other than ds/dt into the above and we will have a value for ds/dt. So what numbers were given in the problems? Only that northbound speed is 40 mph:

```   dy
=  40
dt
```
and the westbound speed is 30 mph:
```  dx
=  30
dt
```
Also that at the moment we want to know the speed, the northbound taxicab is half a mile north of Park and 59th:
```  y  =  0.5
```
In order to complete this step we still have to know x and s. We already determined that x = (3/4)y. Substituting 0.5 for y we find x = 0.375. As for s, we have the Pythagorean distance formula (eq. 8.2-5) to determine s from x and y:
```          _______      _______________
s  =  √x2 + y2  =  √0.140625 + 0.25  =  0.625                  eq. 8.2-11
```
When I put all the numbers into the equation that resulted from answering question 6, I get ds/dt = 50 mph

Example 2: A solution drains through a filter-funnel at a rate of 10 cc/minute. From the center point of its mouth to its apex, the funnel is 10 cm. At its mouth it is 12 cm in diameter. How fast is the solution-level dropping when there are 200 cc left in the funnel? (Assume that the funnel is a right circular cone) For reference, the volume of a cone is given by

 ``` V = ``` ``` 1   3 ``` ``` πr2h eq. 8.2-12 ```
where V is the volume, r is the radius of the base, and h is the cone's height from base to apex.

Answering Question 1: The problem asks, "how fast," so clearly the independent variable is time, t.

Answering Question 2: The problem mentions the solution-level, the volume, and the rate at which volume is being lost. So clearly solution-level, h, and volume, V, are dependent variables. The rate at which volume is being lost is the derivative of volume, so it is dV/dt.

Answering Question 3: Think about the situation. You know that the funnel is a cone. But the volume of solution in the funnel is also a cone, although this cone shrinks as the solution flows out. The top surface of the solution is a circle. And the amount of volume of solution is related to the radius of that circle by the volume formula for a cone (given above). So that radius, r, is also a dependent variable.

Answering Question 4: The volume of a cone formula is a relationship that the problem gave us. Again it is:

 ``` V = ``` ``` 1   3 ``` ``` πr2h eq. 8.2-12 ```

 There is another relationship you can get by what you know about cones. Since the diameter of the base of the funnel is given as 12 cm, clearly the radius of the base is 6 cm. The height is given as 10 cm. So the ratio of height to radius is 10/6 = 5/3. But it is an intrinsic property of cones that the radius of its skirt increases in linear proportion to the distance along the axis from its apex. That means that if the ratio of the big funnel-cone's height to base-radius is 5/3, then the smaller cone that is the volume of solution must have the same ratio of height to base-radius. Figure 8-2 shows how the triangular crossection of a cone leads to this conclusion (recall studying similar triangles in geometry). The height of the smaller cone is the solution level, which we have called, h, and is a dependent variable in this problem. The base radius of that smaller cone we have called, r, and it is also a dependent variable. From this discussion we now know that:
```         3
r  =    h                                                      eq. 8.2-13
5
```

Answering Question 5: There are two ways you could go at this point. One is simply to take the derivative of the volume-of-a-cone formula (eq. 8.2-12), which is the hard way, but does lead to a solution. I'll show you the easy way first, and then demonstrate that the hard way gets you to the same place. The easy way is to observe that you can make a substitution for r before you take the derivative of the volume formula:

 ``` 1 V = π 3 ``` ``` ``` ```3 h 5 ``` ``` ``` ```2 ``` ``` 3 h =   25 ``` ``` πh3 eq. 8.2-14 ```
Now that we have reduced the number of variables by one we take the derivative of the above:

 ``` dV = dt ``` ``` 9 π 25 ``` ``` h2 ``` ``` dh eq. 8.2-15 dt ```
But suppose we had just taken the derivative of volume formula without first making the convenient substitution. Since the volume formula (eq. 8.2-12) contains the product of two dependent variables (that is it contains r2h) we have to use the product rule. When you do you get

 ``` dV 1 = π dt 3 ``` ``` ``` ``` dr 2rh + dt ``` ``` r2 ``` ``` dh   dt ``` ``` ``` ``` eq. 8.2-16 ```
If you made the substitution at this point, that is replaced r with 3h/5 into eq. 8.2-16, what would happen? You'd still be left with a dr/dt term, but you can substitute that as well by taking the derivative of:

```         3
r  =    h                                                      eq. 8.2-17a
5
```
to find that
```   dr     3 dh
=                                                          eq. 8.2-17b
dt     5 dt
```
I'll let you take that substitution the rest of the way. If you make no mistakes, you will end up with the exact same equation we got by substituting first and then taking the derivative.

Answering Question 6: The problem asks for the rate at which the solution level is dropping. The solution level is given by the dependent variable, h. So the rate of change of the solution level is dh/dt. And so that is what we solve for.

 ``` dh = dt ``` ``` dV 25   dt 9π ``` ``` 1 eq. 8.2-18 h2 ```

Answering Question 7: The problem gives dV/dt = -10 cc/minute (the minus sign tells you that the volume in the funnel is decreasing). But the problem doesn't give a value for h. So what do we do? The problem does tell us to find the rate that the level is decreasing when the volume of solution left in the funnel is 200 cc. That is the same as saying, when V = 200 cc. And we know from eq. 8.2-14 that

 ``` V = ``` ``` 3   25 ``` ``` πh3 eq. 8.2-19 ```
From that you can solve for h by:

 ``` h = ``` ``` ``` ```25 V 3π ``` ``` ``` ```1/3 eq. 8.2-20 ```
Since V is given, from here it's just a matter of poking your calculator keys to get an answer. I get h = 8.0953 cm, and dh/dt = -0.13492 cm/minute (note that the minus sign indicates that the level is going down, as we would expect if the solution is draining out).

Example 3: A board that is 96 inches long rests with one end on top of a 12 inch high curb. The other end rests on the pavement beneath the curb, which is assumed to be flat. The lower end is slid along the pavement toward the curb at a rate of 1 inch per second. When there is 36 inches left between the end of the board and the base of the curb, how fast is the free end of the board moving? And at what rate is the angle of the board changing?

This problem is a step up in level of difficulty from the two that preceded it. But the difference is tantamount to the difference between untying one knot and untangling an entire snarled wad of yarn. The techniques are the same, you just have to do more of them in this problem than in the last two. But like the snarled wad of yarn, you find an end then methodically work it back through all the loops and tangles, and if you keep at it long enough it will be unsnarled.

Answering Question 1: Again the problem is asking "how fast" and "at what rate," so it is clear that the independent variable is time, t.

Answering Question 2: The position of the lower end of the board, or more specifically its distance from the curb, is clearly a dependent variable. Let's call that xc. The rate at which that end moves is also the rate at which xc decreases. And that rate is dxc/dt. There is also the position of the other end of the board. But it has both a horizontal and vertical position. So let's call them xe and ye respectively, and we shall reference them from the base of the curb. Finally there is the angle of the board, θ, which we shall consider in radians referenced from horizontal. Figure 8-3 shows all the dimensions of this problem. Observe that the way the diagram is drawn, xe is negative.

Answering Question 3: When you draw a diagram of this problem, it becomes clear that there is an important variable that is not mentioned in the problem. That is the horizontal distance from the lower end of the board to the spot on the ground immediately below the upper end of the board. Call this xb. Also, if the angle of the board is a dependent variable, perhaps we might consider the slope of the board also as a dependent variable, m.

Answering Question 4: So what are the relationships among the variables here? Clearly there is a right triangle present here, so the Pythagorean distance formula applies. From the diagram we can see that

```   962  =  xb2 + ye2                                              eq. 8.2-21
```
Now look at the two base lengths, xb and xc. Notice that the length from (0,0) to (xc,0) added to the length from (0,0) to (xe,0) is exactly equal to the length, xb. But since xe is negative, the relationship is:
```   xc - xe  =  xb                                                 eq. 8.2-22
```
Look at the diagram again. Do you see the similar triangles? The length, xc, forms the smaller triangle, and the length, xb, forms the larger one. By the rule for similar triangles we have:

```   -m  =  tan(θ)                                                  eq. 8.2-24a
```
or equivalently
```   arctan(-m)  =  θ                                               eq. 8.2-24b
```
Again observe that there is a minus sign on the slope, m, because the board is sloping down the way it is drawn in the diagram. You also have by trigonometry:
```   96 cos(θ)  =  x                                                eq. 8.2-24c

96 sin(θ)  =  y                                                eq. 8.2-24c

```

Answering Question 5: All we need to do for this one is take the derivatives of each of the relationships listed in the question 4 paragraph above. So taking the derivative of both sides of eq. 8.2-21:

```             dxb         dye
0  =  2xb      +  2ye                                          eq. 8.2-25a
dt          dt
```
from which you can divide out a factor of 2 to get:
```            dxb        dye
0  =  xb      +  ye                                            eq. 8.2-25b
dt         dt
```
Taking the derivative of
eq. 8.2-22 is pretty easy:
```   dxc     dxe     dxb
-       =                                                 eq. 8.2-26
dt      dt      dt
```
On
eq. 8.2-23, it turns out the we will mostly be interested in taking the derivative of 12/xc = -m:

 ``` 12 −   xc2 ``` ``` dxc = - dt ``` ``` dm eq. 8.2-27a dt ```
and the minus signs cancel to give:

 ``` 12   xc2 ``` ``` dxc = dt ``` ``` dm eq. 8.2-27b dt ```
From the trig equations, the arctan equation (
8.2-24b) turns out to be more useful than the tan equation (8.2-24a). Its derivative (remembering that the derivative of arctan(x) is 1/(1 + x2)) is

```             dθ     dxb
-96 sin(θ)     =                                                eq. 8.2-28c
dt     dt

dθ     dye
96 cos(θ)     =                                                eq. 8.2-28d
dt     dt
```

Answering Question 6: Although this problem is harder than the other two, it is not the calculus part that makes it harder. We have already done all the calculus we need to do on this problem. All that is left to be done is the algrebra and trig. There is more of it to do on this problem than on the others, but it is still stuff you have had before. The problem asked for the speed of the free end, which will be a combination of dxe/dt and dye/dt. And it also asked for the rate at which the angle, θ, is changing, which would be dθ/dt. At this point we have all the equations we need to solve all this. All we need to do is organize them in a way that makes sense and substitute one into another until we get a solution.

It turns out that dθ/dt is easier to solve for than the either of the other two variables. So let's do that one first. Remember that the distance from the low end of the board to the base of the curb, xc, is given. So is the rate at which it is being pushed, dxc/dt. From those two we can get the slope of the board, m, from eq. 8.2-23 and we can get the rate at which that slope is changing, dm/dt, from eq. 8.2-27b:

```   ye        ye        12
=           =       =  -m                                  eq. 8.2-23
xb     xc - xe      xc
```

 ``` 12 xc2 ``` ``` dxc = dt ``` ``` dm eq. 8.2-27b dt ```
Once we know m and dm/dt, eq. 8.2-28b will give us dθ/dt:

 ``` -1   1 + m2 ``` ``` dm = dt ``` ``` dθ eq. 8.2-28b dt ```
We have expressions for dxb/dt and dye/dt in equations 8.2-28c and 8.2-28d. But they involve the sine and cosine of θ, which we do not yet know. But we do know what tan(θ) is from eq. 8.2-24a. And, remembering that tan(x) = sin(x)/cos(x) and sin2(x) + cos2(x) = 1, you can easily derive:

```                   1
cos(x)  =                                                      eq. 8.2-29a
√1 + tan2(x)

tan(x)
sin(x)  =                                                      eq. 8.2-29b
√1 + tan2(x)
```
And don't just take my word for it. Derive these for yourself. Eq. 8.2-24a tells us that  -m = tan(θ).  So from the trig identities in equations 8.2-29a and 8.2-29b we have:
```                 1
cos(θ)  =                                                      eq. 8.2-30a
√1 + m2

-m
sin(θ)  =                                                      eq. 8.2-30b
√1 + m2
```
Substituting those into equations
8.2-28c and 8.2-28d and substituting dθ/dt from eq. 8.2-28b, you have:
```          -m       dm     dxb
96                  =                                          eq. 8.2-30c
(1 + m2)3/2   dt     dt

-1       dm     dye
96                  =                                          eq. 8.2-30d
(1 + m2)3/2   dt     dt
```
Finally we observe that we can get dxe/dt using
eq. 8.2-26. Solving that for dxe/dt gives:
```   dxc     dxb     dxe
-       =                                                 eq. 8.2-31
dt      dt      dt
```

Answering Question 7: The problem tells you that xc = 36 inches. It also tells you that dxc/dt = -1 inch/sec (it's negative because the low end of the board is moving to the left so xc is decreasing). Now we take it step by step. First, eq. 8.2-23 tells us that m = -1/3. From eq. 8.2-27b you get dm/dt = -1/108. From eq. 8.2-28b you get dθ/dt = 1/120 radian/sec. From eq. 8.2-30c you get dxb/dt = -0.2529822 inches/sec (that is, xb is decreasing), and from eq. 8.2-30d you get dye/dt = 0.7589466 inches/sec (that is the free end of the board is getting higher). From eq. 8.2-31 you have dxe/dt = -0.7470178 inches/sec (that is the high end of the board is moving to the left).

Finally, now that we have values for dxe/dt and dye/dt, which are the x and y components of the speed of the free end of the board, how do we find the composite speed of that end? Velocities that are at right angles to each other combine according to the Pythagorean rule (you can prove that by drawing a diagram that shows the movement of the end of the board in the x and y directions during a short interval of time, Δt, and then taking the limit as Δt goes to zero). So if we allow v to be the composite speed, then

 ``` v2 = ``` ``` ``` ```dxe   dt ``` ``` ``` ```2 + ``` ``` ``` ```dye   dt ``` ``` ``` ```2 eq. 8.2-32 ```
Solving for velocity, my calculator tells me that v = 1.0649110 inches/sec.

### Exercises

1) Return to the taxicab problem, but this time give the westbound taxicab a quarter mile head start. Keep everything else the same. When the northbound taxicab has gone half a mile from its starting position, how fast is the distance between the two taxicabs increasing? You should be able to get through this one using the worked example on this page as a model. When you are done, click here to view the solution.

2) We shall remain in New York City for this problem, which is similar to the taxicab problem. The Verazzano-Narrows bridge, which connects Brooklyn with Staten Island, has a span length of about 4200 feet. Assume that the entire roadbed is 200 feet above the water. A small motor boat traveling perpendicular to the bridge is 1000 feet away from passing under the midspan of the bridge. The boat is cruising 20 feet per second toward the bridge. At the same time a Brooklyn-bound runner on the bridge is passing the tower on the Staten Island side (the towers are at either end of the span, so the runner is entering the span). The runner moves at 10 feet per second along the bridge. How fast is the distance between the boat and the runner decreasing?

Although this one has the additional wrinkle of being in three dimensions instead of two (like the taxicabs), the methods are the same. Remember that the Pythagorean distance formula works just as well in three dimensions as it does in two. So go as far as you can with this before clicking here to see the solution.

3) Here's one that is very important in mechanics. A circular track has radius, r. A train moves around the track at ω radians/second. That means that if the center of the circle is at (0,0), the position of the train as a function of time, t, is:

```   ( r cos(ωt), r sin(ωt) )
```
Find the x and y components of the train's velocity (which is the first time derivative of position) and the x and y components of its acceleration (which is the second time derivative of position). What can you observe about the direction of the train's velocity and acceleration in relation to its position on the track? And how would the magnitude of velocity and acceleration be effected if you varied r? or if you varied ω?

## Dimensional Checking (sometimes called Units Checking)

For another point of view on this topic by R. Horan and M. Lavelle, click here (requires Acrobat PDF reader).

In the early paragraphs of this section I suggested that you always do reasonableness checks on your answers to problems like these. The tests I suggested were things like making sure that the sign of the answer made sense and that the answer seemed in the right ballpark. But till now I have completely skipped the most powerful method of checking your answer, and that is checking it dimensionally.

If you have taken or are taking chemistry or physics, you are likely to have already encountered this method. Besides working on chemistry and physics problems, it works on related rate problems as well (and just about all other word problems too). If you are going into engineering, this method will serve you well for many years to come.

Back in grade school when you learned to do word problems, your teacher, if he or she was any good, told you, "Always write in the units." This was and still is excellent advice. In the real world most quantities have units like meters, dollars, pints, degrees C, and so on. If you drive 60 miles per hour, you are also driving 88 feet per second. So if you had to write down how fast you were going, which is correct, 60 or 88? Neither! You could be indicating leagues per sidereal day for all I know. But both "60 miles per hour" and "88 feet per second" are correct answers. And the two quantities are indeed equal, despite the difference in the numerals. This is because the "miles per hour" and the "feet per second" are as much a part of each quantity as the 60 or the 88.

The real beauty of having units attached to your quantities is that you can do an arithmetic of sorts on the units themselves. And when you are done, if the problem asks, "how fast is Jane running?" the answer better come out having units of velocity. If it doesn't then you did something wrong somewhere and your answer is probably wrong.

Let's take a very simple example. Light travels 299,799 kilometers in a second. What is the speed of light in kilometers per year if there are 31,557,600 seconds in a year?

 ``` km 299,799 × sec ``` ``` sec 31,557,600 = year ``` ``` 9.460937 × 1012 ``` ``` km   year ```

Remember that whenever you see the word, "per," it means the same as "divided by" or "over." And as it is with numbers, so it is with units. "Kilometers per second" means the same thing as "kilometers over seconds." That is how we wrote it here. Same thing with "seconds in a year." It also means "seconds per year," which means "seconds over years." The problem asks for an answer in "kilometers per year." You can see that when we multiply "kilometers over seconds" times "seconds over years," the seconds cancel and we are left with "kilometers over years," which are the units the problem calls for. Since the units came out right, there is a good likelihood that we did the problem correctly.

Now suppose a problem asked, how many seconds does it take light to travel 1.5 × 108 kilometers? Here is the setup for that:

```                       1    sec
1.5 × 108 km  ×               =  500 sec
299,799  km
```
The problem asks for an answer in seconds. Observe how, in order to get kilometers to cancel and for seconds to come out on top, we had to take the reciprocal of "kilometers per second" and multiply by it. When you take the reciprocal of the units, you also take the reciprocal of the number attached to them. Or equivalently you could have just divided by the kilometers per second quantity. So the units are cluing you in on what gets multiplied and what gets divided to get the correct answer.

Some quantities have no units. These are called dimensionless quantities. For example, π is the ratio of the circumference of a circle to its diameter. It doesn't matter how large the circle is or in what units you measure the circumference and diameter (as long as you use the same units for both), the value of π is still the same. If c is the circumference in furlongs and d is the diameter in furlongs, then

```   c furlongs
=  π
d furlongs
```
The units cancel in this case, no matter what units you use.

Angles, and particularly angles measured in radians are dimensionless. Remember that the definition of radian measure was that you take the distance around the circle the angle takes you, and then you divide that distance by the radius of the circle. That is the quotient of two lengths, so the length-units cancel leaving only the dimensionless ratio.

Units can be raised to a power. You know very well that if you multiply the height of a rectangle by its width, you get its area. If both the height and width are in inches, then the area comes out in square inches, which is the same as inches squared.

```   14 in × 7 in  =  98 in2
```
Not only length-units can be raised to a power. It can happen to other types of units as well. For example, acceleration measures how many meters per second an object's velocity gains per second. So the units of acceleration would be meters per second per second.
```     meters

sec         meters
=
sec          sec2
```
You can see how that's the same as meters per second squared.

You can only add quantities that have the same units. In other words, you can't add apples to oranges. So if you find yourself adding two quantities whose units differ, you made a mistake somewhere. The units that the sum has are the same as the units of each of its summands.

When you use trig functions, inverse trig functions, logs or exponentials, the argument must be dimensionless. You can't take the sine of 3 centimeters or the exponential of 40 grams. The thing you take the sine or exponential (or any of the other trig or log functions) of must not have any units. And the value the function returns is also dimensionless. For example, if you have an exponential function of time, the time, t, will always be multiplied by a rate, k, whose dimensions are inverse time. Suppose t is in seconds. Then there would be a rate, k, whose dimensions are in per second (that is inverse seconds). Suppose  k = 0.001 /second. Then you could form the exponential function

```   f(t)  =  e-kt  =  e-0.001t
```
Of course this equation gives you a value, f(t), that is dimensionless. If the result had to have dimensions of, say watts, then the exponential would have to have a multiplier that had dimensions of watts. For example, you might have  P = 3 × 107 watts. Then you could have
```   g(t)  =  P e-kt  =  3 × 107 watts × e-0.001t
```
The only possible exception to this rule is the log. Since a quantity is the product of its units and a number, you can argue that

 ``` ln ``` ``` ``` ```1400 cm   14 cm ``` ``` ``` ``` = ln(1400 cm) - ln(14 cm) = ln(1400) - ln(14) + ln(cm) - ln(cm) ```
but only if you are willing to accept the existence of the abstraction, ln(cm), and only if, in the end, such abstractions all cancel out. The log still returns a dimensionless result regardless.

Units conversion factors, deep down, are all equal to one. Consider that

```   1 hour  =  60 minutes
```
That means that if you divide out 1 hour, you get the units conversion factor of
```            minutes
1  =  60
hour
```
When you convert 14 hours to minutes, you do:
```                 minutes
14 hours × 60          =  840 minutes
hour
```
Isn't it true that 14 hours and 840 minutes are precisely the same length of time? The two quantities are equal. Which means that 60 minutes/hour must be equal to 1 and nothing else.

A derivative always takes the ratio of units. For example, when you take the derivative of distance (say in meters) with respect to time (say in seconds), the derivative would have units of meters per second. If you have any dependent variable, v, and you take its derivative with respect to an independent variable, t, to get dv/dt, the result will have units that are v's units divided by t's units.

Units of second derivatives are tricky. If you take the second derivative of a dependent variable, v, with respect to the independent variable, t, to get

```   d2v

dt2
```
this second derivative will have units that are the units of v divided by the square of the units of t. Note that you only square the denominator units and not the numerator units. This makes sense because the second derivative is the derivative of the first derivative. So you take the units of the first derivative and divide it by the units of the independent variable. For example, acceleration is the second time derivative of distance. If you are using meters and seconds, acceleration has units of
```   meters

sec2
```
This same principle goes with higher derivatives too. If you take the nth derivative of v with respect to t, the units you end up with are the units of v divided by the units of t taken to the nth power. Only the denominator units are taken to the nth power, not the units of v.

Let's dimensionally check a solution to one of the problems in this section. We'll do problem 4. Here you were given the rocket's altitude, h in meters, the distance to the radar, x in meters, and the rocket's speed, dh/dt in meters per second. The problem asks for a rate, dθ/dt, in radians per second. Since radians is dimensionless, that is the same as plain old "per second" (when the "per" is not preceded by any units, it is assumed that the numerator of the units is dimensionless, which is to say that the numerator is just 1). The solution equation for the problem is

```   dh    x        dθ
=
dt x2 + h2     dt
```
On the left we have dh/dt, which is in meters per second. The next factor has x in the numerator, so the numerator is meters. The denominator has the sum of x2 and h2. Both x and h have units of meters, so their squares both have units of square meters. Because those units are the same, it is ok to add those two quantities. To the right of the equal sign is dθ/dt, which we already determined has units of per second. So, expressing this equation only in its units we have
```   meters  meters      1
=
sec   meter2     sec
```
I'm sure you can convince yourself that meters and meters squared all cancel out of the left-hand side of this, leaving only per second on both the left and right of the equal. So the units agree. The original equation smells good.