Related rates problems involve two (or more) unknown quantities that are related through an equation. As the two variables depend on each other, also so do their rates - change with respect to some variable which is often time, though exactly how remains to be discovered. Hence the name "related rates."
#### Examples
The following is a typical "book" problem:
> A screen saver displays the outline of a $3$ cm by $2$ cm rectangle and then expands the rectangle in such a way that the $2$ cm side is expanding at the rate of $4$ cm/sec and the proportions of the rectangle never change. How fast is the area of the rectangle increasing when its dimensions are $12$ cm by $8$ cm? [Source.](http://oregonstate.edu/instruct/mth251/cq/Stage9/Practice/ratesProblems.html)
As $t$ increases, the size of the rectangle grows. The ratio of width to height is fixed. If we know the rate of change in time for the width ($dw/dt$) and the height ($dh/dt$) can we tell the rate of change of *area* with respect to time ($dA/dt$)?
"""
n=6
anim = @animate for i=1:n
growing_rects_graph(i)
end
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ImageFile(imgfile, caption)
```
Here we know $A = w \cdot h$ and we know some things about how $w$ and $h$ are related *and* about the rate of how both $w$ and $h$ grow in time $t$. That means that we could express this growth in terms of some functions $w(t)$ and $h(t)$, then we can figure out that the area - as a function of $t$ - will be expressed as:
$$
A(t) = w(t) \cdot h(t).
$$
We would get by the product rule that the *rate of change* of area with respect to time, $A'(t)$ is just:
$$
A'(t) = w'(t) h(t) + w(t) h'(t).
$$
As an aside, it is fairly conventional to suppress the $(t)$ part of the notation $A=wh$ and to use the Leibniz notation for derivatives:
$$
\frac{dA}{dt} = \frac{dw}{dt} h + w \frac{dh}{dt}.
$$
This relationship is true for all $t$, but the problem discusses a certain value of $t$ - when $w(t)=8$ and $h(t) = 12$. At this same value of $t$, we have $w'(t) = 4$ and so $h'(t) = 6$. Substituting these 4 values into the 4 unknowns in the formula for $A'(t)$ gives:
$$
A'(t) = 4 \cdot 12 + 8 \cdot 6 = 96.
$$
Summarizing, from the relationship between $A$, $w$ and $t$, there is a relationship between their rates of growth with respect to $t$, a time variable. Using this and known values, we can compute. In this case, $A'$ at the specific $t$.
We could also have done this differently. We would recognize the following:
* The area of a rectangle is just:
```{julia}
A(w,h) = w * h
```
* The width - expanding at a rate of $4t$ from a starting value of $2$ - must satisfy:
```{julia}
w(t) = 2 + 4*t
```
* The height is a constant proportion of the width:
```{julia}
h(t) = 3/2 * w(t)
```
This means again that area depends on $t$ through this formula:
```{julia}
A(t) = A(w(t), h(t))
```
This is why the rates of change are related: as $w$ and $h$ change in time, the functional relationship with $A$ means $A$ also changes in time.
Now to answer the question, when the width is 8, we must have that $t$ is:
```{julia}
tstar = find_zero(x -> w(x) - 8, [0, 4]) # or solve by hand to get 3/2
```
The question is to find the rate the area is increasing at the given time $t$, which is $A'(t)$ or $dA/dt$. We get this by performing the differentiation, then substituting in the value.
Here we do so with the aid of `Julia`, though this problem could readily be done "by hand."
We have expressed $A$ as a function of $t$ by composition, so can differentiate that:
```{julia}
A'(tstar)
```
---
Now what? Why is $96$ of any interest? It is if the value at a specific time is needed. But in general, a better question might be to understand if there is some pattern to the numbers in the figure, these being $6, 54, 150, 294, 486, 726$. Their differences are the *average* rate of change:
```{julia}
xs = [6, 54, 150, 294, 486, 726]
ds = diff(xs)
```
Those seem to be increasing by a fixed amount each time, which we can see by one more application of `diff`:
```{julia}
diff(ds)
```
How can this relationship be summarized? Well, let's go back to what we know, though this time using symbolic math:
```{julia}
@syms t
diff(A(t), t)
```
This should be clear: the rate of change, $dA/dt$, is increasing linearly, hence the second derivative, $dA^2/dt^2$ would be constant, just as we saw for the average rate of change.
So, for this problem, a constant rate of change in width and height leads to a linear rate of change in area, put otherwise, linear growth in both width and height leads to quadratic growth in area.
##### Example
A ladder, with length $l$, is leaning against a wall. We parameterize this problem so that the top of the ladder is at $(0,h)$ and the bottom at $(b, 0)$. Then $l^2 = h^2 + b^2$ is a constant.
If the ladder starts to slip away at the base, but remains in contact with the wall, express the rate of change of $h$ with respect to $t$ in terms of $db/dt$.
We have from implicitly differentiating in $t$ the equation $l^2 = h^2 + b^2$, noting that $l$ is a constant, that:
$$
0 = 2h \frac{dh}{dt} + 2b \frac{db}{dt}.
$$
Solving, yields:
$$
\frac{dh}{dt} = -\frac{b}{h} \cdot \frac{db}{dt}.
$$
* If when $l = 12$ it is known that $db/dt = 2$ when $b=4$, find $dh/dt$.
We just need to find $h$ for this value of $b$, as the other two quantities in the last equation are known.
But $h = \sqrt{l^2 - b^2}$, so the answer is:
```{julia}
length, bottom, dbdt = 12, 4, 2
height = sqrt(length^2 - bottom^2)
-bottom/height * dbdt
```
* What happens to the rate as $b$ goes to $l$?
As $b$ goes to $l$, $h$ goes to $0$, so $b/h$ blows up. Unless $db/dt$ goes to $0$, the expression will become $-\infty$.
:::{.callout-note}
## Note
Often, this problem is presented with $db/dt$ having a constant rate. In this case, the ladder problem defies physics, as $dh/dt$ eventually is faster than the speed of light as $h \rightarrow 0+$. In practice, were $db/dt$ kept at a constant, the ladder would necessarily come away from the wall. The trajectory would follow that of a tractrix were there no gravity to account for.
:::
##### Example
```{julia}
#| hold: true
#| echo: false
caption = "A man and woman walk towards the light."
