262 lines
7.1 KiB
Plaintext
262 lines
7.1 KiB
Plaintext
# Related rates
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{{< include ../_common_code.qmd >}}
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This section uses these add-on packaages:
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```{julia}
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using CalculusWithJulia
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using Plots
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using Roots
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using SymPy
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```
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---
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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."
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#### Examples
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The following is a typical "book" problem:
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> 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)
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```{julia}
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#| hold: true
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#| echo: false
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#| cache: true
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### {{{growing_rects}}}
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## Secant line approaches tangent line...
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function growing_rects_graph(n)
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w = (t) -> 2 + 4t
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h = (t) -> 3/2 * w(t)
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t = n - 1
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w_2 = w(t)/2
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h_2 = h(t)/2
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w_n = w(5)/2
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h_n = h(5)/2
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plt = plot(w_2 * [-1, -1, 1, 1, -1], h_2 * [-1, 1, 1, -1, -1], xlim=(-17,17), ylim=(-17,17),
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legend=false, size=fig_size)
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annotate!(plt, [(-1.5, 1, "Area = $(round(Int, 4*w_2*h_2))")])
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plt
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end
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caption = L"""
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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 ($dA/dt$)?
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"""
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n=6
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anim = @animate for i=1:n
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growing_rects_graph(i)
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end
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imgfile = tempname() * ".gif"
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gif(anim, imgfile, fps = 1)
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ImageFile(imgfile, caption)
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```
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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:
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$$
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A(t) = w(t) \cdot h(t).
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$$
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We would get by the product rule that the *rate of change* of area with respect to time, $A'(t)$ is just:
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$$
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A'(t) = w'(t) h(t) + w(t) h'(t).
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$$
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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:
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$$
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\frac{dA}{dt} = \frac{dw}{dt} h + w \frac{dh}{dt}.
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$$
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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:
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$$
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A'(t) = 4 \cdot 12 + 8 \cdot 6 = 96.
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$$
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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 $t$.
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We could also have done this differently. We would recognize the following:
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* The area of a rectangle is just:
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```{julia}
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A(w,h) = w * h
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```
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* The width - expanding at a rate of $4t$ from a starting value of $2$ - must satisfy:
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```{julia}
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w(t) = 2 + 4*t
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```
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* The height is a constant proportion of the width:
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```{julia}
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h(t) = 3/2 * w(t)
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```
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This means again that area depends on $t$ through this formula:
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```{julia}
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A(t) = A(w(t), h(t))
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```
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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.
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Now to answer the question, when the width is 8, we must have that $t$ is:
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```{julia}
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tstar = find_zero(x -> w(x) - 8, [0, 4]) # or solve by hand to get 3/2
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```
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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.
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Here we do so with the aid of `Julia`, though this problem could readily be done "by hand."
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We have expressed $A$ as a function of $t$ by composition, so can differentiate that:
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```{julia}
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A'(tstar)
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```
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---
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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:
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```{julia}
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xs = [6, 54, 150, 294, 486, 726]
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ds = diff(xs)
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```
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Those seem to be increasing by a fixed amount each time, which we can see by one more application of `diff`:
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```{julia}
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diff(ds)
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```
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How can this relationship be summarized? Well, let's go back to what we know, though this time using symbolic math:
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```{julia}
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@syms t
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diff(A(t), t)
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```
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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.
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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.
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##### Example
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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.
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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 $db/dt$.
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We have from implicitly differentiating in $t$ the equation $l^2 = h^2 + b^2$, noting that $l$ is a constant, that:
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$$
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0 = 2h \frac{dh}{dt} + 2b \frac{db}{dt}.
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$$
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Solving, yields:
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$$
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\frac{dh}{dt} = -\frac{b}{h} \cdot \frac{db}{dt}.
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$$
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* If when $l = 12$ it is known that $db/dt = 2$ when $b=4$, find $dh/dt$.
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We just need to find $h$ for this value of $b$, as the other two quantities in the last equation are known.
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But $h = \sqrt{l^2 - b^2}$, so the answer is:
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```{julia}
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length, bottom, dbdt = 12, 4, 2
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height = sqrt(length^2 - bottom^2)
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-bottom/height * dbdt
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```
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* What happens to the rate as $b$ goes to $l$?
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As $b$ goes to $l$, $h$ goes to $0$, so $b/h$ blows up. Unless $db/dt$ goes to $0$, the expression will become $-\infty$.
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:::{.callout-note}
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## Note
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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.
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:::
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##### Example
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```{julia}
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#| hold: true
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#| echo: false
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caption = "A man and woman walk towards the light."
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imgfile = "figures/long-shadow-noir.png"
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#ImageFile(:der
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#| hold: true
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#| echo: false
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choices = [
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"``dy/dt = 2x + 1/x``",
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"``dy/dt = 1 - x^2 - \\log(x)``",
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"``dy/dt = -2x - 1/x``",
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"``dy/dt = 1``"
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]
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answ=1
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radioq(choices, answ)
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```
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