Related Rates

Section 3.2 Related Rates

Subsection 3.2.1 Related Rates

Consider the following problem

A spherical balloon is being inflated at a rate of $13cm^3/sec\text{.}$ How fast is the radius changing when the balloon has radius $15cm\text{?}$

There are several pieces of information in the statement:

The balloon is spherical
The volume is changing at a rate of $13cm^3/sec$ — so we need variables for volume (in $cm^3$) and time (in $sec$). Good choices are $V$ and $t\text{.}$
We are asked for the rate at which the radius is changing — so we need a variable for radius and units. A good choice is $r\text{,}$ measured in $cm$ — since volume is measured in $cm^3\text{.}$

Since the balloon is a sphere we know

If you don’t know the formula for the volume of a sphere, now is a good time to revise by looking at Appendix A.11.

that

\begin{align*} V &= \frac{4}{3} \pi r^3 \end{align*}

Since both the volume and radius are changing with time, both $V$ and $r$ are implicitly functions of time; we could really write

\begin{align*} V(t) &= \frac{4}{3}\pi r(t)^3 \end{align*}

We are told the rate at which the volume is changing and we need to find the rate at which the radius is changing. That is, from a knowledge of $\diff{V}{t}\text{,}$ find the related rate

Related rate problems are problems in which you are given the rate of change of one quantity and are to determine the rate of change of another, related, quantity.

$\diff{r}{t}\text{.}$

In this case, we can just differentiate our equation by $t$ to get

\begin{align*} \diff{V}{t} &= 4 \pi r^2 \diff{r}{t}\\ \end{align*}

This can then be rearranged to give

\begin{align*} \diff{r}{t} &= \frac{1}{4\pi r^2} \diff{V}{t} \end{align*}

Now we were told that $\diff{V}{t} = 13\text{,}$ so

\begin{align*} \diff{r}{t} &= \frac{13}{4\pi r^2} \end{align*}

We were also told that the radius is $15cm\text{,}$ so at that moment in time

\begin{align*} \diff{r}{t} &= \frac{13}{\pi 4 \times 15^2} \end{align*}

This is a very typical example of a related rate problem. This section is really just a collection of problems, but all will follow a similar pattern.

The statement of the problem will tell you quantities that must be related (above it was volume, radius and, implicitly, time).
Typically a little geometry (or some physics or…) will allow you to relate these quantities (above it was the formula that links the volume of a sphere to its radius).
Implicit differentiation will then allow you to link the rate of change of one quantity to another.

Another balloon example

Example 3.2.1. A rising balloon.

Consider a helium balloon rising vertically from a fixed point 200m away from you. You are trying to work out how fast it is rising. Now — computing the velocity directly is difficult, but you can measure angles. You observe that when it is at an angle of $\pi/4$ its angle is changing by $0.05$ radians per second.

Start by drawing a picture with the relevant variables
So denote the angle to be $\theta$ (in radians), the height of the balloon (in m) by $h$ and time (in seconds) by $t\text{.}$ Then trigonometry tells us
\begin{align*} h &= 200 \cdot \tan \theta \end{align*}
Differentiating allows us to relate the rates of change
\begin{align*} \diff{h}{t} &= 200 \sec^2 \theta \cdot \diff{\theta}{t} \end{align*}
We are told that when $\theta =\pi/4$ we observe $\diff{\theta}{t} = 0.05\text{,}$ so
\begin{align*} \diff{h}{t} &= 200 \cdot \sec^2(\pi/4) \cdot 0.05\\ &= 200 \cdot 0.05 \cdot \left( \sqrt{2} \right)^2\\ &= 200 \cdot \frac{5}{100} \cdot 2\\ &= 20 m/s \end{align*}
So the balloon is rising at a rate of 20m/s.

The following problem is perhaps the classic related rate problem.

Example 3.2.2. A sliding ladder.

A 5m ladder is leaning against a wall. The floor is quite slippery and the base of the ladder slides out from the wall at a rate of $1m/s\text{.}$ How fast is the top of the ladder sliding down the wall when the base of the ladder is 3m from the wall?

A good first step is to draw a picture stating all relevant quantities. This will also help us define variables and units.
So now define $x(t)$ to be the distance between the bottom of the ladder and the wall, at time $t\text{,}$ and let $y(t)$ be the distance between the top of the ladder and the ground at time $t\text{.}$ Measure time in seconds, but both distances in meters.
We can relate the quantities using Pythagoras:
\begin{align*} x^2 + y^2 &= 5^2 \end{align*}
Differentiating with respect to time then gives
\begin{align*} 2x \diff{x}{t} + 2y \diff{y}{t} &= 0 \end{align*}
We know that $\diff{x}{t} = 1$ and $x=3\text{,}$ so
\begin{align*} 6 \cdot 1 + 2y \diff{y}{t} &= 0 \end{align*}
but we need to determine $y$ before we can go further. Thankfully we know that $x^2+y^2=25$ and $x=3\text{,}$ so $y^2=25-9=16$ and
³
Since the ladder isn’t buried in the ground, we can discard the solution $y=-4\text{.}$
so $y=4\text{.}$
So finally putting everything together
\begin{align*} 6 \cdot 1 + 8 \diff{y}{t} &= 0\\ \diff{y}{t} &= - \frac{3}{4} m/s. \end{align*}
Thus the top of the ladder is sliding towards the floor at a rate of $\frac{3}{4} m/s\text{.}$

The next example is complicated by the rates of change being stated not just as “the rate of change per unit time” but instead being stated as “the percentage rate of change per unit time”. If a quantity $f$ is changing with rate $\diff{f}{t}\text{,}$ then we can say that

$f$ is changing at a rate of $\ds 100 \cdot \frac{\diff{f}{t}}{f}$ percent.

Thus if, at time $t\text{,}$ $f$ has rate of change $r\%\text{,}$ then

\begin{gather*} 100\frac{f'(t)}{f(t)}=r \implies f'(t) =\frac{r}{100} f(t) \end{gather*}

so that if $h$ is a very small time increment

\begin{gather*} \frac{f(t+h) - f(t)}{h} \approx \frac{r}{100} f(t) \implies f(t+h) \approx f(t) + \frac{rh}{100} f(t) \end{gather*}

That is, over a very small time interval $h\text{,}$ $f$ increases by the fraction $\frac{rh}{100}$ of its value at time $t\text{.}$

So armed with this, let’s look at the problem.

Example 3.2.3. Percentage rate of change of $R=PQ$.

The quantities $P,\ Q$ and $R$ are functions of time and are related by the equation $R=PQ\text{.}$ Assume that $P$ is increasing instantaneously at the rate of $8\%$ per year (meaning that $100\frac{P'}{P}=8$) and that $Q$ is decreasing instantaneously at the rate of $2\%$ per year (meaning that $100\frac{Q'}{Q}=-2$). Determine the percentage rate of change for $R\text{.}$

Solution This one is a little different — we are given the variables and the formula, so no picture drawing or defining required. Though we do need to define a time variable — let $t$ denote time in years.

Since $R(t) = P(t)\cdot Q(t)$ we can differentiate with respect to $t$ to get
\begin{align*} \diff{R}{t} &= P Q' + Q P' \end{align*}
But we need the percentage change in $R\text{,}$ namely
\begin{align*} 100 \frac{R'}{R} &= 100 \frac{PQ' +QP'}{R}\\ \end{align*}
but $R = PQ\text{,}$ so rewrite it as
\begin{align*} &= 100 \frac{PQ' +QP'}{PQ}\\ &= 100 \frac{PQ'}{PQ} + 100 \frac{QP'}{PQ}\\ &= 100 \frac{Q'}{Q} + 100 \frac{P'}{P} \end{align*}
so we have stated the instantaneous percentage rate of change in $R$ as the sum of the percentage rate of change in $P$ and $Q\text{.}$
We know the percentage rate of change of $P$ and $Q\text{,}$ so
\begin{align*} 100 \frac{R'}{R} &= -2 + 8 =6 \end{align*}
That is, the instantaneous percentage rate of change of $R$ is 6$\%$ per year.

Yet another falling object example.

Example 3.2.4. The shadow of a falling ball.

A ball is dropped from a height of $49$m above level ground. The height of the ball at time $t$ is $h(t)=49-4.9 t^2$ m. A light, which is also $49$m above the ground, is $10$m to the left of the ball’s original position. As the ball descends, the shadow of the ball caused by the light moves across the ground. How fast is the shadow moving one second after the ball is dropped?

Solution There is quite a bit going on in this example, so read carefully.

First a diagram; the one below is perhaps a bit over the top.
Let’s call $s(t)$ the distance from the shadow to the point on the ground directly underneath the ball.
By similar triangles we see that
\begin{gather*} \frac{4.9 t^2}{10}=\frac{49-4.9 t^2}{s(t)} \end{gather*}
We can then solve for $s(t)$ by just multiplying both sides by $\frac{10}{4.9 t^2}s(t)\text{.}$ This gives
\begin{gather*} s(t)=10\frac{49-4.9 t^2}{4.9 t^2}=\frac{100}{t^2}-10 \end{gather*}
Differentiating with respect to $t$ will then give us the rates,
\begin{gather*} s'(t)=-2\frac{100}{t^3} \end{gather*}
So, at $t=1\text{,}$ $s'(1)=-200$m/sec. That is, the shadow is moving to the left at $200$m/sec.

A more nautical example.

Example 3.2.5. The distance between moving boats.

Two boats spot each other in the ocean at midday — boat $A$ is 15km west of boat $B\text{.}$ Boat $A$ is travelling east at 3km/h and boat $B$ is travelling north at 4km/h. How fast is the distance between the boats changing at 3pm?

First we draw a picture.
Let $x(t)$ be the distance at time $t\text{,}$ in km, from boat $A$ to the original position of boat $B$ (i.e. to the position of boat $B$ at noon). And let $y(t)$ be the distance at time $t\text{,}$ in km, of boat $B$ from its original position. And let $z(t)$ be the distance between the two boats at time $t\text{.}$
Additionally we are told that $x'=-3$ and $y'=4$ — notice that $x' \lt 0$ since that distance is getting smaller with time, while $y' \gt 0$ since that distance is increasing with time.
Further at $3pm$ boat $A$ has travelled 9km towards the original position of boat $B\text{,}$ so $x=15-9 = 6\text{,}$ while boat $B$ has travelled 12km away from its original position, so $y=12\text{.}$
The distances $x,y$ and $z$ form a right-angled triangle, and Pythagoras tells us that
\begin{align*} z^2 &= x^2 + y^2. \end{align*}
At 3pm we know $x=6,y=12$ so
\begin{align*} z^2 &= 36 + 144 = 180\\ z&= \sqrt{180} = 6\sqrt{5}. \end{align*}
Differentiating then gives
\begin{align*} 2z \diff{z}{t} &= 2x \diff{x}{t} + 2y \diff{y}{t}\\ &= 12 \cdot (-3) + 24 \cdot(4)\\ &= 60.\\ \end{align*}
Dividing through by $2z = 12\sqrt{5}$ then gives
\begin{align*} \diff{z}{t} &= \frac{60}{12\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5} \end{align*}
So the distance between the boats is increasing at $\sqrt{5} km/h\text{.}$

One last one before we move on to another topic.

Example 3.2.6. Fuel level in a cylindrical tank.

Consider a cylindrical fuel tank of radius $r$ and length $L$ (in some appropriate units) that is lying on its side. Suppose that fuel is being pumped into the tank at a rate $q\text{.}$ At what rate is the fuel level rising?

Solution If the tank were vertical everything would be much easier. Unfortunately the tank is on its side, so we are going to have to work a bit harder to establish the relation between the depth and volume. Also notice that we have not been supplied with units for this problem — so we do not need to state the units of our variables.

Again — draw a picture. Here is an end view of the tank; the shaded part of the circle is filled with fuel.
Let us denote by $V(t)$ the volume of fuel in the tank at time $t$ and by $h(t)$ the fuel level at time $t\text{.}$
We have been told that $V'(t)=q$ and have been asked to determine $h'(t)\text{.}$ While it is possible to do so by finding a formula relating $V(t)$ and $h(t)\text{,}$ it turns out to be quite a bit easier to first find a formula relating $V$ and the angle $\theta$ shown in the end view. We can then translate this back into a formula in terms of $h$ using the relation
\begin{align*} h(t) &= r - r\cos \theta(t). \end{align*}
Once we know $\theta'(t)\text{,}$ we can easily obtain $h'(t)$ by differentiating the above equation.
The computation that follows below gets a little involved in places, so we will drop the “$(t)$” on the variables $V,h$ and $\theta\text{.}$ The reader must never forget that these three quantities are really functions of time, while $r$ and $L$ are constants that do not depend on time.
The volume of fuel is $L$ times the cross–sectional area filled by the fuel. That is,

$V = $

$L \times$ Area of

While we do not have a canned formula for the area of a chord of a circle like this, it is easy to express the area of the chord in terms of two areas that we can compute.

$V$

$= L\times\text{Area of}$

$= L\times\bigg[ \text{Area of}$

- Area of

$\bigg]$
The piece of pie

is the fraction $\tfrac{2\theta}{2\pi}$ of the full circle, so its area is $\tfrac{2\theta}{2\pi}\pi r^2=\theta r^2\text{.}$
The triangle

has height $r\cos\theta$ and base $2r\sin\theta$ and hence has area $\frac{1}{2}(r\cos\theta)(2r\sin\theta)=r^2\sin\theta\cos\theta = \frac{r^2}{2} \sin(2\theta)\text{,}$ where we have used a double-angle formula (see Appendix A.14).
Subbing these two areas into the above expression for $V$ gives
\begin{align*} V & = L\times\left[\theta r^2- \frac{r^2}{2}\sin2\theta\right] = \frac{Lr^2}{2} \big[2\theta-\sin2\theta \big] \end{align*}
Oof!
Now we can differentiate to find the rate of change. Recalling that $V=V(t)$ and $\theta=\theta(t)\text{,}$ while $r$ and $L$ are constants,
\begin{align*} V' &=\frac{Lr^2}{2} \left[ 2\theta' - 2\cos2\theta \cdot \theta' \right]\\ &= Lr^2 \cdot \theta' \cdot \left[1 - \cos2\theta \right] \end{align*}
Solving this for $\theta'$ and using $V'=q$ gives
\begin{align*} \theta' &= \frac{q}{Lr^2 (1-\cos2\theta)} \end{align*}
This is the rate at which $\theta$ is changing, but we need the rate at which $h$ is changing. We get this from
\begin{align*} h &= r - r\cos \theta & \text{differentiating this gives}\\ h' &= r\sin\theta \cdot \theta' \end{align*}
Substituting our expression for $\theta'$ into the expression for $h'$ gives
\begin{align*} h' &= r\sin\theta \cdot \frac{q}{Lr^2 (1-\cos2\theta)} \end{align*}
We can clean this up a bit more — recall more double-angle formulas
⁴
Take another look at Appendix A.14.
\begin{align*} h' &= r\sin\theta \cdot \frac{q}{Lr^2 (1-\cos2\theta)} & \text{substitute $\cos2\theta = 1-2\sin^2\theta$}\\ &= r\sin\theta \cdot \frac{q}{Lr^2 \cdot 2\sin^2\theta} & \text{now cancel $r$'s and a $\sin\theta$}\\ &= \frac{q}{2Lr\sin\theta} \end{align*}
But we can clean this up even more — instead of writing this rate in terms of $\theta$ it is more natural to write it in terms of $h$ (since the initial problem is stated in terms of $h$). From the triangle

and by Pythagoras we have

\begin{gather*} \sin\theta =\frac{\sqrt{r^2-(r-h)^2}}{r}=\frac{\sqrt{2rh-h^2}}{r} \end{gather*}

and hence

\begin{gather*} h' = \frac{q}{2L\sqrt{2rh-h^2}}. \end{gather*}
As a check, notice that $h'$ becomes undefined when $h \lt 0$ and also when $h \gt 2r\text{,}$ because then the argument of the square root in the denominator is negative. Both make sense — the fuel level in the tank must obey $0\le h\le 2r\text{.}$

Exercises 3.2.2 Exercises

Exercises — Stage 1 .

1.

Suppose the quantities $P$ and $Q$ are related by the formula $P=Q^3\text{.}$ $P$ and $Q$ are changing with respect to time, $t\text{.}$ Given this information, which of the following are problems you could solve?

Given $\ds\diff{P}{t}(0)\text{,}$ find $\ds\diff{Q}{t}(0)\text{.}$ (Remember: the notation $\ds\diff{P}{t}(0)$ means the derivative of $P$ with respect to $t$ at the time $t=0\text{.}$)
Given $\ds\diff{P}{t}(0)$ and the value of $Q$ when $t=0\text{,}$ find $\ds\diff{Q}{t}(0)\text{.}$
Given $\ds\diff{Q}{t}(0)\text{,}$ find $\ds\diff{P}{t}(0)\text{.}$
Given $\ds\diff{Q}{t}(0)$ and the value of $P$ when $t=0\text{,}$ find $\ds\diff{P}{t}(0)\text{.}$

Exercises — Stage 2 .

For problems 3.2.2.2 through 3.2.2.4, the relationship between several variables is explicitly given. Use this information to relate their rates of change.

For Questions 3.2.2.5 through 3.2.2.9, look for a way to use the Pythagorean Theorem.

For Questions 3.2.2.10 through 3.2.2.14, look for tricks from trigonometry.

For Questions 3.2.2.15 through 3.2.2.20, you’ll need to know formulas for volume or area.

2. (✳).

A point is moving on the unit circle $\set{ (x, y)\;:\; x^2 + y^2 = 1 }$ in the $xy$--plane. At $(2/\sqrt{5}, 1/\sqrt{5})\text{,}$ its $y$--coordinate is increasing at rate 3. What is the rate of change of its $x$--coordinate?

3. (✳).

The quantities $P,\ Q$ and $R$ are functions of time and are related by the equation $R=PQ\text{.}$ Assume that $P$ is increasing instantaneously at the rate of $8\%$ per year and that $Q$ is decreasing instantaneously at the rate of $2\%$ per year. That is, $\dfrac{P'}{P}=0.08$ and $\dfrac{Q'}{Q}=-0.02\text{.}$ Determine the percentage rate of change for $R\text{.}$

4. (✳).

Three quantities, $F\text{,}$ $P$ and $Q$ all depend upon time $t$ and are related by the equation

\begin{equation*} F=\frac{P}{Q} \end{equation*}

Assume that at a particular moment in time $P=25$ and $P$ is increasing at the instantaneous rate of 5 units/min. At the same moment, $Q=5$ and $Q$ is increasing at the instantaneous rate of 1 unit/min. What is the instantaneous rate of change in $F$ at this moment?
Assume that at another moment in time $P$ is increasing at the instantaneous rate of $10\%$ and $Q$ is decreasing at the instantaneous rate $5\%\text{.}$ What can you conclude about the rate of change of $F$ at this moment?

5. (✳).

Two particles move in the Cartesian plane. Particle A travels on the $x$-axis starting at $(10,0)$ and moving towards the origin with a speed of $2$ units per second. Particle B travels on the $y$-axis starting at $(0,12)$ and moving towards the origin with a speed of $3$ units per second. What is the rate of change of the distance between the two particles when particle A reaches the point $(4,0)\text{?}$

6. (✳).

Two particles $A$ and $B$ are placed on the Cartesian plane at $(0,0)$ and $(3,0)$ respectively. At time 0, both start to move in the $+y$ direction. Particle $A$ moves at 3 units per second, while $B$ moves at $2$ units per second. How fast is the distance between the particles changing when particle $A$ is at a distance of $5$ units from $B\text{.}$

7. (✳).

Ship A is 400 miles directly south of Hawaii and is sailing south at 20 miles/hour. Ship B is 300 miles directly east of Hawaii and is sailing west at 15 miles/hour. At what rate is the distance between the ships changing?

8. (✳).

Two tall sticks are vertically planted into the ground, separated by a distance of $30$ cm. We simultaneously put two snails at the base of each stick. The two snails then begin to climb their respective sticks. The first snail is moving with a speed of $25$ cm per minute, while the second snail is moving with a speed of $15$ cm per minute. What is the rate of change of the distance between the two snails when the first snail reaches $100$ cm above the ground?

9. (✳).

A $20$m long extension ladder leaning against a wall starts collapsing in on itself at a rate of $2$m/s, while the foot of the ladder remains a constant $5$m from the wall. How fast is the ladder moving down the wall after $3.5$ seconds?

10.

A watering trough has a cross section shaped like an isosceles trapezoid. The trough is 2 metres long, 50 cm high, 1 metre wide at the top, and 60 cm wide at the bottom.

A pig is drinking water from the trough at a rate of 3 litres per minute. When the height of the water is 25 cm, how fast is the height decreasing?

11.

A tank is 5 metres long, and has a trapezoidal cross section with the dimensions shown below.

A hose is filling the tank up at a rate of one litre per second. How fast is the height of the water increasing when the water is 10 centimetres deep?

12.

A rocket is blasting off, 2 kilometres away from you. You and the rocket start at the same height. The height of the rocket in kilometres, $t$ hours after liftoff, is given by

\begin{equation*} h(t)=61750t^2 \end{equation*}

How fast (in radians per second) is your line of sight rotating to keep looking at the rocket, one minute after liftoff?

13. (✳).

A high speed train is traveling at 2 km/min along a straight track. The train is moving away from a movie camera which is located 0.5 km from the track.

How fast is the distance between the train and the camera increasing when they are 1.3 km apart?
Assuming that the camera is always pointed at the train, how fast (in radians per min) is the camera rotating when the train and the camera are 1.3 km apart?

14.

A clock has a minute hand that is 10 cm long, and an hour hand that is 5 cm long. Let $D$ be the distance between the tips of the two hands. How fast is $D$ decreasing at 4:00?

15. (✳).

Find the rate of change of the area of the annulus $\{ (x, y) \;:\; r^2 \le x^2 + y^2 \le R^2 \}\text{.}$ (i.e. the points inside the circle of radius $R$ but outside the circle of radius $r$) if $R = 3 \;\mathrm{cm}\text{,}$ $r = 1\;\mathrm{cm}\text{,}$ $\ds\diff{R}{t} = 2\;\frac{\mathrm{cm}}{\mathrm{s}}\text{,}$ and $\ds\diff{r}{t} = 7\;\frac{\mathrm{cm}}{\mathrm{s}}\text{.}$

16.

Two spheres are centred at the same point. The radius $R$ of the bigger sphere at time $t$ is given by $R(t)=10+2t\text{,}$ while the radius $r$ of the smaller sphere is given by $r(t)=6t\text{,}$ $t \ge 0\text{.}$ How fast is the volume between the spheres (inside the big sphere and outside the small sphere) changing when the bigger sphere has a radius twice as large as the smaller?

17.

You attach two sticks together at their ends, and stick the other ends in the mud. One stick is 150 cm long, and the other is 200 cm.

The structure starts out being 1.4 metres high at its peak, but the sticks slide, and the height decreases at a constant rate of three centimetres per minute. How quickly is the area of the triangle (formed by the two sticks and the level ground) changing when the height of the structure is 120 cm?

18.

The circular lid of a salt shaker has radius 8. There is a cut-out to allow the salt to pour out of the lid, and a door that rotates around to cover the cut-out. The door is a quarter-circle of radius 7 cm. The cut-out has the shape of a quarter-annulus with outer radius 6 cm and inner radius 1 cm. If the uncovered area of the cut-out is $A$ cm$^2\text{,}$ then the salt flows out at $\frac{1}{5}A$ cm$^3$ per second.

Recall: an annulus is the set of points inside one circle and outside another, like a flat doughnut (see Question 3.2.2.15).

While pouring out salt, you spin the door around the lid at a constant rate of $\frac{\pi}{6}$ radians per second, covering more and more of the cut-out. When exactly half of the cut-out is covered, how fast is the flow of salt changing?

19.

A cylindrical sewer pipe with radius 1 metre has a vertical rectangular door that slides in front of it to block the flow of water, as shown below. If the uncovered area of the pipe is $A$ m$^2\text{,}$ then the flow of water through the pipe is $\frac{1}{5}A$ cubic metres per second.

The door slides over the pipe, moving vertically at a rate of 1 centimetre per second. How fast is the flow of water changing when the door covers the top 25 centimetres of the pipe?

20.

A martini glass is shaped like a cone, with top diameter 10 cm and side length 10 cm.

When the liquid in the glass is 7 cm high, it is evaporating at a rate of 5 mL per minute. How fast is the height of the liquid decreasing?

Exercises — Stage 3 .

21.

A floating buoy is anchored to the bottom of a river. As the river flows, the buoy is pulled in the direction of flow until its 2-metre rope is taut. A sensor at the anchor reads the angle $\theta$ between the rope and the riverbed, as shown in the diagram below. This data is used to measure the depth $D$ of water in the river, which depends on time.

If $\theta = \dfrac{\pi}{4}$ and $\ds\diff{\theta}{t}=0.25\;\frac{\mathrm{rad}}{\mathrm{hr}}\text{,}$ how fast is the depth $D$ of the water changing?
A measurement shows $\ds\diff{\theta}{t}=0\text{,}$ but $\ds\diff{D}{t}\neq0\text{.}$ Under what circumstances does this occur?
A measurement shows $\ds\diff{\theta}{t} \gt 0\text{,}$ but $\ds\diff{D}{t} \lt 0\text{.}$ Under what circumstances does this occur?

22.

A point is moving in the $xy$-plane along the quadrilateral shown below.

When the point is at $(0,-2)\text{,}$ it is moving to the right. An observer stationed at the origin must turn at a rate of one radian per second to keep looking directly at the point. How fast is the point moving?
When the point is at $(0,2)\text{,}$ its $x$-coordinate is increasing at a rate of one unit per second. How fast it its $y$-coordinate changing? How fast is the point moving?

23.

You have a cylindrical water bottle 20 cm high, filled with water. Its cross section is a circle of radius 5. You slowly smoosh the sides, so the cross section becomes an ellipse with major axis (widest part) $2a$ and minor axis (skinniest part) $2b\text{.}$

After $t$ seconds of smooshing the bottle, $a=5+t$ cm. The perimeter of the cross section is unchanged as the bottle deforms. The perimeter of an ellipse is actually quite difficult to calculate, but we will use an approximation derived by Ramanujan and assume that the perimeter $p$ of our ellipse is

\begin{equation*} p \approx \pi\left[3(a+b)-\sqrt{(a+3b)(3a+b)}\right]. \end{equation*}

The area of an ellipse is $\pi a b\text{.}$

Give an equation that relates $a$ and $b$ (and no other variables).
Give an expression for the volume of the bottle as it is being smooshed, in terms of $a$ and $b$ (and no other variables).
Suppose the bottle was full when its cross section was a circle. How fast is the water spilling out when $a$ is twice as big as $b\text{?}$

24.

The quantities $A\text{,}$ $B\text{,}$ $C\text{,}$ and $D$ all depend on time, and are related by the formula

\begin{equation*} AB=\log\left(C^2+D^2+1\right). \end{equation*}

At time $t=10\text{,}$ the following values are known:

$\displaystyle A=0$
$\ds\diff{A}{t}=2$ units per second

What is $B$ when $t=10\text{?}$

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CLP-1 Differential Calculus