Exercise 0.1: Differentiate a relation with respect to time: ten direct calculations

In each line, every letter stands for a quantity that depends on time $t$ . Differentiate the relation with respect to $t$ , then substitute the given values and rates to obtain the rate asked for.

Three moves, always in the same order: differentiate both sides with respect to $t$ , writing the rate $\frac{dx}{dt}$ , $\frac{dy}{dt}$ ... next to each differentiated quantity, as an inner derivative; only then substitute the values at that instant; finally isolate the unknown rate. A term that contains two variable quantities is a product or a quotient, and is differentiated as such.

a) $y=x^{3}-2x$ , with $\frac{dx}{dt}=3$ : $\frac{dy}{dt}$ when $x=2$ .
b) $y=\frac{4}{x}$ , with $\frac{dx}{dt}=-2$ : $\frac{dy}{dt}$ when $x=4$ .
c) $x^{2}-y^{2}=16$ , with $\frac{dx}{dt}=6$ : $\frac{dy}{dt}$ at the point $(5, 3)$ .
d) $x^{2}+4y^{2}=25$ , with $\frac{dx}{dt}=4$ : $\frac{dy}{dt}$ at the point $(3, 2)$ .
e) $A=xy$ , with $x=8$ , $y=5$ , $\frac{dx}{dt}=2$ and $\frac{dy}{dt}=-3$ : $\frac{dA}{dt}$ .
f) $A=\frac{bh}{2}$ , with $b=10$ , $h=6$ , $\frac{db}{dt}=1.5$ and $\frac{dh}{dt}=-0.5$ : $\frac{dA}{dt}$ .
g) $A=\pi r^{2}$ , with $\frac{dA}{dt}=12$ : $\frac{dr}{dt}$ when $r=3$ , then the rate of change of the circumference $C=2\pi r$ at the same instant.
h) $V=x^{2}h$ , with $x=3$ , $h=5$ , $\frac{dx}{dt}=0.5$ and $\frac{dh}{dt}=-1$ : $\frac{dV}{dt}$ .
i) $z=\frac{x}{y}$ , with $x=4$ , $y=2$ , $\frac{dx}{dt}=3$ and $\frac{dy}{dt}=-1$ : $\frac{dz}{dt}$ .
j) $\tan\theta=\frac{x}{5}$ , with $\frac{dx}{dt}=4$ : $\frac{d\theta}{dt}$ when $x=5\sqrt{3}$ .

Type your answers, the page tells you right or wrong 0/11

a)

\frac{dy}{dt}

b)

\frac{dy}{dt}

c)

\frac{dy}{dt}

d)

\frac{dy}{dt}

e)

\frac{dA}{dt}

f)

\frac{dA}{dt}

g)

\frac{dr}{dt}

\frac{dC}{dt}

h)

\frac{dV}{dt}

i)

\frac{dz}{dt}

j)

\frac{d\theta}{dt}

Show the solution

Answers

a) $\frac{dy}{dt}=(3x^{2}-2)\frac{dx}{dt}$ ; $30$
b) $\frac{dy}{dt}=-\frac{4}{x^{2}}\frac{dx}{dt}$ ; $\frac{1}{2}$
c) $\frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}$ ; $10$
d) $\frac{dy}{dt}=-\frac{x}{4y}\frac{dx}{dt}$ ; $-1.5$
e) $\frac{dA}{dt}=\frac{dx}{dt}y+x\frac{dy}{dt}$ ; $-14$
f) $\frac{dA}{dt}=\frac{1}{2}\left(\frac{db}{dt}h+b\frac{dh}{dt}\right)$ ; $2$
g) $\frac{dr}{dt}=\frac{1}{2\pi r}\frac{dA}{dt}=\frac{2}{\pi}\approx 0.637$ ; $\frac{dC}{dt}=4$
h) $\frac{dV}{dt}=2xh\frac{dx}{dt}+x^{2}\frac{dh}{dt}$ ; $6$
i) $\frac{dz}{dt}=\frac{\frac{dx}{dt}y-x\frac{dy}{dt}}{y^{2}}$ ; $2.5$
j) $\frac{d\theta}{dt}=\frac{\cos^{2}\theta}{5}\frac{dx}{dt}$ ; $0.2$

The move that decides everything: every variable quantity, once differentiated, leaves its rate as a factor. $x^{3}$ becomes $3x^{2}\frac{dx}{dt}$ and not $3x^{2}$ , $y^{2}$ becomes $2y\frac{dy}{dt}$ , and $xy$ becomes $\frac{dx}{dt}y+x\frac{dy}{dt}$ . Numerical values come in only AFTER the differentiation: a quantity replaced by a number too early would have a zero derivative.

a)

$\frac{dy}{dt}=(3x^{2}-2)\frac{dx}{dt}$ . At $x=2$ : $\frac{dy}{dt}=(12-2)\times 3=30$ . Pitfall: stopping at $3x^{2}-2=10$ , the derivative with respect to $x$ , without the factor $\frac{dx}{dt}$ .

b)

$y=4x^{-1}$ , so $\frac{dy}{dt}=-\frac{4}{x^{2}}\frac{dx}{dt}$ . At $x=4$ : $\frac{dy}{dt}=-\frac{4}{16}\times(-2)=\frac{1}{2}$ . Two minus signs, a positive result: $x$ decreases, so $\frac{4}{x}$ increases.

c)

$2x\frac{dx}{dt}-2y\frac{dy}{dt}=0$ , so $\frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}$ . At the point $(5, 3)$ , which does satisfy $25-9=16$ : $\frac{dy}{dt}=\frac{5}{3}\times 6=10$ . Pitfall: differentiating $y^{2}$ as $2y$ , without $\frac{dy}{dt}$ .

d)

$2x\frac{dx}{dt}+8y\frac{dy}{dt}=0$ , so $\frac{dy}{dt}=-\frac{x}{4y}\frac{dx}{dt}$ . At the point $(3, 2)$ , which satisfies $9+16=25$ : $\frac{dy}{dt}=-\frac{3}{8}\times 4=-1.5$ . The coefficient $4$ is differentiated along with $y^{2}$ : $(4y^{2})'=8y\frac{dy}{dt}$ .

e)

Product rule: $\frac{dA}{dt}=\frac{dx}{dt}\,y+x\frac{dy}{dt}=2\times 5+8\times(-3)=10-24=-14$ . Pitfall: $\frac{dx}{dt}\cdot\frac{dy}{dt}=-6$ , which is the derivative of nothing.

f)

$\frac{dA}{dt}=\frac{1}{2}\left(\frac{db}{dt}h+b\frac{dh}{dt}\right)=\frac{1}{2}\left(1.5\times 6+10\times(-0.5)\right)=\frac{1}{2}(9-5)=2$ . The factor $\frac{1}{2}$ multiplies BOTH terms of the product.

g)

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$ , so $\frac{dr}{dt}=\frac{12}{2\pi\times 3}=\frac{2}{\pi}\approx 0.637$ . Then $\frac{dC}{dt}=2\pi\frac{dr}{dt}=2\pi\times\frac{2}{\pi}=4$ . Here you go back from the rate of the area to the rate of the radius: you isolate $\frac{dr}{dt}$ instead of multiplying it. Notice that $\frac{dC}{dt}=2\pi\frac{dr}{dt}$ does not contain $r$ : when the radius of a ripple on water grows at constant speed, its circumference also grows at constant speed, whatever radius it has reached. Only the area speeds up, since $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$ grows with $r$ .

h)

$\frac{dV}{dt}=2xh\frac{dx}{dt}+x^{2}\frac{dh}{dt}=2\times 3\times 5\times 0.5+9\times(-1)=15-9=6$ . Pitfall: treating $h$ as a constant and keeping only $2xh\frac{dx}{dt}=15$ .

i)

Quotient rule: $\frac{dz}{dt}=\frac{\frac{dx}{dt}y-x\frac{dy}{dt}}{y^{2}}=\frac{3\times 2-4\times(-1)}{4}=\frac{6+4}{4}=2.5$ . Pitfall: the minus sign of the quotient in front of $x\frac{dy}{dt}$ , which becomes a plus because $\frac{dy}{dt}<0$ .

j)

$\sec^{2}\theta\,\frac{d\theta}{dt}=\frac{1}{5}\frac{dx}{dt}$ . At $x=5\sqrt{3}$ , $\tan\theta=\sqrt{3}$ , so $\sec^{2}\theta=1+\tan^{2}\theta=4$ without even finding $\theta$ . Hence $\frac{d\theta}{dt}=\frac{4}{5\times 4}=0.2$ radian per unit of time. Pitfall: forgetting the $\sec^{2}\theta$ , which gives $0.8$ .

Check: move time forward by a very small step $h$ , with $x$ replaced by $x+h\frac{dx}{dt}$ (and likewise for the other quantities whose rate is given); the quantity you are looking for must change by about $h$ times the rate you found.

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Exercise 1: The sliding ladder: the method, step by step

A $5$ m ladder leans against a vertical wall. Its foot slides away from the wall at a constant speed of $0.6$ m/s.

a) Draw the diagram, name the variable quantities and say which ones depend on time.
b) Write the relation that links these quantities and that holds at EVERY instant, not only at the instant asked for.
c) Differentiate this relation with respect to time, and only then substitute the numerical values: how fast is the top of the ladder sliding down when the foot is $3$ m from the wall?
d) What happens to this speed as the foot approaches $5$ m from the wall? Interpret the result and say what the model leaves out.

Show the solution

Answers

a) $x(t)$ and $y(t)$ vary; the length $5$ is constant
b) $x^{2}+y^{2}=25$
c) $\frac{dy}{dt}=-0.45$ m/s: the top slides down
d) $\frac{dy}{dt}\to-\infty$ : the ladder leaves the wall

a)

Let $x(t)$ be the distance from the foot to the wall and $y(t)$ the height of the top of the ladder on the wall, both in metres and both functions of time $t$ in seconds. The LENGTH of the ladder, on the other hand, does not depend on time: it is $5$ at every instant. This is the distinction that organizes the whole chapter. A quantity that varies gets a function name and will be differentiated; a constant quantity gets a number and will come out of the derivative. Given: $\frac{dx}{dt}=0.6$ m/s. Unknown: $\frac{dy}{dt}$ at the instant when $x=3$ .

b)

The ladder, the wall and the ground form a right triangle, so $x^{2}+y^{2}=25$ . This relation holds at every instant, which is exactly the condition for being allowed to differentiate it with respect to $t$ . A relation that holds only at the instant in question, for example “ $x=3$ ”, cannot be differentiated: it would give $\frac{dx}{dt}=0$ , which is false.

c)

Differentiate both sides with respect to $t$ , treating $x$ and $y$ as composite functions: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ , hence $\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}$ . NOW substitute. When $x=3$ , the relation gives $y=\sqrt{25-9}=4$ . So $\frac{dy}{dt}=-\frac{3}{4}\times 0.6=-0.45$ m/s. The negative sign is the answer on its own: the top of the ladder is going DOWN, at $0.45$ m/s. Answer with the unit and the physical direction, not with a bare number.

d)

As $x\to 5^{-}$ , we get $y\to 0^{+}$ and the quotient $-\frac{x}{y}\cdot 0.6$ tends to $-\infty$ : the model predicts an infinite speed of descent. Numerical check: at $x=4.9$ we find $y=\sqrt{25-24.01}\approx 0.995$ and $\frac{dy}{dt}\approx-2.955$ m/s, already six times more than at $x=3$ . What the model leaves out is that the ladder eventually leaves the wall: the relation $x^{2}+y^{2}=25$ assumes permanent contact at the top and at the bottom, an assumption that stops being physically tenable well before $x=5$ . A related rate that blows up is not a computational error, it is the sign that the geometric constraint no longer holds.

Exercise 2: Substituting before differentiating: the error that cancels everything

A spherical balloon is inflated at a rate of $100$ cm $^{3}$ /s. The figure below shows the situation.

a) How fast is the radius increasing at the instant when it is $5$ cm?
b) A student writes: “the radius is $5$ , so $V=\frac{4}{3}\pi(5)^{3}$ , so $\frac{dV}{dt}=0$ ”. Explain precisely where the absurdity comes from.
c) How fast is the surface area of the balloon increasing at the same instant?
d) Is the radius growing faster and faster or more and more slowly? Justify with a formula, not with an impression.

Show the solution

Answers

a) $\frac{dr}{dt}=\frac{1}{\pi}\approx 0.318$ cm/s
b) Substituting $r=5$ before differentiating gives a constant
c) $\frac{dS}{dt}=40$ cm²/s
d) $\frac{dr}{dt}\propto\frac{1}{r^{2}}$ : more and more slowly

a)

The relation that holds at every instant is $V=\frac{4}{3}\pi r^{3}$ , where $V$ and $r$ both depend on $t$ . Differentiate: $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$ . Then substitute $\frac{dV}{dt}=100$ and $r=5$ : $100=4\pi(25)\frac{dr}{dt}$ , so $\frac{dr}{dt}=\frac{100}{100\pi}=\frac{1}{\pi}\approx 0.318$ cm/s.

b)

The absurdity fits in one sentence: by substituting $r=5$ BEFORE differentiating, you replace a function of time by a number. $\frac{4}{3}\pi(5)^{3}=\frac{500\pi}{3}$ is a constant, and the derivative of a constant is zero. The student has therefore proved that the volume does not change, whereas the problem says it increases by $100$ cm $^{3}$ /s. The rule that follows holds for the whole chapter: differentiate the LITERAL relation first, and substitute the numerical values only afterwards. The value $r=5$ describes one instant, the relation $V=\frac{4}{3}\pi r^{3}$ describes all instants; only the second can be differentiated.

c)

$S=4\pi r^{2}$ , so $\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi(5)\left(\frac{1}{\pi}\right)=40$ cm $^{2}$ /s. The $\pi$ cancels, which is a good consistency signal: $\frac{dr}{dt}$ contained $\frac{1}{\pi}$ and $\frac{dS}{dt}$ contains a factor $\pi$ . Another route, faster and without going through $\frac{dr}{dt}$ : since $\frac{dS}{dt}=8\pi r\frac{dr}{dt}$ and $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$ , we have $\frac{dS}{dt}=\frac{2}{r}\cdot\frac{dV}{dt}=\frac{2}{5}\times 100=40$ ✓.

d)

From $\frac{dr}{dt}=\frac{1}{4\pi r^{2}}\cdot\frac{dV}{dt}$ with $\frac{dV}{dt}$ constant, we read that $\frac{dr}{dt}$ is proportional to $\frac{1}{r^{2}}$ : the radius grows more and more slowly. In numbers: at $r=5$ we have $\frac{1}{\pi}\approx 0.318$ cm/s, at $r=10$ we have $\frac{1}{4\pi}\approx 0.0796$ cm/s, FOUR times less for a radius twice as large. The physical explanation is that the same volume of air has to spread over a surface four times larger.

Exercise 3: The three usual solids: cube, sphere, cylinder

Seven independent situations. In each one, write the relation, differentiate it, then substitute.

a) The edge of a cube grows by $2$ cm/min. How fast are its volume and its total surface area growing when the edge is $10$ cm?
b) A ball of ice melts, losing $20$ cm $^{3}$ /min. How fast is its radius decreasing when it is $4$ cm?
c) A cylinder of fixed height $10$ cm has its radius growing by $0.3$ cm/min. How fast is its volume growing when the radius is $6$ cm?
d) In a cylinder where the radius AND the height vary, we have $r=6$ , $h=10$ , $\frac{dr}{dt}=0.3$ and $\frac{dh}{dt}=-0.5$ (in cm and cm/min). Is the volume increasing or decreasing?
e) The volume of a cube increases by $12$ cm $^{3}$ /min. How fast are its edge, then its total surface area, increasing when the edge is $20$ cm?
f) A snowball is being rolled: its radius grows by $0.5$ cm/s. How fast is its volume increasing at the moment its DIAMETER is $30$ cm?
g) A cylindrical tank with a vertical axis and an inside diameter of $4$ m is filled at the constant rate of $6$ m $^{3}$ /h. How fast is the water level rising when it reaches $1$ m? And when it reaches $2$ m? Compare the result with statement c) of exercise 8.

Type your answers, the page tells you right or wrong 0/11

a)

\frac{dV}{dt}

cm³/min

\frac{dS}{dt}

cm²/min

b)

\frac{dr}{dt}

cm/min

c)

\frac{dV}{dt}

cm³/min

d)

\frac{dV}{dt}

cm³/min

The volume

e)

\frac{da}{dt}

cm/min

\frac{dS}{dt}

cm²/min

f)

\frac{dV}{dt}

cm³/s

g)

\frac{dh}{dt}

at

1

m m/h

At

2

m, the level rises

Show the solution

Answers

a) $600$ cm³/min and $240$ cm²/min
b) $-\frac{5}{16\pi}\approx-0.0995$ cm/min
c) $36\pi\approx 113.1$ cm³/min
d) $18\pi\approx 56.5$ cm³/min: the volume increases
e) $\frac{da}{dt}=0.01$ cm/min, $\frac{dS}{dt}=2.4$ cm²/min
f) $450\pi\approx 1413.7$ cm³/s
g) $\frac{3}{2\pi}\approx 0.477$ m/h, at any height

a)

$V=a^{3}$ gives $\frac{dV}{dt}=3a^{2}\frac{da}{dt}=3(100)(2)=600$ cm $^{3}$ /min. $S=6a^{2}$ gives $\frac{dS}{dt}=12a\frac{da}{dt}=12(10)(2)=240$ cm $^{2}$ /min. Notice that the two answers have different units: a volume per minute and an area per minute. An answer with no unit, or with the wrong one, costs the same half mark as the computation itself.

b)

$V=\frac{4}{3}\pi r^{3}$ gives $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$ . Here $\frac{dV}{dt}=-20$ : the word “losing” imposes the negative sign, and it is the only place in the problem where the wording translates into a sign. So $-20=4\pi(16)\frac{dr}{dt}$ and $\frac{dr}{dt}=-\frac{20}{64\pi}=-\frac{5}{16\pi}\approx-0.0995$ cm/min. The radius decreases by about $1$ mm per minute.

c)

The height is CONSTANT, so it is not differentiated: $V=\pi r^{2}h=10\pi r^{2}$ and $\frac{dV}{dt}=20\pi r\frac{dr}{dt}=20\pi(6)(0.3)=36\pi\approx 113.1$ cm $^{3}$ /min. This is the easy case, and that is precisely why you must ask yourself in every exercise which quantities vary: treating $h$ as a variable here adds a ghost term.

d)

Both vary, so the product is differentiated with the product rule: $V=\pi r^{2}h$ gives $\frac{dV}{dt}=\pi\left(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}\right)$ . Numerically: $2(6)(10)(0.3)=36$ and $(36)(-0.5)=-18$ , so $\frac{dV}{dt}=\pi(36-18)=18\pi\approx 56.5$ cm $^{3}$ /min. The volume INCREASES, despite a height that decreases: the widening wins over the sinking. This is the situation of a dough being flattened, and the computation says exactly beyond which ratio of the two rates the volume would start to decrease, namely $\frac{dh}{dt}<-\frac{2h}{r}\frac{dr}{dt}=-1$ cm/min here.

e)

Here the known rate is not that of the edge, and the relation between $V$ and $S$ cannot be written directly: you go BACK through the edge. $V=a^{3}$ gives $\frac{dV}{dt}=3a^{2}\frac{da}{dt}$ , so $12=3(400)\frac{da}{dt}$ and $\frac{da}{dt}=0.01$ cm/min. Then $S=6a^{2}$ gives $\frac{dS}{dt}=12a\frac{da}{dt}=12(20)(0.01)=2.4$ cm $^{2}$ /min. Eliminating $\frac{da}{dt}$ between the two lines, $\frac{dS}{dt}=\frac{4}{a}\frac{dV}{dt}$ : for a fixed volume rate, the area grows more and more slowly as the cube gets bigger (only $1.2$ cm $^{2}$ /min at $a=40$ cm). Pitfall: differentiating $S=6a^{2}$ as $12a$ and stopping there, which gives $240$ , the derivative with respect to the edge and not with respect to time.

f)

The volume formula is written with the radius: translate the diameter first, $r=\frac{30}{2}=15$ cm. Then $V=\frac{4}{3}\pi r^{3}$ gives $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}=4\pi(225)(0.5)=450\pi\approx 1413.7$ cm $^{3}$ /s. Pitfall: substituting $r=30$ gives $1800\pi\approx 5654.9$ , four times too much, because the radius is squared. If the problem had given the rate of change of the DIAMETER, it would also have had to be divided by two: $\frac{dr}{dt}=\frac{1}{2}\frac{dD}{dt}$ .

g)

The radius is FIXED, $r=2$ m, and only the height $h$ varies: $V=\pi r^{2}h=4\pi h$ , so $\frac{dV}{dt}=4\pi\frac{dh}{dt}$ and $\frac{dh}{dt}=\frac{6}{4\pi}=\frac{3}{2\pi}\approx 0.477$ m/h, about $0.8$ cm/min. The height does not appear in the result: the level rises at the same speed at $1$ m, at $2$ m and at any height. This is the case where statement 8 c) is TRUE, because the horizontal cross-section $\pi r^{2}$ does not change with $h$ ; in the cone of exercise 4 or the spherical tank of exercise 15 it does change, and so does the speed of the level. Pitfalls: taking $4$ m as the radius, which gives $\frac{6}{16\pi}\approx 0.119$ m/h, four times too little; and looking for a value of $h$ to substitute, when none is needed.

Exercise 4: The inverted cone: eliminate a variable BEFORE differentiating

A tank has the shape of an inverted cone, vertex down: its radius at the top is $3$ m and its total height $6$ m. Water is poured in at a rate of $2$ m $^{3}$ /min.

a) Express the radius $r$ of the water surface in terms of its height $h$ , and say which geometric argument allows it.
b) How fast is the level rising when the water reaches a height of $4$ m?
c) What happens if you differentiate $V=\frac{1}{3}\pi r^{2}h$ WITHOUT eliminating $r$ first? Do the computation and conclude.
d) Compare the rising speed at $h=2$ m and at $h=4$ m, then explain the ratio obtained.

Show the solution

Answers

a) $r=\frac{h}{2}$ by similar triangles
b) $\frac{dh}{dt}=\frac{1}{2\pi}\approx 0.159$ m/min
c) Two unknowns: you need $\frac{dr}{dt}=\frac{1}{2}\frac{dh}{dt}$
d) $\frac{2}{\pi}$ at $h=2$ : four times faster

a)

The water surface and the side of the cone cut out two SIMILAR triangles, both right triangles sharing the angle at the vertex. The ratio of radius to height is therefore the same for the water and for the whole cone: $\frac{r}{h}=\frac{3}{6}=\frac{1}{2}$ , hence $r=\frac{h}{2}$ . This similarity relation is the real content of the exercise, and it is what is missing in the solutions that get stuck.

b)

Substitute $r=\frac{h}{2}$ into the volume BEFORE differentiating: $V=\frac{1}{3}\pi\left(\frac{h}{2}\right)^{2}h=\frac{\pi h^{3}}{12}$ . Only $h$ remains. Differentiate: $\frac{dV}{dt}=\frac{\pi h^{2}}{4}\cdot\frac{dh}{dt}$ . Substitute $\frac{dV}{dt}=2$ and $h=4$ : $2=\frac{\pi(16)}{4}\frac{dh}{dt}=4\pi\frac{dh}{dt}$ , so $\frac{dh}{dt}=\frac{1}{2\pi}\approx 0.159$ m/min, about $16$ cm per minute.

c)

Without elimination, differentiation gives $\frac{dV}{dt}=\frac{\pi}{3}\left(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}\right)$ . This equality is correct, but it contains TWO unknowns, $\frac{dr}{dt}$ and $\frac{dh}{dt}$ , for a single equation: the problem cannot be solved as it stands. You then have to add the differentiated relation $\frac{dr}{dt}=\frac{1}{2}\frac{dh}{dt}$ , and substituting it you do get back $\frac{dV}{dt}=\frac{\pi}{3}\left(2\cdot 2\cdot 4\cdot\frac{1}{2}\frac{dh}{dt}+4\frac{dh}{dt}\right)=\frac{\pi}{3}(12)\frac{dh}{dt}=4\pi\frac{dh}{dt}$ ✓, the same result as in b). The lesson for the method: eliminating before differentiating is not compulsory, it is simply twice as short and removes the risk of forgetting the relation between the two rates.

d)

At $h=2$ : $2=\frac{\pi(4)}{4}\frac{dh}{dt}=\pi\frac{dh}{dt}$ , so $\frac{dh}{dt}=\frac{2}{\pi}\approx 0.637$ m/min. The ratio with the value found at $h=4$ is $\frac{2/\pi}{1/(2\pi)}=4$ : the level rises FOUR times more slowly at $4$ m than at $2$ m, for exactly the same flow rate. The explanation is in the formula $\frac{dh}{dt}=\frac{4}{\pi h^{2}}\cdot\frac{dV}{dt}$ : the rising speed is inversely proportional to the SQUARE of the height, because the cross-section of the cone at height $h$ is proportional to $h^{2}$ . A constant flow rate never gives a constant rise, except in a container with a constant cross-section.

Exercise 5: Similar triangles: the street lamp and the shadow

A street lamp is $6$ m tall. A $1.8$ m person walks away from it in a straight line at $1.5$ m/s.

a) Let $x$ be the distance from the person to the foot of the lamp and $s$ the length of their shadow. Establish the relation between $x$ and $s$ .
b) How fast is the LENGTH of the shadow increasing?
c) How fast is the TIP of the shadow moving away from the lamp? It is not the same question.
d) Do these two speeds depend on the person's position? What becomes of the answer if the person is $2$ m tall?

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Answers

a) $s=\frac{3}{7}x$
b) $\frac{ds}{dt}\approx 0.643$ m/s
c) $\frac{d(x+s)}{dt}=\frac{15}{7}\approx 2.143$ m/s
d) Independent of $x$ ; $0.75$ and $2.25$ m/s for $2$ m

a)

The large triangle, formed by the lamp and the tip of the shadow, and the small triangle, formed by the person and that same tip, are similar: same angles, two verticals and a common ground. The ratio of height to base is therefore the same: $\frac{6}{x+s}=\frac{1.8}{s}$ . Cross-multiplying: $6s=1.8(x+s)$ , that is $4.2s=1.8x$ and $s=\frac{3}{7}x$ . The writing trap is here: the base of the LARGE triangle is $x+s$ , not $x$ . A solution that writes $\frac{6}{x}=\frac{1.8}{s}$ treats the shadow as if the lamp stood where the person is.

b)

Differentiate the relation obtained, which is linear: $\frac{ds}{dt}=\frac{3}{7}\cdot\frac{dx}{dt}=\frac{3}{7}(1.5)=\frac{4.5}{7}\approx 0.643$ m/s. The shadow grows longer by about $64$ cm per second.

c)

The tip of the shadow is at a distance $x+s$ from the foot of the lamp, so its speed is $\frac{d}{dt}(x+s)=1.5+0.643=\frac{15}{7}\approx 2.143$ m/s. This is the question that costs the most marks in the chapter, because the two wordings look alike: “how fast is the shadow lengthening” asks for $\frac{ds}{dt}$ , “how fast is the tip of the shadow moving” asks for $\frac{d(x+s)}{dt}$ . Underlining in the problem the exact quantity whose rate is asked for, before even starting, avoids this loss for sure.

d)

No: the relation $s=\frac{3}{7}x$ is LINEAR, so its rate does not depend on $x$ . The person can be $2$ m or $200$ m from the lamp, the shadow always lengthens at $\frac{4.5}{7}$ m/s. That is what sets this problem apart from the cone, where the relation was cubic and the rate depended on the height. With a $2$ m person: $\frac{6}{x+s}=\frac{2}{s}$ gives $6s=2x+2s$ , that is $s=\frac{x}{2}$ , then $\frac{ds}{dt}=0.75$ m/s and $\frac{d(x+s)}{dt}=2.25$ m/s. The taller the person, the faster their shadow lengthens: the ratio $\frac{p}{6-p}$ , where $p$ is their height, grows with $p$ and becomes infinite for a $6$ m person, whose shadow would have no tip at all.

Exercise 6: Angles: the lighthouse sweeping the coast

A lighthouse stands $3$ km from a straight coastline. Its beam turns at a rate of $4$ revolutions per minute. Let $\theta$ be the angle between the beam and the perpendicular to the coast, and $x$ the distance between the lit point and the foot of that perpendicular.

a) Convert the rotation speed into the only unit that can be used here, and say why degrees are not allowed.
b) Establish the relation between $x$ and $\theta$ , then compute the speed of the lit point when $\theta=30°$ .
c) Same question for $\theta=60°$ . Is the ratio of the two speeds equal to the ratio of the two angles?
d) What happens to the speed as $\theta\to 90°$ ? Interpret.

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Answers

a) $8\pi$ rad/min
b) $32\pi\approx 100.5$ km/min
c) $96\pi\approx 301.6$ km/min: ratio $3$
d) $+\infty$ : the beam becomes parallel to the coast

a)

One revolution is $2\pi$ radians, so $\frac{d\theta}{dt}=4\times 2\pi=8\pi$ rad/min. Degrees are not allowed because the differentiation formulas for trigonometric functions, starting with $(\tan\theta)'=\sec^{2}\theta$ , hold only in radians: they rest on $\lim_{u\to 0}\frac{\sin u}{u}=1$ , which is false in degrees, where this limit equals $\frac{\pi}{180}$ . An answer in degrees per minute is off by a factor of about $57.3$ , and the marker sees it immediately from the order of magnitude.

b)

In the right triangle formed by the lighthouse, the foot of the perpendicular and the lit point, the side opposite $\theta$ is $x$ and the adjacent side is $3$ : $\tan\theta=\frac{x}{3}$ , that is $x=3\tan\theta$ . Differentiate: $\frac{dx}{dt}=3\sec^{2}\theta\cdot\frac{d\theta}{dt}$ . At $\theta=\frac{\pi}{6}$ : $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ , so $\sec^{2}\frac{\pi}{6}=\frac{4}{3}$ and $\frac{dx}{dt}=3\cdot\frac{4}{3}\cdot 8\pi=32\pi\approx 100.5$ km/min. The beam sweeps the coast at about $6000$ km/h, which is not absurd at all: no object is moving, it is an intersection point.

c)

At $\theta=\frac{\pi}{3}$ : $\cos\frac{\pi}{3}=\frac{1}{2}$ , so $\sec^{2}\frac{\pi}{3}=4$ and $\frac{dx}{dt}=3(4)(8\pi)=96\pi\approx 301.6$ km/min. The ratio of the speeds is $\frac{96\pi}{32\pi}=3$ , whereas the ratio of the angles is $2$ : the rates are NOT proportional to the angles, because the relation $x=3\tan\theta$ is not linear. Doubling the angle does not double the speed, here it triples it.

d)

As $\theta\to\frac{\pi}{2}^{-}$ , $\cos\theta\to 0$ so $\sec^{2}\theta\to+\infty$ and $\frac{dx}{dt}\to+\infty$ . The beam becomes parallel to the coast and the lit point runs off to infinity: beyond $90°$ there is simply no intersection point with the coast any more, which is the geometric translation of the vertical asymptote of the tangent. The computation and the drawing say the same thing, and that is the best consistency check available on this type of problem.

Exercise 7: Two moving objects: Pythagoras, then the law of cosines

Two cars leave the same intersection at the same instant. The first drives east at $60$ km/h, the second north at $80$ km/h. The figure below shows the situation of question a).

a) How fast are they moving apart after $2$ hours?
b) Show that this speed does not depend on the instant chosen, and give the geometric reason.
c) Redo question a) if the two roads make an angle of $60°$ instead of $90°$ .
d) Generalize: for an angle $\alpha$ between the roads, what is the speed at which they move apart? Check your formula on the two previous cases and on $\alpha=0$ .

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Answers

a) $100$ km/h
b) $z=100t$ : constant speed
c) $z\approx 144.22$ km; $\frac{dz}{dt}\approx 72.11$ km/h
d) $\sqrt{10\,000-9600\cos\alpha}$ ; $20$ km/h for $\alpha=0$

a)

Let $x(t)$ and $y(t)$ be the distances travelled and $z(t)$ the distance between the cars. Since the roads are perpendicular, $z^{2}=x^{2}+y^{2}$ , so $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$ , that is $\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$ . After $2$ h: $x=120$ , $y=160$ , so $z=\sqrt{120^{2}+160^{2}}=\sqrt{40000}=200$ km. Hence $\frac{dz}{dt}=\frac{120(60)+160(80)}{200}=\frac{7200+12800}{200}=\frac{20000}{200}=100$ km/h.

b)

Substitute $x=60t$ , $y=80t$ : then $z=\sqrt{3600t^{2}+6400t^{2}}=100t$ , so $\frac{dz}{dt}=100$ km/h at every instant. The reason is that both motions are uniform and in a straight line: the relative velocity vector is constant, with components $(60, 80)$ , and its magnitude is $\sqrt{60^{2}+80^{2}}=100$ . A right triangle whose two legs grow in proportion to time has a hypotenuse that also grows in proportion to time, hence at a constant rate. Remember this shortcut: when the two objects start from the SAME point at constant speed, the speed at which they move apart is the magnitude of the relative velocity, and no related rates computation is even needed.

c)

With an angle of $60°$ , Pythagoras no longer applies and the law of cosines replaces it: $z^{2}=x^{2}+y^{2}-2xy\cos 60°=x^{2}+y^{2}-xy$ . Differentiate: $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}-\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right)$ , the last term coming from the product rule on $xy$ . At $t=2$ : $z^{2}=14400+25600-19200=20800$ , so $z=\sqrt{20800}\approx 144.22$ km. The numerator is $2(120)(60)+2(160)(80)-\left[120(80)+160(60)\right]=14400+25600-19200=20800$ . So $\frac{dz}{dt}=\frac{20800}{2(144.22)}\approx 72.11$ km/h. Check: this is exactly $\frac{z}{t}=\frac{144.22}{2}$ , which confirms b).

d)

For an angle $\alpha$ : $z^{2}=x^{2}+y^{2}-2xy\cos\alpha$ with $x=60t$ and $y=80t$ gives $z=t\sqrt{3600+6400-9600\cos\alpha}$ , so $\frac{dz}{dt}=\sqrt{10000-9600\cos\alpha}$ , a constant. Checks: $\alpha=90°$ gives $\sqrt{10000}=100$ ✓; $\alpha=60°$ gives $\sqrt{10000-4800}=\sqrt{5200}\approx 72.11$ ✓; $\alpha=0$ gives $\sqrt{400}=20=80-60$ , which is indeed the speed at which two cars on the SAME road move apart ✓. This last check is worth doing every time: a degenerate case whose answer is obvious tests the general formula better than rereading does.

Exercise 8: Five statements to correct

Each statement is false. Say in one sentence why, then write the correct statement.

a) “The radius is $5$ cm, so I replace $r$ by $5$ in $V=\frac{4}{3}\pi r^{3}$ , then I differentiate.”
b) “ $\frac{dV}{dt}$ and $\frac{dV}{dr}$ are the same thing written differently.”
c) “The volume increases at a constant rate, so the height rises at constant speed.”
d) “I found a negative rate, so there must be a sign error somewhere.”
e) “The angle turns by $30$ degrees per second, so $\frac{d\theta}{dt}=30$ .”

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Answers

a) Differentiate the literal relation, then substitute
b) $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$
c) Constant speed only with a constant cross-section
d) A negative rate signals a quantity that decreases
e) $\frac{\pi}{6}\approx 0.524$ rad/s

a)

FALSE. Substituting before differentiating turns a function of time into a constant, whose derivative is zero: the solution then proves that the volume does not change. Correct statement: “I differentiate the literal relation $V=\frac{4}{3}\pi r^{3}$ , which gives $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$ , THEN I substitute $r=5$ .” The numerical value describes one instant, the relation describes all instants.

b)

FALSE. $\frac{dV}{dr}=4\pi r^{2}$ is the change in volume per unit of RADIUS, an area; $\frac{dV}{dt}$ is its change per unit of TIME, a flow rate. The two do not even have the same units. Correct statement: “They are linked by the chain rule $\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}$ ”, which is in fact the whole theory of the chapter in one line: a related rate is nothing more than a chain rule where the intermediate variable is time.

c)

FALSE as soon as the container does not have a constant cross-section. In the cone of exercise 4, a constant flow rate of $2$ m $^{3}$ /min gives $\frac{dh}{dt}=\frac{4}{\pi h^{2}}\times 2$ , which decreases like $\frac{1}{h^{2}}$ : four times slower at $4$ m than at $2$ m. Correct statement: “The height rises at constant speed if and only if the cross-sectional area is constant”, that is, in a cylinder or a right prism. Exercise 3 g) gives the true case: in a cylindrical tank, $\frac{dh}{dt}=\frac{1}{\pi r^{2}}\frac{dV}{dt}$ is constant.

d)

FALSE. The sign is INFORMATION, not an anomaly: it says that the quantity decreases. The top of the ladder slides down, the ice melts, the angle of elevation of a plane flying away decreases. Correct statement: “A negative rate means that the quantity decreases; I only check that this direction is consistent with the situation described.” The opposite error also exists and costs just as much: writing $\frac{dV}{dt}=20$ when the problem says “loses $20$ cm $^{3}$ per minute”.

e)

FALSE as long as the unit is not converted. The trigonometric differentiation formulas hold only in radians. Correct statement: “ $\frac{d\theta}{dt}=30\times\frac{\pi}{180}=\frac{\pi}{6}$ rad/s.” The size of the error is a factor of $\frac{180}{\pi}\approx 57.3$ , which makes the answer absurd and easy to spot: a sweeping speed of several thousand kilometres per minute for a lighthouse that makes one revolution every fifteen seconds.

Exercise 9: Problem: the V-shaped trough that fills up

A horizontal trough $3$ m long has a cross-section shaped like an isosceles triangle, vertex down: it is $1$ m wide at the top and $0.5$ m deep. Water is poured in at $0.04$ m $^{3}$ /min.

a) Express the volume of water in terms of its height $h$ .
b) How fast is the level rising when $h=0.3$ m? And when $h=0.1$ m?
c) How long does it take to fill the trough?
d) Give $h$ as a function of time and explain, from this expression, why the rise slows down.

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Answers

a) $V=3h^{2}$
b) $0.0222$ m/min and $0.0667$ m/min
c) $18.75$ min
d) $h(t)=\sqrt{\frac{0.04t}{3}}$ : growth like $\sqrt{t}$

a)

At height $h$ , the width $w$ of the free surface is read by similar triangles: $\frac{w}{h}=\frac{1}{0.5}=2$ , so $w=2h$ . The wet cross-section is a triangle with base $w$ and height $h$ , of area $\frac{1}{2}wh=\frac{1}{2}(2h)(h)=h^{2}$ . Since the trough is a prism of length $3$ m, $V=3h^{2}$ (in m $^{3}$ , with $h$ in m). Check: at $h=0.5$ , $V=3(0.25)=0.75$ m $^{3}$ , which is indeed the total volume of the full trough.

b)

$\frac{dV}{dt}=6h\frac{dh}{dt}$ , so $\frac{dh}{dt}=\frac{0.04}{6h}$ . At $h=0.3$ : $\frac{0.04}{1.8}\approx 0.0222$ m/min, that is $2.2$ cm per minute. At $h=0.1$ : $\frac{0.04}{0.6}\approx 0.0667$ m/min, that is $6.7$ cm per minute, exactly THREE times faster. The ratio of the speeds is the inverse of the ratio of the heights, because the area of the free surface, $3\times 2h=6h$ , is proportional to $h$ : $\frac{dV}{dt}=6h\frac{dh}{dt}$ , and the same flow spreads over a surface that keeps widening.

c)

The total volume is $0.75$ m $^{3}$ and the flow rate is $0.04$ m $^{3}$ /min, so $t=\frac{0.75}{0.04}=18.75$ min, that is $18$ min $45$ s. No differentiation here: when the flow rate is constant, the filling time is a simple quotient. Mistaking it for a related rates computation is a pure waste of time.

d)

The volume accumulated at time $t$ is $0.04t$ , so $3h^{2}=0.04t$ and $h(t)=\sqrt{\frac{0.04t}{3}}$ . The square root says it all: the height grows like $\sqrt{t}$ , hence more and more slowly, and its derivative $\frac{dh}{dt}=\frac{0.02}{\sqrt{0.12t}}$ tends to $0$ . Consistency check between the two points of view: at $h=0.3$ we have $t=\frac{3(0.09)}{0.04}=6.75$ min, and $\frac{0.02}{\sqrt{0.12\times 6.75}}=\frac{0.02}{0.9}\approx 0.0222$ m/min ✓, the same value as in b). A V-shaped trough fills quickly at first and slowly at the end, which is the opposite of the common impression and is proved in two lines.

Exercise 10: Problem: the plane, the radar and the decreasing angle

A plane flies level at an altitude of $6$ km and at $500$ km/h. It passes directly over a radar station, then flies away. Let $x$ be its horizontal distance from the station, $z$ its direct distance from the station and $\theta$ the angle of elevation at which the radar sees it. The figure below shows the situation.

a) How fast is the distance $z$ increasing when it is $10$ km?
b) What is $\frac{dz}{dt}$ at the instant when the plane is exactly above the radar? Justify without computation, then with the formula.
c) What value does $\frac{dz}{dt}$ tend to as the plane flies away indefinitely?
d) At the instant of a), how fast is the angle of elevation decreasing? Give the result in radians per hour, then in degrees per minute.

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Answers

a) $x=8$ km; $400$ km/h
b) $0$ directly above
c) $500$ km/h
d) $-30$ rad/h, that is $-28.6$ °/min

a)

The relation that holds at every instant is $z^{2}=x^{2}+36$ , the altitude $6$ being constant. Differentiate: $2z\frac{dz}{dt}=2x\frac{dx}{dt}$ , that is $\frac{dz}{dt}=\frac{x}{z}\cdot\frac{dx}{dt}$ . When $z=10$ : $x=\sqrt{100-36}=8$ km. So $\frac{dz}{dt}=\frac{8}{10}(500)=400$ km/h. The direct distance increases more slowly than the speed of the plane, and the factor $\frac{x}{z}=\cos\theta$ says why: only the component of the velocity along the line of sight makes $z$ grow.

b)

Directly above, $x=0$ : the formula $\frac{dz}{dt}=\frac{x}{z}\frac{dx}{dt}$ gives $0$ . Without computation: at that instant the plane moves perpendicular to the radar-plane line, so to first order it neither approaches nor moves away, exactly like an object at its point of closest approach. It is also the minimum of $z$ , which gives the result through another door: the derivative of a differentiable function is zero at its minimum.

c)

As $x\to+\infty$ , $\frac{x}{z}=\frac{x}{\sqrt{x^{2}+36}}\to 1$ , so $\frac{dz}{dt}\to 500$ km/h. Far from the radar, the altitude becomes negligible compared with the horizontal distance, and everything happens as if the plane were flying straight away from the radar. Numerical check: at $x=100$ km, $\frac{dz}{dt}=\frac{100}{\sqrt{10036}}(500)\approx 499.1$ km/h.

d)

The angle of elevation satisfies $\tan\theta=\frac{6}{x}$ . Differentiate: $\sec^{2}\theta\cdot\frac{d\theta}{dt}=-\frac{6}{x^{2}}\cdot\frac{dx}{dt}$ . At the instant of a), $x=8$ and $z=10$ , so $\cos\theta=\frac{x}{z}=0.8$ and $\sec^{2}\theta=\frac{1}{0.64}=1.5625$ . Hence $\frac{d\theta}{dt}=\frac{-6(500)}{64\times 1.5625}=\frac{-3000}{100}=-30$ rad/h. In degrees per minute: $-30\times\frac{180}{\pi}\div 60\approx-28.6$ °/min. The negative sign is consistent: the plane flies away, so the radar must tilt down. A useful shortcut for checking: since $\sec^{2}\theta=\frac{z^{2}}{x^{2}}$ , the formula simplifies to $\frac{d\theta}{dt}=-\frac{6}{z^{2}}\frac{dx}{dt}=-\frac{6(500)}{100}=-30$ rad/h ✓, without ever computing an angle.

Exercise 11: The rope pulling a boat to the dock

A boat is pulled toward a dock by a rope tied to its bow and passed through a ring fixed $2$ m above the level of the bow. The rope is pulled in at a constant speed of $1$ m/s. Let $s$ be the length of rope between the ring and the bow, and $x$ the horizontal distance between the boat and the dock.

a) Write the relation between $s$ and $x$ that holds at every instant.
b) How fast is the boat approaching the dock when $8$ m of rope remain?
c) Same question when $2.5$ m of rope remain. What do you notice?
d) Show that the boat always moves faster than the rope, and express the ratio of the two speeds in terms of the angle $\varphi$ that the rope makes with the horizontal.
e) Compute this angle, in degrees, when $8$ m of rope remain.

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Answers

a) $s^{2}=x^{2}+4$
b) $1.033$ m/s
c) $1.667$ m/s: the boat speeds up
d) Speed of the boat $=\frac{1}{\cos\varphi}\times$ speed of the rope
e) $\varphi\approx 14.5°$

a)

The ring, the point of the dock at the level of the bow, and the bow form a right triangle: $s^{2}=x^{2}+4$ . The height $2$ is constant, $s$ and $x$ depend on time.

b)

Differentiate: $2s\frac{ds}{dt}=2x\frac{dx}{dt}$ , so $\frac{dx}{dt}=\frac{s}{x}\frac{ds}{dt}$ . Pulling the rope SHORTENS it: $\frac{ds}{dt}=-1$ . For $s=8$ : $x=\sqrt{64-4}=\sqrt{60}\approx 7.746$ , and $\frac{dx}{dt}=\frac{8}{7.746}\times(-1)\approx-1.033$ m/s. The boat approaches at $1.033$ m/s.

c)

For $s=2.5$ : $x=\sqrt{6.25-4}=1.5$ and $\frac{dx}{dt}=\frac{2.5}{1.5}\times(-1)\approx-1.667$ m/s. The rope still moves at $1$ m/s, but the boat SPEEDS UP as it gets closer: near the dock, the rope pulls more and more upward and the smallest centimetre of rope translates into a large horizontal displacement.

d)

$\left|\frac{dx}{dt}\right|=\frac{s}{x}\left|\frac{ds}{dt}\right|$ and $s>x$ since $s$ is the hypotenuse: the ratio $\frac{s}{x}$ is always greater than $1$ . Now $\frac{x}{s}=\cos\varphi$ , so the speed of the boat is $\frac{1}{\cos\varphi}$ times that of the rope. The opposite intuition, “the boat follows the rope, so it goes at most as fast”, confuses the speed along the rope with the horizontal speed.

e)

$\sin\varphi=\frac{2}{8}=0.25$ , so $\varphi\approx 14.5°$ . Check: $\frac{1}{\cos 14.48°}\approx 1.033$ , the speed found in b).

Exercise 12: The shadow cast on a wall by a spotlight on the ground

A spotlight on the ground lights up a wall $12$ m away. A $2$ m person starts at the spotlight and walks straight toward the wall at $1.6$ m/s. Let $x$ be their distance from the spotlight and $h$ the height of their shadow on the wall. The figure below shows the situation.

a) Establish the relation between $h$ and $x$ .
b) How fast is the height of the shadow changing when the person is $4$ m from the wall?
c) At what distance from the WALL is the shadow shrinking at a speed of $1$ m/s?
d) How fast is the shadow changing at the moment the person reaches the wall?
e) Unlike the shadow on the ground under a street lamp, does this speed depend on the position? Why?

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Answers

a) $h=\frac{24}{x}$
b) $-0.6$ m/s
c) $x\approx 6.20$ m, so $5.80$ m from the wall
d) $-0.267$ m/s
e) Yes: the relation is not linear

a)

The light ray leaves the spotlight, grazes the top of the head and hits the wall. Two similar triangles with a common vertex, the spotlight: $\frac{h}{12}=\frac{2}{x}$ , that is $h=\frac{24}{x}$ .

b)

$\frac{dh}{dt}=-\frac{24}{x^{2}}\frac{dx}{dt}$ . At $4$ m from the wall, $x=8$ : $\frac{dh}{dt}=-\frac{24}{64}\times 1.6=-0.6$ m/s. The shadow SHRINKS by $60$ cm per second, and at that instant it is $h=3$ m tall.

c)

$\frac{24\times 1.6}{x^{2}}=1$ gives $x^{2}=38.4$ and $x\approx 6.20$ m from the spotlight, that is $12-6.20\approx 5.80$ m from the wall. Watch the distance asked for: $x$ is measured from the spotlight.

d)

At the wall, $x=12$ : $h=2$ , the shadow is the size of the person, and $\frac{dh}{dt}=-\frac{24}{144}\times 1.6\approx-0.267$ m/s.

e)

Yes: the relation $h=\frac{24}{x}$ is not linear, its derivative $-\frac{24}{x^{2}}$ depends on $x$ . The shadow shrinks very fast near the spotlight and more and more slowly toward the wall. Under a street lamp, the relation between the shadow and the distance was linear, and the rate constant: it is the FORM of the relation, not the context, that tells you whether the rate varies.

Exercise 13: The volume that does not change: rolling a cylinder of clay

A cylinder of modelling clay of constant volume $V=10\pi$ cm $^{3}$ is rolled on a table. At the instant considered, its radius is $1$ cm, its height $10$ cm, and the radius is decreasing by $0.02$ cm/s. The figure below shows the situation.

a) Write the relation between the rates of change of the radius and of the height.
b) How fast is the height increasing at this instant?
c) Later, the radius is $0.5$ cm and is still decreasing by $0.02$ cm/s. Compute the height and its rate of change.
d) How fast is the lateral surface area $S=2\pi rh$ changing at the first instant?
e) If the radius decreases at constant speed, does the height grow at constant speed?

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Answers

a) $\frac{dh}{dt}=-\frac{2h}{r}\frac{dr}{dt}$
b) $0.4$ cm/s
c) $h=40$ cm; $3.2$ cm/s
d) $0.4\pi\approx 1.257$ cm²/s
e) $\frac{dh}{dt}=\frac{0.4}{r^{3}}$ : faster and faster

a)

$V=\pi r^{2}h$ is CONSTANT, so $\frac{dV}{dt}=0$ : $\pi\left(2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}\right)=0$ , that is $\frac{dh}{dt}=-\frac{2h}{r}\frac{dr}{dt}$ . The zero on the left-hand side is the main piece of information in the problem, given by a single word: “constant”.

b)

$\frac{dh}{dt}=-\frac{2\times 10}{1}\times(-0.02)=0.4$ cm/s. The radius loses two tenths of a millimetre per second, the height gains four millimetres.

c)

$h=\frac{V}{\pi r^{2}}=\frac{10}{0.25}=40$ cm. $\frac{dh}{dt}=-\frac{2\times 40}{0.5}\times(-0.02)=3.2$ cm/s, eight times faster than at the start.

d)

Since $h=\frac{V}{\pi r^{2}}$ , $S=2\pi rh=\frac{2V}{r}=\frac{20\pi}{r}$ . So $\frac{dS}{dt}=-\frac{20\pi}{r^{2}}\frac{dr}{dt}=-20\pi\times(-0.02)=0.4\pi\approx 1.257$ cm $^{2}$ /s. By the product rule without eliminating: $2\pi\left(h\frac{dr}{dt}+r\frac{dh}{dt}\right)=2\pi(-0.2+0.4)=0.4\pi$ ✓. The area increases: a thin coil of clay has more surface than a thick one, for the same volume.

e)

No. Replacing $h=\frac{10}{r^{2}}$ : $\frac{dh}{dt}=-\frac{20}{r^{3}}\frac{dr}{dt}=\frac{0.4}{r^{3}}$ . As $r$ decreases, $\frac{1}{r^{3}}$ blows up: the height grows faster and faster, as seen between b) and c), where dividing the radius by $2$ multiplied the speed by $2^{3}=8$ .

Exercise 14: Problem: the piston, the connecting rod and the law of cosines

In an engine, a crank of radius $r=5$ cm turns about the axis $O$ and, through a connecting rod of length $L=15$ cm, drives a piston $P$ that slides along a line through $O$ . Let $\theta$ be the angle of the crank measured from this line and $x=OP$ . The crank turns at $3000$ revolutions per minute. The figure below shows the situation.

a) Using the law of cosines in the triangle formed by $O$ , the crank-rod joint and $P$ , show that $L^{2}=r^{2}+x^{2}-2rx\cos\theta$ , then that $x=r\cos\theta+\sqrt{L^{2}-r^{2}\sin^{2}\theta}$ .
b) Compute $x$ for $\theta=0$ and for $\theta=\pi$ , then the stroke of the piston.
c) Convert the rotation speed into radians per second.
d) By differentiating the relation of a) implicitly, compute the speed of the piston for $\theta=\frac{\pi}{2}$ .
e) Compute the speed of the piston for $\theta=\frac{\pi}{3}$ . Is the piston fastest when the crank is perpendicular to its line?

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Answers

a) $x=r\cos\theta+\sqrt{L^{2}-r^{2}\sin^{2}\theta}$
b) $20$ cm, $10$ cm; stroke $10$ cm
c) $100\pi\approx 314.16$ rad/s
d) $15.7$ m/s
e) $16.0$ m/s: faster before $\frac{\pi}{2}$

a)

The triangle has sides $OA=r$ , $AP=L$ and $OP=x$ , with the angle at $O$ equal to $\theta$ . The law of cosines at $O$ gives $L^{2}=r^{2}+x^{2}-2rx\cos\theta$ . This is a quadratic equation in $x$ : $x^{2}-2r\cos\theta\,x+r^{2}-L^{2}=0$ , whose positive root is $x=r\cos\theta+\sqrt{r^{2}\cos^{2}\theta-r^{2}+L^{2}}=r\cos\theta+\sqrt{L^{2}-r^{2}\sin^{2}\theta}$ .

b)

$\theta=0$ : $x=5+15=20$ cm, piston at its farthest. $\theta=\pi$ : $x=-5+15=10$ cm, at its closest. Stroke: $10$ cm, twice the crank radius.

c)

$3000$ revolutions per minute is $50$ revolutions per second, so $\frac{d\theta}{dt}=50\times 2\pi=100\pi\approx 314.16$ rad/s.

d)

Differentiate $L^{2}=r^{2}+x^{2}-2rx\cos\theta$ with respect to time, $L$ and $r$ constant: $0=2x\frac{dx}{dt}-2r\cos\theta\frac{dx}{dt}+2rx\sin\theta\frac{d\theta}{dt}$ , hence $\frac{dx}{dt}=-\frac{rx\sin\theta}{x-r\cos\theta}\frac{d\theta}{dt}$ . For $\theta=\frac{\pi}{2}$ : $x=\sqrt{225-25}\approx 14.142$ and $\cos\theta=0$ , so $\frac{dx}{dt}=-r\frac{d\theta}{dt}=-500\pi\approx-1571$ cm/s, that is $15.7$ m/s toward $O$ .

e)

For $\theta=\frac{\pi}{3}$ : $x=2.5+\sqrt{225-18.75}\approx 2.5+14.361=16.861$ cm, and $x-r\cos\theta\approx 14.361$ . $\frac{dx}{dt}=-\frac{5\times 16.861\times 0.8660}{14.361}\times 314.16\approx-1597$ cm/s, that is $16.0$ m/s: FASTER than at $\frac{\pi}{2}$ . The slant of the connecting rod moves the maximum speed ahead of the perpendicular position. A related rate computed at two points is enough to refute an intuition that the symmetry of the crank made plausible.

Exercise 15: Problem: the spherical tank that empties

A spherical tank with inside radius $R=2$ m empties through the bottom at the constant rate of $0.5$ m $^{3}$ /min. When the water depth is $h$ , the volume of water is $V=\frac{\pi h^{2}(3R-h)}{3}=\frac{\pi h^{2}(6-h)}{3}$ . The figure shows a cross-section of the tank.

a) Check that the formula gives the volume of the sphere when the tank is full.
b) Compute $\frac{dV}{dh}$ and show that it equals the area of the free surface of the water.
c) How fast is the level dropping when $h=1$ m?
d) Compare with $h=3$ m and $h=2$ m. Where does the level drop most slowly?
e) Compute the speed at which the level drops when $h=0.1$ m, and the total time to empty the tank.

Show the solution

Answers

a) $\frac{32\pi}{3}\approx 33.51$ m³
b) $\frac{dV}{dh}=\pi(4h-h^{2})$ , area of the free surface
c) $-0.053$ m/min
d) Same speed at $3$ m; $-0.040$ m/min at $2$ m, the slowest
e) $-0.408$ m/min; $67.0$ min

a)

Full, $h=4$ : $V=\frac{\pi\times 16\times 2}{3}=\frac{32\pi}{3}\approx 33.51$ m $^{3}$ , and $\frac{4}{3}\pi R^{3}=\frac{4}{3}\pi\times 8=\frac{32\pi}{3}$ ✓.

b)

$V=\frac{\pi}{3}\left(6h^{2}-h^{3}\right)$ , so $\frac{dV}{dh}=\frac{\pi}{3}\left(12h-3h^{2}\right)=\pi(4h-h^{2})$ . The free surface is a disk whose radius $\rho$ satisfies, by Pythagoras in the cross-section, $\rho^{2}=R^{2}-(R-h)^{2}=4-(2-h)^{2}=4h-h^{2}$ . Its area is therefore $\pi(4h-h^{2})$ : exactly $\frac{dV}{dh}$ . Adding a thin layer of thickness $dh$ adds a volume equal to the surface area times the thickness.

c)

$\frac{dV}{dt}=\pi(4h-h^{2})\frac{dh}{dt}$ with $\frac{dV}{dt}=-0.5$ . For $h=1$ : $\pi\times 3\times\frac{dh}{dt}=-0.5$ , so $\frac{dh}{dt}=-\frac{0.5}{3\pi}\approx-0.053$ m/min, that is $5.3$ cm per minute.

d)

For $h=3$ : $4h-h^{2}=3$ as well, same speed: the free surface has the same radius $1$ m from the bottom and $1$ m from the top. For $h=2$ : $4h-h^{2}=4$ , the surface is largest, and $\frac{dh}{dt}=-\frac{0.5}{4\pi}\approx-0.040$ m/min. The level drops most slowly at mid-height, where the sphere is widest.

e)

For $h=0.1$ : $4h-h^{2}=0.39$ and $\frac{dh}{dt}=-\frac{0.5}{0.39\pi}\approx-0.408$ m/min: near the bottom, the surface shrinks to almost nothing and the level plunges. Total time: $\frac{33.51}{0.5}\approx 67.0$ min. For this last question no derivative is needed: constant flow rate, known volume, a simple quotient.

Corrected exercises: related rates (201-SN2-RE)

Course recap

Calculations: related rates to compute

Exercise 0.1: Differentiate a relation with respect to time: ten direct calculations

Part A: translate, differentiate, then substitute (/50)

Exercise 1: The sliding ladder: the method, step by step

Exercise 2: Substituting before differentiating: the error that cancels everything

Exercise 3: The three usual solids: cube, sphere, cylinder

Exercise 4: The inverted cone: eliminate a variable BEFORE differentiating

Exercise 5: Similar triangles: the street lamp and the shadow

Part B: exam-level problems (/50)

Exercise 6: Angles: the lighthouse sweeping the coast

Exercise 7: Two moving objects: Pythagoras, then the law of cosines

Exercise 8: Five statements to correct

Exercise 9: Problem: the V-shaped trough that fills up

Exercise 10: Problem: the plane, the radar and the decreasing angle

Part C: the classics (/50)

Exercise 11: The rope pulling a boat to the dock

Exercise 12: The shadow cast on a wall by a spotlight on the ground

Exercise 13: The volume that does not change: rolling a cylinder of clay

Exercise 14: Problem: the piston, the connecting rod and the law of cosines

Exercise 15: Problem: the spherical tank that empties

See also

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