Single-Variable-Calculus · Unit 10 · Video 3 · Interactive Practice

Related Rates: The Police Radar and the Conical Tank

IKey Formulas

Formula	Name	Video values
$30^2 + x^2 = D^2$	Radar triangle	$x = 40$ , $D = 50$ now
$2x\,x' = 2D\,D' \;\Rightarrow\; x' = \dfrac{D}{x}\,D'$	Rates of the radar triangle	$D' = -80 \Rightarrow x' = -100\ \text{ft/s}$
$V = \tfrac{1}{3}\pi r^2 h,\ \ r = \tfrac{2}{5}h \;\Rightarrow\; V = \tfrac{4}{75}\pi h^3$	Tank volume (similar triangles)	rim radius $4$ , height $10$
$\dfrac{dV}{dt} = \tfrac{4}{25}\pi h^2\,\dfrac{dh}{dt}$	Rates of the tank	$\dfrac{dV}{dt} = 2,\ h = 5 \Rightarrow \dfrac{dh}{dt} = \dfrac{1}{2\pi}\ \text{ft/min}$

Key Insight: Picture, then $\frac{d}{dt}$ , then plug in. Give a letter to every quantity that changes; a number substituted before differentiating is treated as a constant, and its rate is lost.

IIRadar Reading vs. Road Speed

At a road speed of $100$ ft/s the radar reads only $\frac{x}{D}$ of it, because the road leg $x$ is shorter than $D$ .

Car position x (ft) 40

IIIWhy $D$ Needs a Letter

$30^2 + x^2 = 50^2$ is true at one instant only; $30^2 + x^2 = D^2$ is true at every instant, so only it may be differentiated.

Equation Time t (s) 0.00

IVThe Conical Tank Filling

Water enters at a steady $2\ \text{ft}^3/\text{min}$ , yet the level slows as it rises: $\frac{dh}{dt} = \frac{2}{\pi r^2}$ .

Time t (min) 10.5

VQuiz Questions

Problem 1 · A Second Radar Reading

Given: Police are parked $60$ ft from a straight road. Their radar reads that a car is $100$ ft away and closing at $90$ ft/s. Find the car's speed along the road.

✅ Correct!

x' = \frac{D}{x}D' = \frac{100}{80}(-90) = -112.5

, so the speed is

112.5

ft/s, more than the radar's

90

❌ The ratio is upside down. From

2x\,x' = 2D\,D'

x' = \frac{D}{x}D'

; since

D > x

, the road speed is larger than the radar reading, not smaller.

❌ That is the radar reading itself. The radar measures

D'

, the rate along its line of sight, which differs from

x'

along the road.

❌ Wrong leg.

\frac{100}{60}

uses the fixed

60

-ft side; the formula needs the road leg

x = \sqrt{100^2 - 60^2} = 80

Show solution

Let $x$ be the distance along the road to the foot of the perpendicular and $D$ the radar distance. The $60$ -ft side is constant:

60^2 + x^2 = D^2.

Differentiate with respect to $t$ : $2x\,x' = 2D\,D'$ , so $x' = \frac{D}{x}D'$ .

Now plug in. $D = 100$ gives $x = \sqrt{10000 - 3600} = 80$ , and $D' = -90$ :

x' = \frac{100}{80}(-90) = -112.5\ \text{ft/s}.

The car's speed is $112.5$ ft/s; the negative sign means $x$ is shrinking.

Problem 2 · Which Numbers May Go In Early?

Given: A $13$ -ft ladder leans against a wall. Its foot is $5$ ft from the wall and sliding away at $2$ ft/s. A student writes $x^2 + y^2 = 13^2$ (with $x$ the foot's distance from the wall and $y$ the top's height) before differentiating. Which statement is correct?

✅ Correct! Only quantities that change need letters. The ladder's length is a true constant, like the police's

30

ft;

y

is not.

❌ Too strict. The rule is about quantities that change. A genuine constant, such as the

30

-ft side in the radar problem, gets no letter.

❌ That freezes a changing quantity.

y = 12

only at this instant; substituting it makes

y'

vanish, the same mistake as plugging in

D = 50

first.

❌ The ratio is upside down. From

2x\,x' + 2y\,y' = 0

y' = -\frac{x}{y}\,x'

, with

x = 5

on top.

Show solution

The ladder's length is $13$ at every instant, so $x^2 + y^2 = 13^2$ is true for all $t$ and may be differentiated:

2x\,x' + 2y\,y' = 0 \quad\Rightarrow\quad y' = -\frac{x}{y}\,x'.

Plug in only now: $x = 5$ , $y = \sqrt{169 - 25} = 12$ , $x' = 2$ :

y' = -\frac{5}{12}(2) = -\frac{5}{6}\ \text{ft/s}.

The top slides down at $\frac{5}{6}$ ft/s. Substituting $y = 12$ before differentiating would wrongly give $y' = 0$ .

Problem 3 · A Different Conical Tank

Given: A conical tank stands point down, with rim radius $3$ ft and height $6$ ft. Water flows in at $3\ \text{ft}^3/\text{min}$ . Find the volume in terms of the depth $h$ , then how fast the level rises when $h = 2$ ft.

Volume of water in terms of $h$ :

Rate the level rises at $h = 2$ :

✅ Correct!

V = \frac{\pi}{12}h^3

gives

\frac{dV}{dt} = \frac{\pi}{4}h^2\frac{dh}{dt}

, so

3 = \pi\,\frac{dh}{dt}

and

\frac{dh}{dt} = \frac{3}{\pi} \approx 0.95

ft/min.

❌ The radius is not the depth. Similar triangles give

\frac{r}{h} = \frac{3}{6}

, so

r = \frac{h}{2}

, not

r = h

❌ The similar-triangle ratio is upside down.

\frac{r}{h} = \frac{\text{rim radius}}{\text{height}} = \frac{3}{6}

, so

r = \frac{h}{2}

, not

r = 2h

❌ The cone's $\tfrac{1}{3}$ is missing.

V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi\big(\tfrac{h}{2}\big)^2 h

❌ That plugs in $h = 2$ too early.

V = \frac{\pi}{12}(2)^3

is a constant, so its derivative is

0

; differentiate first, then substitute.

❌ Check the power rule.

\frac{d}{dt}h^3 = 3h^2\frac{dh}{dt}

, so at

h = 2

the factor is

h^2 = 4

, not

h = 2

❌ Check the volume. Express

r

in terms of

h

with similar triangles, then substitute into

V = \tfrac{1}{3}\pi r^2 h

❌ Check the rate. Differentiate

V(h)

with the chain rule, then set

\frac{dV}{dt} = 3

and

h = 2

Show solution

Step 1: picture and equation. The water is a smaller cone of the same shape, so $\frac{r}{h} = \frac{3}{6}$ and $r = \frac{h}{2}$ at every depth:

V = \frac{1}{3}\pi\Big(\frac{h}{2}\Big)^2 h = \frac{\pi}{12}h^3.

Step 2: differentiate.

\frac{dV}{dt} = \frac{3\pi}{12}h^2\frac{dh}{dt} = \frac{\pi}{4}h^2\frac{dh}{dt}.

Step 3: plug in $\frac{dV}{dt} = 3$ and $h = 2$ :

3 = \frac{\pi}{4}(4)\frac{dh}{dt} = \pi\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} = \frac{3}{\pi} \approx 0.955\ \text{ft/min}.

Check: at $h = 2$ the surface radius is $1$ , so the surface area is $\pi$ ft², and $3 \div \pi = \frac{3}{\pi}$ .

Problem 4 · Inflating a Balloon

Given: A spherical balloon is inflated at $100\ \text{cm}^3/\text{s}$ . Find how fast its radius grows when $r = 5$ cm.

✅ Correct!

\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}

: the inflow spreads over the surface area

4\pi(5)^2 = 100\pi

, just as the tank's inflow spread over

\pi r^2

❌ The power rule's $3$ is missing.

\frac{d}{dt}\big(\tfrac{4}{3}\pi r^3\big) = \tfrac{4}{3}\pi\cdot 3r^2\frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}

❌ That plugs in $r = 5$ too early.

\tfrac{4}{3}\pi(5)^3

is a constant; differentiate while

r

is still a letter.

❌ That differentiates twice.

8\pi r

is the derivative of

4\pi r^2

; the first derivative of

\tfrac{4}{3}\pi r^3

4\pi r^2

Show solution

The radius changes, so it keeps its letter: $V = \tfrac{4}{3}\pi r^3$ .

Differentiate with respect to $t$ (power rule and chain rule):

\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}.

Plug in $\frac{dV}{dt} = 100$ and $r = 5$ :

100 = 4\pi(25)\frac{dr}{dt} = 100\pi\frac{dr}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = \frac{1}{\pi} \approx 0.318\ \text{cm/s}.

As in the tank, the rate of volume divided by the area of the growing surface gives the rate of growth.

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Related Rates: The Police Radar and the Conical Tank

IKey Formulas

IIRadar Reading vs. Road Speed

IIIWhy DDD Needs a Letter

IVThe Conical Tank Filling

VQuiz Questions

IIIWhy $D$ Needs a Letter