20b. Related Rates — Practice

Drill sheet for 20. Related Rates. Book-style solved ideas as guided checks and regenerable worksheets.

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1. Spherical balloon

Gas escapes at $2$ ft³/min. Surface shrink rate when $r=12$:

$dV/dt=-2=4\pi r^{2}\,dr/dt$, $dS/dt=8\pi r\,dr/dt=-4/r=-1/3$ ft²/min.

Sphere / circle area rates

V, S, A linked to r

(1)When is $dS/dt=dr/dt$ for a sphere $S=4\pi r^{2}$?

$8\pi r\,dr/dt=dr/dt$ ⇒ $r=1/(8\pi)$ (if $dr/dt\neq 0$).

$r=1/(8\pi)$.

(2)Oil slick: $dr/dt=3$ m/min, $r=200$. Find $dA/dt$ for $A=\pi r^{2}$.

$dA/dt=2\pi r\,dr/dt=1200\pi$ m²/min.

$1200\pi$ m²/min.

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2. Conical funnel

Similar triangles $r=h/2$; $V=\pi h^{3}/12$. With $dV/dt=-1$ and $h=6$ (2 in from top of an 8 in funnel):

$$\frac{dh}{dt}=-\frac{1}{9\pi}\ \mathrm{in/s}.$$

Fig. 20-2 — water cone in a funnel

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3. Sand pile

$h=(4/3)r$ ⇒ $V=(4/9)\pi r^{3}$, $dV/dt=(4/3)\pi r^{2}\,dr/dt$.

(a) $r=3$, $dr/dt=1/4$ (ft/min) ⇒ $dV/dt=3\pi$.

(b) $r=6$, $dV/dt=24$ ⇒ $dr/dt=1/(2\pi)$.

Cone / sand / reservoir

One-variable volume via similar triangles

(1)Same pile: $r=6$, $dV/dt=24$. Find $dr/dt$.

$24=(4/3)\pi\cdot 36\,dr/dt$ ⇒ $dr/dt=1/(2\pi)$.

$1/(2\pi)$ ft/min.

(2)Conical reservoir $R=3$, $H=10$; $dV/dt=-4$. Depth $h=6$: find $dh/dt$ and $dr/dt$.

$r=(3/10)h$, $V=\pi r^{2}h/3=\pi h^{3}/100$… book: $dh/dt=-100/(81\pi)$, $dr/dt=-10/(27\pi)$.

$100/(81\pi)$ down; $10/(27\pi)$ radius shrinking.

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4. Two ships

$D^{2}=(32-16t)^{2}+(12t)^{2}$, $dD/dt=(400t-512)/D$.

$t$	behavior
$1$	approaching $5.6$ mi/h
$2$	separating $12$ mi/h
$1.28$	nearest ($D=19.2$ mi)

Fig. 20-3 — A south, B east

Distance between moving points

Ships, kite, balloon

(1)Kite at height $150$; moves horizontally $20$ ft/s. String rate when $250$ ft away?

$z^{2}=x^{2}+150^{2}$; $z=250$ ⇒ $x=200$; $z'=16$ ft/s.

$16$ ft/s.

(2)Kite at height $150$; moves horizontally $20$ ft/s. String rate when $250$ ft away?

$z^{2}=x^{2}+150^{2}$; $z=250$ ⇒ $x=200$; $z'=16$ ft/s.

$16$ ft/s.

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5. Rectangle of constant area

$xy=50$, $dx/dt=2$. Then $x\,dy/dt+y\cdot 2=0$ and $dP/dt=2(dx/dt+dy/dt)$.

• $x=5$: $dP/dt=-4$ (decreasing)
• $x=10$: $dP/dt=+2$ (increasing)
• Ceases decreasing when the figure is a square of side $5\sqrt{2}$

Book wording

Some keys say “decreasing” at $x=10$; the rate $+2$ means the perimeter is increasing.

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6. When $dS/dt=dr/dt$

For $S=4\pi r^{2}$: $8\pi r=1$ ⇒ $r=1/(8\pi)$.

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7. Pulley and truck

Rope $50$ ft; pulley $20$ ft up; truck hitch $2$ ft up. When weight raised $x=6$ and truck speed $dy/dt=9$:

$$\frac{dx}{dt}=\frac92\sqrt{3}\ \mathrm{ft/s}.$$

Fig. 20-4 — weight and truck

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8. Shadow tip

Object height $h$, light height $H$, object speed $v$: tip speed

$$V=\frac{Hv}{H-h}.$$

As $h\to 0$, $V\to v$; as $h\to H$, $V\to+\infty$.

Fig. 20-5 — tip of the shadow

Ladder · shadow · trough · cylinder

Classic geometry setups

(1)Rectangle area $50$ constant; $dx/dt=2$. Find $dP/dt$ when $x=5$ and when $x=10$.

$xy=50$, $x y'+y\cdot 2=0$.

$x=5$: $y'=-4$, $P'=-4$; $x=10$: $y'=-1$, $P'=2$.

$-4$; $+2$ in/s.

(2)Trough $8\times 2$ top (rectangular), water in at $2$ ft³/min. Surface rise rate when depth is $1$ ft?

Horizontal surface area is constant $8\cdot 2=16$ ft².

$dV/dt=16\,dh/dt=2$ ⇒ $dh/dt=1/8$ ft/min.

$1/8$ ft/min.

(3)Trough $8\times 2$ top (rectangular), water in at $2$ ft³/min. Surface rise rate when depth is $1$ ft?

Horizontal surface area is constant $8\cdot 2=16$ ft².

$dV/dt=16\,dh/dt=2$ ⇒ $dh/dt=1/8$ ft/min.

$1/8$ ft/min.

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Supplementary Problems

9–11. Trough, cylinder, walking shadow

Trough rise $1/8$ ft/min; cylinder rise $2/(9\pi)$ ft/min; man/light: tip $16/3$, shadow length $4/3$ ft/s.

12–16. Balloon, ladder, cone, barge, kite

Balloon–ground distance $12$ ft/s; $20$ ft ladder at $x=12$: top falls $3/2$ ft/s; cone reservoir answers involve $\pi$ in the denominator; barge cable $9/13$ ft/s; kite string $16$ ft/s.

17–21. Trains, falling ball shadow, ships, leaky cistern, filter

Trains separate at $\frac{105\sqrt{2}}{2}$ mi/h at 3 p.m.; ball shadow $200$ ft/s after $1$ s; leak rate $(10-3\pi)$ ft³/min; cylinder rises $4/9$ in/min under the filter.

22–24. Oil slick and curve points

$dA/dt=1200\pi$ at $r=200$; hyperbola point $dy/dt=50$; on $y=x^{2}-2x$, $dy/dt=2\,dx/dt$ at $(2,0)$.

Misc related rates

Pulley · barge · curves

(1)Shadow tip: object height $h$, light $H$, object speed $v$. Formula for tip speed $V$?

$V=Hv/(H-h)$.

$V=\dfrac{Hv}{H-h}$.

(2)Shadow tip: object height $h$, light $H$, object speed $v$. Formula for tip speed $V$?

$V=Hv/(H-h)$.

$V=\dfrac{Hv}{H-h}$.

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Mixed review

Related-rates mixed set

Sphere · cone · ships · ladder · more

(1)Gas escapes a sphere at $2$ ft³/min. How fast is surface area shrinking when $r=12$?

$dV/dt=-2=4\pi r^{2}\,dr/dt$.

$dS/dt=8\pi r\,dr/dt=-4/r=-1/3$ at $r=12$.

Shrinking at $1/3$ ft²/min.

(2)Funnel $R=4$, $H=8$; $dV/dt=-1$ in³/s. How fast is the level dropping when 2 in from the top?

$r=h/2$, $V=\pi h^{3}/12$, $dh/dt=-4/(\pi h^{2})$.

At $h=6$: $dh/dt=-1/(9\pi)$ in/s.

$1/(9\pi)$ in/s downward.

(3)When is $dS/dt=dr/dt$ for a sphere $S=4\pi r^{2}$?

$8\pi r\,dr/dt=dr/dt$ ⇒ $r=1/(8\pi)$ (if $dr/dt\neq 0$).

$r=1/(8\pi)$.

(4)Oil slick: $dr/dt=3$ m/min, $r=200$. Find $dA/dt$ for $A=\pi r^{2}$.

$dA/dt=2\pi r\,dr/dt=1200\pi$ m²/min.

$1200\pi$ m²/min.

(5)Balloon rises $15$ ft/s over $A$; $B$ is $30$ ft from $A$. Rate of change of distance from $B$ when height is $40$?

$D^{2}=30^{2}+h^{2}$; $D\,D'=h\,h'$ ⇒ $D'=12$ at $h=40$, $D=50$.

$12$ ft/s.

(6)On $y=x^{2}-2x$, where is $dy/dt=2\,dx/dt$?

$(2x-2)x'=2x'$ ⇒ $x=2$ (if $x'\neq 0$); point $(2,0)$.

$(2,0)$.

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Self-check

Quiz

Related rates review

1 / 5

Balloon r=12, dV/dt=−2: |dS/dt| equals…

Pick an answer to see if you are right.

Further Reading

• 20. Related Rates — method and sliding ladder
• 21. Differentials · Newton's Method — next: linear approx & Newton
• 19b. Motion — Practice — rates along a line
• 19. Rectilinear & Circular Motion — ds/dt language
• 16. Review of Trigonometry — similar triangles