Related rates — setting up and solving: Question bank

Every item is a Question ::: Answer reveal. Cover the answer, decide, then check.

To keep the traps concrete, here are the two geometries the items keep returning to, drawn and labelled the way Step 1 of the recipe demands. Refer back to them whenever an item names $x, y$ or $h, r$ .

Figure — Related rates — setting up and solving

True or false — justify

If you plug in the instant's numbers before differentiating, you still get the right rate as long as you differentiate carefully afterwards.

False. Once a changing variable becomes a fixed number, its derivative is

0

, so differentiating afterwards silently drops the rate you were trying to find.

In the ladder problem, the top of the ladder moves down at a constant speed because the base moves out at a constant speed.

False.

\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}

depends on the ratio

x / y

, which changes as the ladder slides — so even a constant base speed gives an accelerating top.

For

A = π r^{2}

, the rate

\frac{d A}{d t}

is the same at every instant if

\frac{d r}{d t}

is constant.

False.

\frac{d A}{d t} = 2 π r \frac{d r}{d t}

grows with

r

: a bigger circle adds more rim per second, so the area rate keeps climbing even at fixed radial speed.

Related rates and implicit differentiation are essentially the same procedure.

True. Both differentiate a relation while treating variables as functions of a hidden parameter — implicit differentiation uses

x

, related rates uses time

t

. See Implicit differentiation.

A negative value of

\frac{d h}{d t}

in the cone problem would mean the tank is draining.

True. The sign of a rate is its direction; negative depth-rate means the water level is falling, so volume is decreasing at that instant.

The chain rule is optional in related rates — you could always solve for one variable in terms of

t

and differentiate directly.

False in practice. You almost never have a variable as an explicit function of

t

; the Chain rule is precisely the tool that lets you differentiate without that formula.

If two quantities are linked by an equation but neither is explicitly time-dependent in that equation, then neither changes with time.

False. The equation constrains their values, not their time-behaviour; both can be functions of

t

— that hidden time-dependence is the whole engine of related rates.

Spot the error

"The oil circle has

r = 10

, so

A = π (100) = 100 π

; therefore

\frac{d A}{d t} = 100 π

The student differentiated a number, not a rate.

100 π

is the area value; the rate needs

\frac{d A}{d t} = 2 π r \frac{d r}{d t}

, which uses the given

\frac{d r}{d t}

, not just

r

"Ladder:

x^{2} + y^{2} = 25

, differentiate to get

2 x + 2 y = 0

They forgot the

\frac{d x}{d t}

and

\frac{d y}{d t}

factors from the Chain rule. Correct:

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0

"Cone:

V = \frac{1}{3} π r^{2} h

, differentiate to

\frac{d V}{d t} = \frac{1}{3} π r^{2} \frac{d h}{d t}

They treated

r

as constant. Both

r

and

h

change, so either eliminate

r

first via Similar triangles, or apply the product rule to

r^{2} h

"The ladder's top rate came out negative, so I made it positive since speed can't be negative."

Dropping the sign destroys the physics. The negative sign is information: it says the top is moving downward. Speed is

∣ \frac{d y}{d t} ∣

, but the signed rate must keep its sign.

\frac{d}{d t} (x^{2}) = 2 x

Missing the inner rate. Since

x = x (t)

\frac{d}{d t} (x^{2}) = 2 x \frac{d x}{d t}

; writing just

2 x

pretends we're differentiating with respect to

x

, not

t

"For the shadow of a walking person: let

x

be the person's distance from a lamppost and

s

the shadow's tip distance, so the light, head and tip make two similar triangles. I set

x = 5

m first, then wrote the similar-triangle relation and differentiated."

Fixing the distance

x = 5

before building and differentiating the relation freezes the very variable whose rate

\frac{d x}{d t}

(the walking speed) matters — same deadly early-substitution error, just disguised.

\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}

, and since

\frac{d V}{d t} = 2

is constant,

\frac{d h}{d t}

is constant too."

\frac{d h}{d t} = \frac{8}{π h ^{2}}

shrinks as

h

grows — a constant inflow does not give a constant rise in a widening cone.

Why questions

Why do we differentiate the linking equation with respect to

t

rather than with respect to

x

r

Because we want time rates (

\frac{d}{d t}

). Time is the shared parameter every quantity depends on, so differentiating in

t

produces relations between the actual speeds we're asked about.

Why must we eliminate

r

in the cone problem before differentiating?

Keeping both

r

and

h

forces an unknown

\frac{d r}{d t}

into the equation. Using Similar triangles to write

r = \frac{h}{2}

first leaves only the rate we can solve for.

Why does the sliding-ladder top speed blow up to infinity as the ladder nears flat?

\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}

, the height

y \to 0

sits in the denominator, so the rate

\to - \infty

— a limiting behaviour, not a real infinite speed (the model breaks first).

Why is drawing a labelled picture Step 1, before any algebra?

The picture reveals which geometric law links the variables (Pythagoras, similar triangles, a volume formula) — you cannot pick a linking equation you haven't seen.

Why does the chain rule, not the product rule, get top billing in the parent's recipe?

Every single variable is a composite — a quantity of a quantity of time — so the Chain rule fires on each term; the product rule only shows up when two changing variables multiply.

Why can a related-rates answer be exactly

0

at some instant even though everything is moving?

A rate is

0

when the geometry is momentarily stationary in that coordinate — e.g. a point at the top of an arc has zero vertical rate while still moving horizontally.

Edge cases

What happens to

\frac{d A}{d t} = 2 π r \frac{d r}{d t}

at the very start, when

r = 0

It equals

0

: a point-sized spill has no rim to push out, so the area rate is momentarily zero even though the radius is already growing.

In the ladder problem, what is

\frac{d y}{d t}

at the instant the base is exactly at the wall (

x = 0

\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} = 0

: when the ladder is vertical the top is momentarily motionless before it begins to fall.

What does the cone formula

\frac{d h}{d t} = \frac{8}{π h ^{2}}

predict for depth-rate when the tank is nearly empty,

h \to 0^{+}

The rate

\to + \infty

— a tiny amount of water at the sharp apex fills a very thin layer very fast, so the level shoots up initially.

\frac{d r}{d t} = 0

(radius momentarily frozen) but the spill has area, is

\frac{d A}{d t} = 0

Yes:

\frac{d A}{d t} = 2 π r \cdot 0 = 0

. At an instant of zero radial speed the area is momentarily not changing, regardless of how large the circle is.

Can

\frac{d V}{d t}

be positive while

\frac{d h}{d t}

is negative in the cone?

No — for this cone

\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}

with

\frac{π h ^{2}}{4} > 0

for

h > 0

, so the two rates always share the same sign; they can only both vanish together (in the limit

h \to 0

What if the linking equation involves a constant (like the fixed ladder length

L

)? What is its time-derivative?

The derivative of any constant is

0

\frac{d}{d t} (L^{2}) = 0

. That vanishing is what produces the "

= 0

" on the right side of the ladder equation.

Recall One-line survival summary

Traps ::: keep variables alive until after differentiating, never drop a sign, and always let the geometry (not a number) supply the linking equation first.

Connections

Related rates — setting up and solving — the parent this bank stress-tests.
Chain rule — the engine every trap here circles back to.
Implicit differentiation — the twin technique in disguise.
Similar triangles — the elimination trick behind the cone traps.
Pythagorean theorem — the ladder's linking law.
Derivatives as rates of change — why signs mean directions.
Optimization — where stationary rates ( $= 0$ ) return with new meaning.