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Introduction to Related Rates AB & BC

A related rates setup ties changing quantities together and differentiates the relationship with respect to $t$, so every varying letter picks up its own rate factor by the chain rule: $A = \pi r^2$ becomes $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$, and $V = \pi r^2 h$ with both dimensions moving needs the product rule and two factors. A constraint such as $r = h/3$ in a cone removes a variable you have no rate for, which is what makes the equation solvable.

Two failures dominate the setup. The chain rule goes missing, so the derivative is taken with respect to a length rather than time and $\frac{dr}{dt}$ never appears, or it is attached to one term and forgotten on another. And the model is wrong: the Pythagorean relation gets used where similar triangles are needed, a genuinely moving quantity is held fixed, or a constraint that links two dimensions is never used at all.

THE RELATIONSHIP, WITH TIME MADE VISIBLE A(t) = π [ r(t) ]² both letters move as t moves DIFFERENTIATE d/dt DIFFERENTIATE d/dr BY MISTAKE dA/dt = 2πr · dr/dt dA/dt = 2πr units: (length²/time) = (length)(length/time) units: (length²/time) = (length). No. dr/dt the missing factor is the derivative of the INSIDE function; a units check catches it
The same equation differentiated against two different variables. Only one of them answers a question about time.
r(t) h(t) full radius 4 full height 12 both FIXED similar triangles force r = h/3, at every instant GENUINELY CONSTANT the tank's own 4 and 12 a rigid ladder's length a lamppost's height these differentiate to 0 ONLY LOOKS FIXED the water's depth right now the slick's radius right now these carry a rate factor without the constraint V = (1/3)πr²h has two unknown rates and one equation; with it, one.
The tank's dimensions never change. The water's do, which is why only one pair of letters carries a rate.

The work

3 ways in · any order
Lesson
Introduction to Related Rates ›

Sets up related rates before any number is substituted: writing every varying quantity as a function of time, differentiating the relationship so each letter picks up its own rate factor, telling a rigid constant from a length that is merely known at one instant, and using a similar-triangle constraint to eliminate a variable.

Skill check · 10 scenarios
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Diagnostic
10-item topic check ›

Ten items spanning the two failure modes of this topic: differentiating with respect to a length instead of time so the rate factors never appear, and modeling the situation with the wrong relationship or holding a genuinely changing quantity fixed.

Not yet available · 10 items
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Targeted Practice
Drill a single misconception ›

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.

Take the diagnostic to identify your misconceptions
Diagnostic first