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Solving Related Rates Problems AB & BC

Solving a related rates problem means differentiating the relationship with respect to $t$ and only then substituting the instant's values. Reversing those two steps freezes a moving quantity: a sphere whose radius grows at 3 cm/s, with $r$ replaced by 4 before differentiating, reports $\frac{dV}{dt} = 0$ instead of $192\pi \approx 603.2$ cubic centimeters per second. The missing length at the instant usually comes from the relationship itself, as $y = 5$ does in the 13-foot ladder at $x = 12$.

Three failures show up in the execution. The instant's numbers go in too early, freezing what was supposed to move. The model is wrong or a changing quantity is held fixed, as when a cone tank's surface radius is pinned at the tank's full radius. And the chain rule goes missing, so rate factors never appear. A fourth habit is worth naming: the shadow problem has two moving quantities, and the length of the shadow and the position of its tip are different answers to different questions.

A BALLOON WHOSE RADIUS GROWS AT 3 cm/s. FIND dV/dt WHEN r = 4. SUBSTITUTE FIRST DIFFERENTIATE FIRST V = (4/3)π(4)³ = 256π/3 V = (4/3)πr³ now it is a NUMBER, and dV/dt = 4πr² · dr/dt a number cannot change NOW put in r = 4 and dr/dt = 3 dV/dt = 0 dV/dt = 192π ≈ 603.2 cm³/s an inflating balloon, not inflating 4 is a snapshot of a moving radius A letter names something that MOVES. Replace it with a value and the motion is gone before you ever differentiate, which is why the wrong answer is always exactly zero.
The same problem, two orders of operations. Only the order on the right leaves the radius free to move.
13-FT LADDER, FOOT SLIDING OUT AT 2 ft/s. HOW FAST IS THE TOP FALLING WHEN x = 12? 1 LABEL x(t) out, y(t) up. 13 is fixed. 2 RATES given dx/dt = 2 wanted dy/dt 3 RELATE x² + y² = 169 4 d/dt 2x(dx/dt) + 2y(dy/dt) = 0 5 SUB NOW x = 12 ⇒ y = 5 from the SAME relation 2(12)(2) + 2(5)(dy/dt) = 0 6 ANSWER dy/dt = -4.8 ft/s: the top is FALLING y was never given: step 5 gets it from the relationship, which is where it usually hides.
Steps 4 and 5 in that order. The negative sign in the last line is the answer, not a slip to be cleaned up.

The work

3 ways in · any order
Lesson
Solving Related Rates Problems ›

Carries related rates through to an answer with the ladder, cone, and shadow problems worked end to end, shows exactly what substituting the instant's values before differentiating destroys, and closes on checking the units, the sign, and whether the rate found is the rate the question asked for.

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

Ten items spanning the three failure modes of this topic: substituting the instant's numbers before differentiating, modeling with the wrong relationship or freezing a quantity that moves, and losing the chain-rule factors that make the rates appear.

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