Differentiate first. Substitute last AB & BC

The procedure, in the order it has to happen:

1. Draw and label. Every changing length gets a letter.
2. Write down the given rate and the wanted rate, with signs.
3. Find the relationship, and use any constraint to eliminate variables.
4. Differentiate both sides with respect to $t$.
5. Now substitute the instant's values, and solve.
6. Answer the question that was asked, with units and a sign.

Steps 4 and 5 are not interchangeable. A variable is a name for something that moves; a number is not. Substituting first converts a moving quantity into a frozen one, and the derivative then reports, correctly, that a frozen quantity is not changing.

Take a sphere with radius growing at 3 centimeters per second, and ask for $\frac{dV}{dt}$ when $r = 4$. Substituting first:

$$V = \frac{4}{3}\pi (4)^{3} = \frac{256\pi}{3} \;\Rightarrow\; \frac{dV}{dt} = 0.$$

A balloon inflating at 3 centimeters of radius per second is reported as not changing volume at all. Differentiating first:

$$\frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt} = 4\pi (4)^{2}(3) = 192\pi \approx 603.2 \text{ cm}^{3}\text{/s}.$$

The 4 goes in at the end, into the derivative, where it is a snapshot of a moving radius rather than a replacement for it.

A 13-foot ladder leans against a vertical wall. Its foot slides away from the wall at 2 feet per second. How fast is the top sliding down when the foot is 12 feet from the wall?

1. Label. $x$ is the distance from the wall to the foot, $y$ the height of the top. Both change; 13 does not.
2. Rates. Given $\frac{dx}{dt} = 2$. Wanted $\frac{dy}{dt}$ when $x = 12$.
3. Relationship. $x^{2} + y^{2} = 169$.
4. Differentiate. $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.
5. Substitute. At $x = 12$, the relationship gives $y = 5$. Then $2(12)(2) + 2(5)\frac{dy}{dt} = 0$, so $\frac{dy}{dt} = -\frac{48}{10} = -4.8$.
6. Answer. The top is sliding down at 4.8 feet per second.

Two details worth noticing. The value $y = 5$ was not given; it came from the relationship at that instant, which is normally where the missing length comes from. And the answer is negative, which is the model telling you the height is decreasing. Reporting it as a positive 4.8 feet per second downward is fine; reporting $\frac{dy}{dt} = +4.8$ is not.

Water pours into a cone-shaped tank, point down, with full radius 4 feet and full height 12 feet, at 8 cubic feet per minute. How fast is the depth rising when the water is 6 feet deep?

The volume formula holds two variables and only one rate is known, so the constraint comes first. Similar triangles give $\frac{r}{h} = \frac{1}{3}$, so

$$V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi\left(\frac{h}{3}\right)^{2}h = \frac{\pi h^{3}}{27}.$$

Differentiating with respect to $t$ and then substituting:

$$\frac{dV}{dt} = \frac{\pi h^{2}}{9}\frac{dh}{dt} \;\Rightarrow\; 8 = \frac{\pi (36)}{9}\frac{dh}{dt} = 4\pi \frac{dh}{dt} \;\Rightarrow\; \frac{dh}{dt} = \frac{2}{\pi} \approx 0.637 \text{ ft/min}.$$

Three ways this one gets damaged, all fatal:

1. Holding $r$ at 4. That is the tank's radius, not the water's surface radius, and the water's radius grows as it fills.
2. Substituting $r = 2$ first, its value at depth 6. Same error as the sphere: a moving length replaced by a snapshot.
3. Dropping the $\frac{1}{3}$, which is a cylinder's formula wearing a cone's problem.

A correct derivative solved for the wrong quantity earns nothing, and this happens most often in the shadow problem, where two different things are moving.

A 6-foot person walks away from a 15-foot lamppost at 4 feet per second. Similar triangles give $\frac{15}{x + s} = \frac{6}{s}$, which simplifies to $9s = 6x$, so $\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt} = \frac{8}{3}$ feet per second.

1. The length of the shadow grows at $\frac{8}{3} \approx 2.67$ feet per second.
2. The tip of the shadow moves at $\frac{d}{dt}(x + s) = 4 + \frac{8}{3} = \frac{20}{3} \approx 6.67$ feet per second, since the tip moves away from the post, not from the person.

Both are correct answers to different questions, and the problem asks only one of them. Reread the last sentence of the prompt before writing the number down.

Finish with the same three checks every time: units from the quantity per the time, sign matching the physical direction, and the quantity matching what was asked.