AP Calculus AB and BC

Related Rates Scene: Sliding Ladder and Cone Tank

Related rates link two changing quantities with a geometric relation, then differentiate to connect their rates. The trick is order: differentiate while the quantities are still variables, and plug in the current numbers only at the end. A sliding ladder and a filling cone tank recompute the unknown rate as you drag.

Drag the amber handle or focus it and use the arrow keys (Shift for larger steps).

x (ft)

6.0

y = sqrt(100 - x^2)

8.000

dy/dt (ft/s)

-1.5000

Relation

x^2 + y^2 = 10^2 = 100

Differentiate the relation with respect to t

2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0

Given rate

\frac{dx}{dt} = +2 \text{ ft/s}

Solve for the unknown rate

\frac{dy}{dt} = -\frac{2x}{2y}\,\frac{dx}{dt} = -\frac{x}{y}\,\frac{dx}{dt}

Substitute the current state

\frac{dy}{dt} = -\frac{2(6)(2)}{2(8)} = -1.5

Differentiate the relation before you substitute. The relation holds for every instant, so its derivative in t links the rates. If you plug the current numbers in first, x becomes a fixed constant, its derivative is zero, and the equation can no longer solve for how fast the top slides down the wall.

The drag stops at x = 9. Since dy/dt = -2x / sqrt(100 - x^2) and the height y = sqrt(100 - x^2) heads to 0 as x approaches 10, the speed |dy/dt| grows without bound near the wall. A ladder problem that asks for the rate at the instant the ladder is flat has no finite answer, an AP favorite.

Every related rates problem starts with a relation, an equation that ties the changing quantities together and holds at every instant. Because it holds for all time, you can differentiate both sides with respect to $t$ . The chain rule pulls a rate out of each quantity, and the single equation you get links all of those rates at once. This is implicit differentiation with time as the hidden variable (Topic 3.1, The Chain Rule and Topic 3.2, Implicit Differentiation, plus Topic 4.4, Introduction to Related Rates).

x^2 + y^2 = 10^2 = 100

Differentiating with respect to $t$ brings a $\frac{dx}{dt}$ and a $\frac{dy}{dt}$ down through the chain rule, one term per changing length:

2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0

The order is the whole game

Notice that $x$ and $y$ are still symbols when you differentiate. If you substitute the current numbers first, say $x = 6$ and $y = 8$ , then $x$ is a fixed constant, $\frac{dx}{dt}$ is zero, and the equation collapses to $0 = 0$ with nothing left to solve. Differentiate the relation while the quantities are variables, and plug in the state only at the end. The tool freezes the top four derivation rows for exactly this reason: the calculus is done before any number appears, and only the final substitution line moves when you drag.

How to read it

The scene puts the picture and its live derivation side by side. Drag the amber handle to slide the ladder's base along the floor, or switch to the cone tank and drag the water surface up and down, and the derivation's final line recomputes the unknown rate at that exact state. Everything above that line, the relation, its derivative, the given rate, and the solved formula, stays frozen, because all of the calculus is finished before any number is plugged in.

A 10-foot ladder leans against a wall while its base slides away at $\frac{dx}{dt} = +2$ ft/s. The base distance $x$ and the wall height $y$ satisfy $x^2 + y^2 = 100$ , so $y = \sqrt{100 - x^2}$ . The question asks for $\frac{dy}{dt}$ , the speed at which the top slides down. The scene walks the same five steps you use on any related rates free-response prompt:

Name the quantities and write the relation that ties them together.
Record the given rate and the rate you are solving for.
Differentiate the relation with respect to $t$ , keeping every quantity a variable.
Substitute the known rate and the current values of the quantities.
Solve for the unknown rate and attach its units and sign.

Differentiating and solving the rate as a formula in $x$ and $y$ gives:

\frac{dy}{dt} = -\frac{x}{y}\,\frac{dx}{dt} = -\frac{2x}{y}

Only now does the current state go in. At $x = 6$ the height is exactly $y = \sqrt{100 - 36} = 8$ , so:

\frac{dy}{dt} = -\frac{2(6)}{8} = -\frac{3}{2} = -1.5 \text{ ft/s}

The sign is information, not decoration: $\frac{dy}{dt} < 0$ says the top is moving down. And the answer depends on where the ladder is, so sliding the base recomputes the last line while the four rows above it stay put.

Reading the ladder as you drag

$\frac{dy}{dt}$ is negative at every position (the top slides down) and grows in magnitude as the base moves out, from about $-0.2$ ft/s when the base is near the wall to about $-4.1$ ft/s at $x = 9$ . At $x = 6$ the height is exactly $y = 8$ , so the line reads an exact $\frac{dy}{dt} = -1.5$ ft/s with an equals sign; at most other positions $y$ is an irrational square root and the readout switches to the approximately-equal sign.

The cone scene fills an inverted cone (height 10, top radius 5) at $\frac{dV}{dt} = +3$ cubic units per second and asks how fast the depth $h$ rises. A cone's volume is $V = \frac{1}{3}\pi r^2 h$ , which carries two changing lengths, $r$ and $h$ . Similar triangles pin them together: the water forms a smaller cone with the same proportions as the tank, so $\frac{r}{h} = \frac{5}{10}$ , that is $r = \frac{h}{2}$ .

That constraint holds at every instant, so you may substitute it before differentiating. Doing so leaves the volume in a single variable, which is what makes the derivative clean:

V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2 h = \frac{\pi}{12}h^3

Differentiate with respect to $t$ and solve for the unknown rate:

\frac{dV}{dt} = \frac{\pi}{4}h^2\,\frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = \frac{12}{\pi h^2}

At $h = 2$ this gives $\frac{dh}{dt} = \frac{12}{4\pi} = \frac{3}{\pi} \approx 0.955$ units per second. Because $h^2$ sits in the denominator, the level climbs quickly when the cone is narrow near the tip and slows as the tank widens, even though $\frac{dV}{dt}$ never changes. Drag the surface up and watch $\frac{dh}{dt}$ fall.

Substituting the constraint is not the same as substituting the state

It is legitimate to plug $r = \frac{h}{2}$ in early because that relation is true for all time. What you must not do is plug in the instantaneous value $h = 2$ before differentiating, because that particular number is only true at one instant. The rule is about the changing state variable, not about relations that hold forever.

Reading the cone as you drag

$\frac{dh}{dt} = \frac{12}{\pi h^2}$ shrinks as $h$ rises: the surface climbs fast when the tank is narrow near the tip and slows as it widens, even though the fill rate $\frac{dV}{dt}$ stays fixed at $+3$ . Every cone value carries the approximately-equal sign because the answer always contains $\pi$ . The radius readout stays locked at $r = \frac{h}{2}$ for every depth, the similar-triangles constraint that lets you rewrite the volume in one variable before differentiating.

Slide the ladder base toward the wall's far limit and $\frac{dy}{dt}$ races off the scale. The formula shows why: with $\frac{dx}{dt} = 2$ , the rate is $\frac{dy}{dt} = \frac{-2x}{\sqrt{100 - x^2}}$ , and as $x \to 10$ the height $y = \sqrt{100 - x^2}$ heads to $0$ , so the denominator vanishes and the magnitude of $\frac{dy}{dt}$ grows without bound.

\lim_{x \to 10^-} \frac{-2x}{\sqrt{100 - x^2}} = -\infty

The drag stops at $x = 9$ on purpose. A classic exam question asks for the rate at the instant the ladder lies flat, and the honest answer is that no finite rate exists there. Physically the top would have to move infinitely fast, the tell that the modeling assumption of a constant $\frac{dx}{dt}$ has broken down at the wall. Solving Related Rates Problems (Topic 4.5) rewards recognizing that the geometry, not the algebra, is what makes an instant impossible.

As you drag, only the bottom 'Substitute the current state' line changes; the relation, its derivative, the given rate, and the solved formula stay fixed, because all of the calculus is finished before any number is plugged in.
On the ladder, $\frac{dy}{dt}$ is negative at every position (the top slides down) and grows in magnitude as the base moves out, from about $-0.2$ ft/s when the base is near the wall to about $-4.1$ ft/s at $x = 9$ .
At $x = 6$ the height is exactly $y = 8$ , so the line reads an exact $\frac{dy}{dt} = -1.5$ ft/s with an equals sign; at most other positions $y$ is an irrational square root and the readout switches to the approximately-equal sign.
In the cone, $\frac{dh}{dt} = \frac{12}{\pi h^2}$ shrinks as $h$ rises: the surface climbs fast when the tank is narrow near the tip and slows as it widens, even though the fill rate $\frac{dV}{dt}$ stays fixed at $+3$ . Every cone value carries the approximately-equal sign because the answer always contains $\pi$ .
The radius readout stays locked at $r = \frac{h}{2}$ for every depth, the similar-triangles constraint that lets you rewrite the volume in one variable before differentiating.
The given rate is constant, but the unknown rate is a function of the current state: the same setup gives a different answer at every position, which is exactly why the substitution has to come last.