Why do we use $\frac{d}{dt}$ instead of $\frac{d}{dx}$?

In these problems, the variables are changing as time passes. We are interested in their speeds or rates of change per unit of time (like seconds or minutes), so we must take the derivative with respect to time, $t$.

Why did my answer come out negative?

A negative rate of change simply means that the quantity is decreasing over time. In the ladder example, the distance from the top of the ladder to the ground is shrinking, so its rate of change is negative.

How do I know which equation to use to relate the variables?

Look at the geometry of the situation you drew. Right triangles almost always use the Pythagorean theorem or trigonometric ratios. Spheres, cones, and cylinders will use their respective volume or surface area formulas. The problem context dictates the formula.

What is a related rates problem and how do you set one up?

A related rates problem asks you to figure out how fast one thing is changing based on how fast another connected thing is changing. Think about blowing air into a spherical balloon. You are adding air at a constant rate, so the volume increases steadily. But the balloon's radius grows quickly at first, then slows down as the balloon gets bigger. Both volume and radius are changing over time, and their rates of change are inextricably linked.

In calculus, we solve these problems by finding an equation that connects the variables, like the geometry formula for the volume of a sphere. Then, we use implicit differentiation with respect to time, $t$ , to create a new equation that links their rates of change. Once we have this derivative equation, we can plug in our known values and solve for the unknown rate.

What is a related rate?

In a related rates scenario, multiple variables are changing at the same time. You can think of them like gears in a machine: if you turn one gear at a certain speed, the connected gear must also turn, but its speed depends on the size of the gears. In math terms, we treat all the changing quantities as functions of time, $t$ . Their rates of change are expressed as derivatives with respect to time, such as $\frac{dx}{dt}$ or $\frac{dV}{dt}$ .

The setup strategy

Setting up a related rates problem is usually the hardest part. First, draw a picture and label the quantities that are changing with variables (like $x$ , $y$ , or $r$ ) and the quantities that are constant with their numerical values. Second, write down what you are given and what you need to find using derivative notation, like 'Given $\frac{dx}{dt} = 5$ , find $\frac{dy}{dt}$ when $x = 3$ '. Third, find an equation that connects the variables, such as the Pythagorean theorem or a volume formula. Finally, differentiate both sides of that equation with respect to time $t$ using the chain rule.

Where students slip up

The most common mistake is plugging in the specific snapshot-in-time values before taking the derivative. If you plug a number into your equation for a changing variable before you differentiate, that variable becomes a constant. The derivative of a constant is zero, which will completely ruin your rate equation. Always keep changing quantities as variables, take the derivative, and only plug in the specific numbers at the very end.

Worked through

A 10-foot ladder is leaning against a vertical wall. The bottom of the ladder is being pulled away from the wall at a rate of 2 feet per second. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall?

Let $x$ be the distance from the wall to the bottom of the ladder, and $y$ be the distance from the ground to the top of the ladder. We are given $\frac{dx}{dt} = 2$ ft/s and the ladder's length is constantly 10 ft. We want to find $\frac{dy}{dt}$ when $x = 6$ .

The equation connecting $x$ and $y$ is the Pythagorean theorem: $x^2 + y^2 = 10^2$ . Before differentiating, we can find the specific value of $y$ at this moment: $6^2 + y^2 = 100$ , which gives $y = 8$ .

Now, differentiate the main equation with respect to $t$ : $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ . Divide by 2 to simplify: $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ .

Plug in the snapshot values: $(6)(2) + (8) \frac{dy}{dt} = 0$ . This gives $12 + 8 \frac{dy}{dt} = 0$ , so $\frac{dy}{dt} = -\frac{12}{8} = -1.5$ . The top of the ladder is sliding down the wall at a rate of 1.5 feet per second.

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Where this comes from: OpenStax Calculus Volume 1 · Stewart Calculus, 8th Edition · Khan Academy: Applications of derivatives