Welcome to one of the most important application sections in Calculus I. So far, we've thought of derivatives as slopes of tangent lines. Today, we shift our perspective to think of derivatives as rates of change with respect to time. This allows us to answer powerful, dynamic questions about the world around us.
You have likely heard that this topic can be challenging. The goal of this lecture is to demystify it by giving you a clear, repeatable strategy that will work on every single problem. But first, let's explore the core idea.
Imagine dropping a stone into a calm pond. Ripples spread out in perfect circles. As the ripples expand, both the radius ($r$) and the area ($A$) of the outer ripple are changing over time.
(Click the play button on the slider in the graph below to see the animation.)
It's clear that the rate at which the area grows, $\frac{dA}{dt}$, depends on the rate at which the radius grows, $\frac{dr}{dt}$. These two rates are related. Why? Because the variables $A$ and $r$ are themselves related by a geometric formula that is always true: $A = \pi r^2$.
The Core Concept: In a related rates problem, several quantities are changing over time. We are given the rate of change of one quantity, and we use the relationship between the variables (the "static" equation) to find the rate of change of another.
The 6-step strategy we will learn is the "how," but a deeper understanding gives us the "why." Let's think of our static equation, $A = \pi r^2$, as a machine. You provide an input, $r$, and the machine produces an output, $A$.
The derivative of this equation with respect to its input variable, $\frac{dA}{dr} = 2\pi r$, acts like a "sensitivity factor" or a gear ratio for the machine. It tells us how much the output ($A$) will change for a small change in the input ($r$). Notice this gear ratio isn't fixed; it changes depending on the value of $r$. When the radius is large, the machine is highly sensitive, and a small change in $r$ results in a very large change in $A$.
This insight reveals that Related Rates is just a powerful application of the Chain Rule. The relationship is:
In our ripple example, this translates directly to:
This is the entire theory of related rates. It says: "The rate of change of the Area over time equals the machine's current gear ratio ($2\pi r$) multiplied by the rate of change of the Radius over time."
Our 6-step process is a way to construct and solve this Chain Rule relationship for any given scenario.
Every related rates problem can be solved by following the same core process. Memorize this strategy, practice it, and it will guide you to the solution every time.
Think of the "static" equation as a blueprint for the relationship between the variables for all time. The differentiation in Step 5 turns that blueprint into a new blueprint for the relationship between the rates, also for all time. You must create this "rate blueprint" first. Only then can you plug in the numbers from a single "snapshot in time" (like when $x=12$) to find the rate at that specific instant.
Let's apply our 6-step strategy to the most classic related rates problem. I will model the entire process from start to finish.
A 13-foot ladder is leaning against a vertical wall. The base of the ladder is sliding away from the wall at a constant rate of 5 ft/s. How fast is the top of the ladder sliding down the wall at the instant the base is 12 feet from the wall?
Step 1 & 2: Understand and Draw a Diagram.
We have a right triangle formed by the ladder, the wall, and the ground. The length of the ladder is constant (13 ft), but the distance from the base to the wall ($x$) and the height of the ladder on the wall ($y$) are changing.
(Click the play button on the slider in the graph below to see the animation.)
Let $x$ be the distance from the wall to the base of the ladder. Let $y$ be the height of the top of the ladder on the wall. The ladder itself is the hypotenuse, which is a constant 13.
Step 3: State the "Given" and "Find".
Given: The base is sliding away at 5 ft/s. This means $x$ is increasing. So, $\frac{dx}{dt} = +5$ ft/s.
Find: How fast the top is sliding down. This means we want the rate of change of $y$, which is $\frac{dy}{dt}$. We expect the answer to be negative since $y$ is decreasing. We need to find this at the specific instant when $x=12$ feet.
Step 4: The "Static" Equation.
The relationship between $x$, $y$, and 13 in our right triangle is given by the Pythagorean Theorem. This is true at any moment in time.
$$x^2 + y^2 = 13^2$$Step 5: Differentiate with Respect to Time.
Now we differentiate every term in the static equation with respect to $t$. Remember to use the chain rule for both $x$ and $y$. The derivative of the constant 169 is 0.
$$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(169)$$ $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$Step 6: Substitute and Solve.
We are interested in the instant when $x=12$. We know $\frac{dx}{dt}=5$. The equation has a $y$ in it, so we need to find the value of $y$ at this specific instant. We can use the static equation: $x^2 + y^2 = 13^2 \implies 12^2 + y^2 = 169 \implies 144 + y^2 = 169 \implies y^2 = 25 \implies y=5$.
Now we have all the pieces. Let's plug them into our differentiated equation:
$$2(12)(5) + 2(5)\frac{dy}{dt} = 0$$ $$120 + 10\frac{dy}{dt} = 0$$ $$10\frac{dy}{dt} = -120$$ $$\frac{dy}{dt} = -12 \text{ ft/s}$$The negative sign confirms our intuition: the ladder is sliding down the wall at a rate of 12 ft/s at that specific moment.
A common mistake is to plug in the value $x=12$ before differentiating. If you did that, the static equation would become $12^2 + y^2 = 13^2$. If you then differentiate this with respect to $t$, you get $0 + 2y\frac{dy}{dt} = 0$, which is wrong! You must differentiate the general relationship first, because $x$ is a variable, not a constant.
Now it's your turn to build the solution piece by piece. We will tackle a new problem using a series of "Check Your Understanding" prompts. Try each step on your own before revealing the solution.
Problem Statement: Air is pumped into a spherical balloon at a rate of 100 cm³/s. We want to find how fast the radius is increasing at the instant when the radius is 10 cm.
Read the problem, draw a labeled diagram of the sphere, and write down the "Given" and "Find" rates in calculus notation.
What is the "static" equation that relates the two variables of interest, the volume ($V$) and the radius ($r$) of a sphere?
Now, perform the key calculus step. Differentiate both sides of our static equation ($V = \frac{4}{3}\pi r^3$) with respect to time ($t$). Remember the Chain Rule!
You have the differentiated equation ($\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$) and all the "Given" information. Substitute the known values into the equation and solve for the unknown rate, $\frac{dr}{dt}$.
Sometimes, the initial static equation has too many variables. The key in these problems is to use the geometry of the situation to eliminate a variable before you differentiate.
Problem: A water tank is shaped like an inverted circular cone with a base radius of 5 m and a height of 10 m. Water is being drained from the tank at a rate of 2 m³/min. How fast is the water level falling when the water is 6 m deep?
Following our steps: The diagram is an inverted cone.
(Click the play button on the slider in the graph below to see the animation.)
Let $h$ be the height of the water and $r$ be the radius of the water's surface at time $t$. We are given $\frac{dV}{dt} = -2$ (negative because it's draining) and we want to find $\frac{dh}{dt}$ when $h=6$.
Step 4 (Static Equation): The volume of a cone is $V = \frac{1}{3}\pi r^2 h$.
The "Aha!" Moment: This equation has three variables ($V$, $r$, $h$). Differentiating this would be messy. We want an equation relating just $V$ and $h$. We can do this by finding a relationship between $r$ and $h$ using similar triangles from a cross-section of the cone.
The ratio of the radius to the height must always be the same for the water and for the tank itself:
$$\frac{r}{h} = \frac{\text{tank radius}}{\text{tank height}} = \frac{5}{10} = \frac{1}{2}$$This gives us a simple relationship: $r = \frac{h}{2}$.
The Hidden Step (Simplify the Static Equation): Now, we substitute this into our volume formula before differentiating.
$$V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{4}\right) h = \frac{\pi}{12}h^3$$Step 5 (Differentiate): This is much easier to differentiate with respect to $t$.
$$\frac{dV}{dt} = \frac{\pi}{12} (3h^2 \frac{dh}{dt}) = \frac{\pi}{4}h^2 \frac{dh}{dt}$$Step 6 (Substitute and Solve): Now plug in $\frac{dV}{dt}=-2$ and $h=6$.
$$-2 = \frac{\pi}{4}(6)^2 \frac{dh}{dt}$$ $$-2 = \frac{\pi}{4}(36) \frac{dh}{dt}$$ $$-2 = 9\pi \frac{dh}{dt}$$ $$\frac{dh}{dt} = -\frac{2}{9\pi} \text{ m/min}$$The water level is falling at a rate of $\frac{2}{9\pi}$ m/min.
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 2 ft. If the trough is being filled with water at a rate of 3 ft³/min, how fast is the water level rising when the water is 1 ft deep?
A searchlight on the ground tracks a rocket that takes off vertically. The searchlight is located 2000 feet from the launchpad. When the rocket is 2000 feet high, its velocity is 500 ft/s. How fast is the searchlight's angle of elevation changing at that instant?
The cost $C$ (in dollars) to produce $x$ widgets is given by $C = 5000 + 10x + 0.05x^2$. If the production rate is increasing by 100 widgets per week, find the rate at which the production cost is increasing when 2000 widgets have been produced.
Related rates problems test your ability to connect geometry, trigonometry, and calculus. They seem difficult because they are not simple formula plug-ins. However, every problem can be approached with the same systematic strategy. The calculus (Step 5) is usually straightforward; the real challenge is setting up the problem correctly (Steps 2-4). Practice is the key to mastering this setup process.
After this lecture, you should be able to: