Mastering Related Rates: A Step-by-Step Guide to Calculus Problems

Related rates problems are a classic application of differential calculus. They involve finding the rate at which one quantity is changing by relating it to other quantities whose rates of change are known. These problems often seem daunting at first, but by following a structured approach, you can solve them systematically and confidently. This comprehensive guide will walk you through the steps, provide detailed explanations, and illustrate with examples.

Understanding Related Rates

The core idea behind related rates is the chain rule of differentiation. If we have several variables that are functions of time (t), and these variables are related by an equation, we can differentiate both sides of the equation with respect to time to find the relationship between their rates of change.

For instance, if `x` and `y` are both functions of time `t`, and they are related by the equation `x^2 + y^2 = r^2` (where `r` is a constant), then differentiating both sides with respect to `t` gives:

`2x (dx/dt) + 2y (dy/dt) = 0`

This equation relates the rates of change of `x` and `y` with respect to time, `dx/dt` and `dy/dt`, respectively.

The Problem-Solving Strategy: A Step-by-Step Guide

To effectively tackle related rates problems, follow these steps:

Step 1: Read and Understand the Problem

This is perhaps the most crucial step. Carefully read the problem statement, several times if necessary, to ensure you fully understand what’s being asked. Identify the following:

* **The variables:** What quantities are changing with time? Assign variables to these quantities (e.g., `x`, `y`, `r`, `V`, `A`).
* **The given rates:** What rates of change are provided? Express these as derivatives with respect to time (e.g., `dx/dt = 5 cm/s`). Be mindful of units.
* **The unknown rate:** What rate of change are you asked to find? Express this as a derivative with respect to time (e.g., `dV/dt = ?`).
* **The specific instant:** At what specific time or value of the variables are you supposed to find the rate? This is crucial information to use *after* differentiating.
* **Draw a Diagram (if applicable):** Many related rates problems involve geometric figures. Drawing a diagram can help you visualize the problem and identify relationships between the variables. Label the diagram with the variables you’ve defined.

Example: A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

* Variables: `x` (distance of the bottom of the ladder from the wall), `y` (distance of the top of the ladder from the ground).
* Given rate: `dx/dt = 1 ft/s`.
* Unknown rate: `dy/dt = ?`
* Specific instant: `x = 6 ft`
* Diagram: Draw a right triangle with the ladder as the hypotenuse, the wall as one leg (y), and the ground as the other leg (x).

Step 2: Find an Equation Relating the Variables

The next step is to find an equation that relates the variables identified in Step 1. This equation should express the relationship between the quantities whose rates of change are involved in the problem. This equation is typically derived from:

* **Geometric formulas:** Area of a circle (`A = πr^2`), volume of a sphere (`V = (4/3)πr^3`), volume of a cone (`V = (1/3)πr^2h`), Pythagorean theorem (`a^2 + b^2 = c^2`).
* **Trigonometric relationships:** `sin(θ) = opposite/hypotenuse`, `cos(θ) = adjacent/hypotenuse`, `tan(θ) = opposite/adjacent`.
* **Similar triangles:** If the problem involves similar triangles, use the proportionality of corresponding sides.
* **Other relationships:** The problem statement might provide a specific relationship between the variables.

**Important:** Ensure the equation involves only the variables whose rates of change are relevant to the problem. If other variables are present, try to express them in terms of the relevant variables using other given information or relationships.

In the ladder example, the relationship between `x` and `y` is given by the Pythagorean theorem:

`x^2 + y^2 = 10^2 (x^2 + y^2 = 100)`

Step 3: Differentiate Both Sides of the Equation with Respect to Time (t)

This is where the calculus comes in. Differentiate both sides of the equation you found in Step 2 with respect to time `t`. Remember to use the chain rule whenever you differentiate a variable that is a function of `t`.

For example, if you have the term `x^2`, its derivative with respect to `t` is `2x (dx/dt)`. If you have `y^3`, its derivative is `3y^2 (dy/dt)`. If you have a product of two functions of `t`, like `xy`, use the product rule: `d(xy)/dt = x(dy/dt) + y(dx/dt)`. If you have a quotient, use the quotient rule.

Differentiating both sides of `x^2 + y^2 = 100` with respect to `t`, we get:

`2x (dx/dt) + 2y (dy/dt) = 0`

Step 4: Substitute Known Values

After differentiating, substitute the values of the variables and their rates of change that are known at the *specific instant* in question. This includes the given rates and any other information about the variables at that particular moment. Do *not* substitute values for variables *before* differentiating, as this will treat them as constants and lead to an incorrect result.

In the ladder example, we know that `dx/dt = 1 ft/s` and `x = 6 ft`. We need to find `y` at that instant. Using the Pythagorean theorem:

`6^2 + y^2 = 100`
`36 + y^2 = 100`
`y^2 = 64`
`y = 8 ft`

Now, substitute `x = 6`, `y = 8`, and `dx/dt = 1` into the differentiated equation:

`2(6)(1) + 2(8)(dy/dt) = 0`

Step 5: Solve for the Unknown Rate

Finally, solve the equation obtained in Step 4 for the unknown rate of change you are trying to find. This will give you the value of the derivative you’re looking for at the specific instant.

In the ladder example, solving for `dy/dt`:

`12 + 16(dy/dt) = 0`
`16(dy/dt) = -12`
`dy/dt = -12/16`
`dy/dt = -3/4 ft/s`

Step 6: Interpret the Result and Include Units

The final step is to interpret the result and express it with the correct units. The sign of the rate indicates whether the quantity is increasing or decreasing. A positive rate means the quantity is increasing, and a negative rate means it is decreasing.

In the ladder example, `dy/dt = -3/4 ft/s`. This means that the top of the ladder is sliding down the wall at a rate of 3/4 ft/s. The negative sign indicates that the distance `y` is decreasing.

Example Problems with Detailed Solutions

Let’s work through some more examples to solidify your understanding of related rates.

Example 1: Inflating a Balloon

A spherical balloon is being inflated. If air is pumped into the balloon at a rate of 100 cubic centimeters per second, how fast is the radius of the balloon increasing when the diameter is 50 centimeters?

1. **Understand the problem:**
* Variables: `V` (volume of the balloon), `r` (radius of the balloon).
* Given rate: `dV/dt = 100 cm^3/s`
* Unknown rate: `dr/dt = ?`
* Specific instant: diameter is 50 cm, so `r = 25 cm`
2. **Find an equation:**
* The volume of a sphere is `V = (4/3)πr^3`
3. **Differentiate with respect to time:**
* `dV/dt = (4/3)π * 3r^2 (dr/dt)`
* `dV/dt = 4πr^2 (dr/dt)`
4. **Substitute known values:**
* `100 = 4π(25)^2 (dr/dt)`
* `100 = 2500π (dr/dt)`
5. **Solve for the unknown rate:**
* `dr/dt = 100 / (2500π)`
* `dr/dt = 1 / (25π) cm/s`
6. **Interpret the result:**
* The radius of the balloon is increasing at a rate of `1 / (25π) cm/s` when the diameter is 50 cm. This is approximately 0.0127 cm/s.

Example 2: Approaching an Intersection

Car A is traveling west at 50 mi/h, and car B is traveling north at 60 mi/h. Both are headed for the intersection of two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?

1. **Understand the problem:**
* Variables: `x` (distance of car A from the intersection), `y` (distance of car B from the intersection), `z` (distance between the cars).
* Given rates: `dx/dt = -50 mi/h` (negative because the distance is decreasing), `dy/dt = -60 mi/h` (negative for the same reason).
* Unknown rate: `dz/dt = ?`
* Specific instant: `x = 0.3 mi`, `y = 0.4 mi`
2. **Find an equation:**
* By the Pythagorean theorem: `x^2 + y^2 = z^2`
3. **Differentiate with respect to time:**
* `2x (dx/dt) + 2y (dy/dt) = 2z (dz/dt)`
* `x (dx/dt) + y (dy/dt) = z (dz/dt)`
4. **Substitute known values:**
* First, find `z` when `x = 0.3` and `y = 0.4`:
* `z^2 = (0.3)^2 + (0.4)^2 = 0.09 + 0.16 = 0.25`
* `z = 0.5 mi`
* Now substitute: `(0.3)(-50) + (0.4)(-60) = (0.5)(dz/dt)`
* `-15 – 24 = 0.5(dz/dt)`
* `-39 = 0.5(dz/dt)`
5. **Solve for the unknown rate:**
* `dz/dt = -39 / 0.5`
* `dz/dt = -78 mi/h`
6. **Interpret the result:**
* The cars are approaching each other at a rate of 78 mi/h. The negative sign indicates that the distance between them is decreasing.

Example 3: Filling a Conical Tank

Water is leaking out of an inverted conical tank at a rate of 10,000 cm³/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

1. **Understand the problem:**
* Variables: `V` (volume of water in the tank), `h` (height of water in the tank), `r` (radius of water surface in the tank).
* Given rates: `dV_out/dt = 10000 cm^3/min`, `dh/dt = 20 cm/min` when `h = 2m = 200cm`
* Unknown rate: `dV_in/dt = ?` (the rate at which water is being pumped in)
* Specific instant: `h = 200 cm`
2. **Find an equation:**
* Volume of a cone: `V = (1/3)πr^2h`
* We need to relate `r` and `h` because we only know `dh/dt`. Use similar triangles.
* The tank has height 6m = 600cm and radius 2m = 200cm (half of the diameter).
* So, `r/h = 200/600 = 1/3`, thus `r = h/3`
* Substitute `r = h/3` into the volume equation: `V = (1/3)π(h/3)^2h = (1/3)π(h^2/9)h = (π/27)h^3`
3. **Differentiate with respect to time:**
* `dV/dt = (π/27) * 3h^2 (dh/dt) = (π/9)h^2 (dh/dt)`
* Note: `dV/dt = dV_in/dt – dV_out/dt` , the net rate of change of volume is the inflow rate minus the outflow rate.
4. **Substitute known values:**
* `h = 200 cm`, `dh/dt = 20 cm/min`, `dV_out/dt = 10000 cm^3/min`
* `dV/dt = (π/9)(200)^2(20) = (π/9)(40000)(20) = 800000π/9 cm^3/min`
* Therefore, `dV_in/dt – 10000 = 800000π/9`
5. **Solve for the unknown rate:**
* `dV_in/dt = 800000π/9 + 10000`
* `dV_in/dt = (800000π + 90000)/9 cm^3/min`
* `dV_in/dt ≈ 288176 cm^3/min`
6. **Interpret the result:**
* Water is being pumped into the tank at a rate of approximately 288176 cm³/min.

Tips and Common Mistakes

* **Units:** Pay close attention to units. Ensure all quantities are expressed in consistent units before substituting them into equations. If rates are given in different units (e.g., miles per hour and feet per second), convert them to the same units.
* **Constants vs. Variables:** Distinguish between constants and variables. A constant remains the same throughout the problem, while a variable changes with time. Substitute values for constants only *after* differentiation.
* **Drawing Diagrams:** If the problem involves geometry, draw a clear diagram and label it with variables. This will help you visualize the relationships between the quantities.
* **Chain Rule:** Remember to apply the chain rule correctly when differentiating variables that are functions of time. If `y` is a function of `t`, then the derivative of `y^n` with respect to `t` is `n*y^(n-1)*(dy/dt)`.
* **Negative Rates:** A negative rate indicates that the quantity is decreasing. Be careful to interpret the sign of your answer correctly.
* **Read the Problem Carefully:** The most common mistake is misinterpreting the problem statement. Take your time to read the problem carefully and identify the given rates, the unknown rate, and the specific instant.
* **Similar Triangles:** When dealing with cones or other shapes where the dimensions change proportionally, look for similar triangles to relate the variables.
* **Check Your Work**: Always double-check your algebra and calculus. A small mistake can lead to a completely wrong answer.

Practice Problems

To truly master related rates, practice is essential. Here are some practice problems to test your understanding:

1. A hot air balloon is rising vertically. An observer stands on the ground 100 meters away from the point where the balloon was released. If the balloon is rising at a rate of 3 meters per second, how fast is the distance between the observer and the balloon increasing when the balloon is 50 meters high?
2. A trough is 10 feet long and its ends have the shape of isosceles triangles that are 3 feet across at the top and have a height of 1 foot. If the trough is being filled with water at a rate of 12 cubic feet per minute, how fast is the water level rising when the water has a depth of 6 inches?
3. A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock?

Conclusion

Related rates problems can be challenging, but with a systematic approach and plenty of practice, you can master them. Remember to carefully read the problem, identify the variables and rates, find an equation relating the variables, differentiate with respect to time, substitute known values, and solve for the unknown rate. By following these steps and avoiding common mistakes, you’ll be well on your way to success in calculus!