#Table of Contents

Introduction to Related Rates
Steps to Solve Related Rates Problems
Forming Equations for Related Rates
Using Implicit Differentiation
Additional Considerations
Practice Questions
Glossary
Summary and Key Takeaways

Key Concept

Related rates problems involve finding the rate at which one quantity changes with respect to time, given the rate at which another quantity changes with respect to time.

Chain Rule: The fundamental tool for solving related rates problems. $\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}$

**Example**: If the volume

V

of a balloon changes over time

t

, and the radius

r

of the balloon also changes over time, the chain rule helps relate

\frac{dV}{dt}

and

\frac{dr}{dt}

.

Form an equation linking the rates:
- Identify the rate you are trying to find.
- Determine the rates you know.
- Identify any rates you can calculate through differentiation.
Substitute known values to find the desired rate.

Identify the rate you are trying to find:
- Example: "The rate of change of volume with respect to time" is $\frac{dV}{dt}$ .
Identify known rates:
- Example: You may know the rate of change of height with respect to time, $\frac{dh}{dt}$ .
Identify rates you can calculate:
- Example: You can calculate $\frac{dV}{dh}$ using a formula that links volume with height.
List out all the rates:
- Example: $\frac{dV}{dt}, \frac{dh}{dt}, \frac{dV}{dh}$ .
Start with the rate you want to find and leave blank spaces for helpful rates: $\frac{dV}{dt} = \frac{dV}{} \times \frac{}{dt}$
Fill in the numerator and denominator: $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$

Exam Tip

Ensure that all terms cancel out except for the numerator and denominator you are looking for.

#Using Implicit Differentiation

Differentiate a known equation with respect to the variable of interest:
- Example: If $V = \frac{1}{3}\pi r^3$ , and you need $\frac{dV}{dt}$ : $\frac{dV}{dt} = \pi r^2 \frac{dr}{dt}$
Substitute known values to find the desired rate.

**Worked Example**: Given

V = \frac{1}{3}\pi r^3

and

\frac{dV}{dt} = 7

, find

\frac{dr}{dt}

when

r = 2

.

Identify the rate to find: $\frac{dr}{dt}$ .
Known rate: $\frac{dV}{dt} = 7$ .
Calculate $\frac{dV}{dr}$ : $\frac{dV}{dr} = \pi r^2$
Form the equation: $\frac{dr}{dt} = \frac{1}{\pi r^2} \times 7$
Substitute $r = 2$ : $\frac{dr}{dt} = \frac{7}{4\pi}$
Exact rate: $\frac{dr}{dt} = \frac{7}{4\pi}$ meters per minute.

#Additional Considerations

Multiple Terms: Related rates equations may involve more than three terms. $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{du} \times \frac{du}{dt}$
Differentiation Techniques:
- Chain rule
- Product rule
- Quotient rule
- Implicit differentiation
Constant vs. Changing Variables: Be cautious about which variables are constant and which are changing.
- Example: In a shrinking cone, both height and radius change.
Diagrams: Drawing diagrams can help visualize the problem, especially with similar shapes.

**Worked Example**: A 17-meter ladder slides down a wall. The top slides at 6 meters/second. Find the speed of the bottom when 15 meters from the wall.

Label rates: $\frac{dy}{dt} = -6$ (negative because it's downward).
Hypotenuse remains 17 meters.
Speed at the bottom: $\frac{dx}{dt}$ .

Using Pythagoras: $x^2 + y^2 = 17^2 \implies x^2 + y^2 = 289$

Differentiate implicitly: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \implies \frac{dx}{dt} = -\frac{y}{x} \times \frac{dy}{dt}$

Substitute $x = 15$ and solve for $y$ : $15^2 + y^2 = 289 \implies y = 8$

Substitute known values: $\frac{dx}{dt} = -\frac{8}{15} \times -6 = \frac{48}{15} = 3.2$

Thus, $\frac{dx}{dt} = 3.2$ meters/second.

#Practice Questions

A balloon is being inflated at a rate of 100 cubic centimeters per minute. How fast is the radius increasing when the radius is 10 cm?
A conical tank is being filled with water at a rate of 50 cubic meters per minute. The tank has a height of 10 meters and a base radius of 5 meters. How fast is the water level rising when the water is 2 meters deep?
A car is driving away from a streetlight at a rate of 20 meters per second. The streetlight is 5 meters tall. How fast is the tip of the car's shadow moving when the car is 10 meters away from the streetlight?

#Glossary

Chain Rule: A formula to compute the derivative of a composite function.
Implicit Differentiation: A method to differentiate equations not explicitly solved for one variable in terms of another.
Pythagoras' Theorem: A relation in a right-angled triangle: $a^2 + b^2 = c^2$ .

#Summary and Key Takeaways

Form Equations: Relate the rates of change using the chain rule.
Differentiate Implicitly: When dealing with equations involving multiple variables.
Substitute Known Values: To find the desired rate of change.
Draw Diagrams: To visualize the problem and identify relationships.

Key Concept

Understanding and applying related rates problems involve breaking down the problem, forming the right equations, and carefully differentiating and substituting values.

Exam Tip

Always check which variables are changing and which are constants, and use diagrams to help visualize complex problems.