๐ What are Related Rates?
Related Rates problems in calculus involve finding the rate at which one quantity is changing by relating it to other quantities whose rates of change are known. They typically involve implicit differentiation and a careful understanding of how the variables are related.
- ๐ Definition: Related Rates problems explore how the rates of change of different variables are related to each other.
- ๐๏ธ History: The concept of related rates emerged alongside the development of calculus in the 17th century, pioneered by mathematicians like Isaac Newton and Gottfried Wilhelm Leibniz.
โ๏ธ Key Principles
Solving related rates problems involves a series of steps:
- โ๏ธ Read the problem carefully and identify the given rates and the rate you need to find.
- ๐ Draw a diagram if applicable and label the relevant quantities.
- โ๏ธ Write an equation that relates the variables.
- โ Differentiate both sides of the equation with respect to time ($t$). Remember to use the chain rule when differentiating variables that are functions of $t$.
- ๐ข Substitute the given values into the differentiated equation.
- โ
Solve for the unknown rate.
๐ Real-World Examples
Let's explore some common scenarios where related rates are applied:
๐ง Example 1: Filling a Conical Tank
Water is flowing into a conical tank at a rate of 8 cubic feet per minute. If the height of the tank is 12 feet and the radius of the base is 6 feet, how fast is the water level rising when the water is 4 feet deep?
Solution:
- ๐ Let $V$ be the volume of the water, $h$ be the height, and $r$ be the radius.
- ๐ We know $\frac{dV}{dt} = 8$ ft$^3$/min, and we want to find $\frac{dh}{dt}$ when $h = 4$ ft.
- ๐ The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. Since the radius and height are related ($\frac{r}{h} = \frac{6}{12}$), $r = \frac{1}{2}h$.
- โ Substituting, $V = \frac{1}{3}\pi (\frac{1}{2}h)^2 h = \frac{1}{12}\pi h^3$.
- โฑ๏ธ Differentiating with respect to $t$, $\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}$.
- โ Plugging in values, $8 = \frac{1}{4}\pi (4)^2 \frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{8}{4\pi} = \frac{2}{\pi}$ ft/min.
๐ Example 2: Inflating a Balloon
Air is being pumped into a spherical balloon at a rate of 4.5 cubic inches per minute. Find the rate of change of the radius when the radius is 2 inches.
Solution:
- ๐ Let $V$ be the volume of the sphere and $r$ be the radius.
- โ๏ธ We know $\frac{dV}{dt} = 4.5$ in$^3$/min, and we want to find $\frac{dr}{dt}$ when $r = 2$ in.
- ๐ The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
- โฑ๏ธ Differentiating with respect to $t$, $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
- โ Plugging in values, $4.5 = 4\pi (2)^2 \frac{dr}{dt}$, so $\frac{dr}{dt} = \frac{4.5}{16\pi} = \frac{9}{32\pi}$ in/min.
๐ช Tips for Success
- ๐ก Always draw a diagram if it helps visualize the problem.
- โ๏ธ Clearly identify what is given and what needs to be found.
- ๐ง Write down the relevant formulas.
- โ ๏ธ Use the chain rule correctly when differentiating.
- ๐งฎ Double-check your algebra and calculations.
๐ Conclusion
Related rates problems require a solid understanding of calculus concepts and careful attention to detail. By following the steps outlined above and practicing regularly, you can master this challenging topic. Good luck!