Calculus - Related Rates

Calculus

Related Rates

​Related Rates Problems

A "Related Rates" problem is a situation where the question, and its answer, involve relating two or more derivatives to one another. To solve them we must assess a scenario and determine a function which can describe the situation; we must then differentiate that function and use known quantities to find an instantaneous rate of change. This rate of change is our solution.

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Example 1

A (Somewhat) Practical Example

Finding Key Information

After initially reading the problem, read it a second time to determine what information will be useful in finding the solution. Highlight anything that may be important.

Multiple Select

Of the given values in Example 1, which one represents rates of change?

• $$6.96\times10^5\ \text{km}$$

• $$1.40\times10^{17}\ \frac{\text{km}^3}{\text{year}}$$

• Neither

Sketch

After you know what information is at your disposal, you should try to sketch the situation to the best of your ability. It doesn't need to be pretty -- it just needs to serve as a visual aide to understand the situation better.

Draw

Using the information given in Example 1, draw a rough sketch of the situation. You should label the sketch with all of the quantities that you know, and the quantity that you're trying to find.

Example 1 (Reference)

Suppose that a certain star with a radius of $$6.96\text{ }\times10^5\ \text{km}\$$ is expanding. The volume of the star is increasing by $$1.4\text{ }\times10^{17}\ \frac{\text{km}^3}{\text{year}}$$ . At what rate is the radius of the star increasing?

Relating the Known Quantities

Multiple Choice

Calculate the derivative of $$V=\frac{4}{3}\pi\ \cdot r^3$$ . Assume both $$V$$ and $$r$$ are implicit functions of $$t$$ .

• $$\frac{\text{d}\ V}{\text{d}\ t}=4\pi\cdot r^2\cdot\frac{\text{d}\ r}{\text{d}\ t}$$

• $$\frac{\text{d}\ V}{\text{d}\ t}=4\pi\cdot r\cdot\frac{\text{d}\ r}{\text{d}\ t}$$

• $$\frac{\text{d}\ V}{\text{d}\ t}=4\pi\cdot r$$

• $$\frac{\text{d}\ V}{\text{d}\ t}=\frac{4}{6}\pi\cdot r$$

Wrapping Up

Multiple Choice

Given $$\frac{\text{d}\ V}{\text{d}\ t}=4\pi\cdot r^2\cdot\frac{\text{d}\ r}{\text{d}\ t}$$ , solve for $$\frac{\text{d}\ r}{\text{d}\ t}$$

• $$\frac{\text{d}\ r}{\text{d}\ t}=4\pi\cdot r^2\cdot\frac{\text{d}\ V}{\text{d}\ t}$$

• $$\frac{\text{d}\ r}{\text{d}\ t}=\frac{4\pi\cdot r^2}{\frac{\text{d}\ V}{\text{d}\ t}}$$

• $$\frac{\text{d}\ r}{\text{d}\ t}=\frac{1}{4\pi\cdot r^2}\cdot\frac{\text{d}\ V}{\text{d}\ t}$$

• None of the Above

Solve for

Multiple Choice

Evaluate $$\frac{\text{d}\ r}{\text{d}\ t}$$ using the known quantities.

• $$22,998.5\ \frac{\text{km}^3}{\text{year}}$$

• $$22,998.5\ \frac{\text{km}}{\text{year}}$$

• $$220,998.5\ \frac{\text{km}}{\text{year}}$$

• $$220,998.5\ \frac{\text{km}^3}{\text{year}}$$

End of Example 1

Poll

On a scale of 1-5, rate the quality of this slideshow and example problem. Do you feel like it was a helpful explanation and introduction?

• 1 - Poor Quality

  I did not find this particularly helpful or informative.

• 2 - Below Average

  The slideshow was helpful, but could be improved greatly.

• 3 - Average

  Not great, not horrible.

• 4 - Above Average

  The slideshow and example has room for improvement, but was overall helpful.

• 5 - Excellent

  I found the slideshow and example problem to be a very helpful introduction to related rates problems.

Practice / Take Home

A police car traveling at 60 mph while pursuing a speeding vehicle. The police car is 0.6 miles north, and is approaching a right-angled intersection. The speeding vehicle is 0.8 miles east.

The distance between the two vehicles is increasing at a rate of 20 mph. How fast is the speeding vehicle traveling?

(Hint: The sketch should help you, but you may have to recall some geometry for this problem)