SUNY Geneseo Department of Mathematics

Introduction to Related Rates Problems

Wednesday, October 21

Math 221 02
Fall 2020
Prof. Doug Baldwin

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Previous Lecture

Anything You Want to Talk About?

(No.)

Misc

SI

The next session will be Sunday 6:00 - 7:30. It will review this week, and talk about problem set questions if time permits.

Mid-Semester Feedback

Some themes that showed up in responses (sometimes from multiple classes):

Rejuvenation Day

Next Tuesday is the second rejuvenation day.

This doesn’t directly affect us, but...

...I’m discouraging individual meetings with me that day.

...I’ll also ease off on homework that week, so you can use the week to catch up on grading, questions, etc.

Related Rates

From section 4.1 in the textbook and the related rates discussion.

Squares

(From the discussion)

Suppose you’re drawing a square in such a way that the height and width are both changing at a rate of 2 cm/sec. How fast is the area changing when the sides are both 5 cm long?

What makes this a “related rates” problem? The essence of the problem is to figure out one rate of change (in this case, the area’s) from another (in this case, the side’s).

How would you go about answering it? Write down the equation for area in terms of the length of a side, and then differentiate both sides of that equation. This is, in fact, a good example of the general process for solving related rates problems:

  1. Draw a picture of the situation!
  2. Set up an equation between the quantities in the problem
  3. Differentiate that equation
  4. Rearrange to isolate the derivative you want
  5. Plug in numbers from the problem

So what’s the answer for the square problem?

Following the process outline, start by drawing a picture that shows the relevant quantities and the relationships between them:

A square with sides of length L equals 5 and rate of change d L over d t equals 2

The write down the equation for area in terms of side length:

A equals L squared

Differentiate both sides of that equation…

d A over d t equals 2 L times d L over d t

Those derivatives already have the derivative of area isolated on the left, so there’s no need for the fourth step in this problem. Go straight to plugging in the numbers for L and dL/dt from the problem. Notice that when we do so the units on the answer come out right, which is nice reassurance that we’ve done everything approximately right:

d A over d t equals 20 square centimeters per second

Street Crossings

Suppose you and a car are both approaching a crosswalk. How fast are you and the car closing in on each other if you are 20 feet from the crosswalk walking at a speed of 3 ft/sec, and the car is 100 feet away moving at 22 ft/sec?

Once again, start by drawing the situation:

Pedestrian and car approaching crosswalk

Then use the Pythagorean Theorem to relate the car and pedestrian distances to the crosswalk and the distance between car and pedestrian:

L squared equals x squared plus y squared

Next, differentiate both sides of the equation. Notice that there are a lot of variables in the equation that aren’t the one we’re differentiating with respect to, and we need to use the chain rule when differentiating expressions involving those variables. This is typical of related rates problems.

2 times L times d L d t equals 2 x times d x d t plus 2 y times d y d t

Next, people wanted to plug known numbers into the equation. This is a different order for doing things than in the outline above, but it actually doesn’t matter which order you do the “rearrange” and “plug in numbers” steps.

Rate of change in distance from pedestrian to car with actual numbers in equation

Finally, do the arithmetic and rearrange to isolate dL/dt:

d L d t is approximately 22.16 feet per second

Next

Continue related rates examples.

Read “Examples of the Process” in section 4.1 of the textbook by class time tomorrow.

Keep posting questions, ideas, etc. in the related rates discussion.

Next Lecture