SUNY Geneseo Department of Mathematics

More Related Rates Examples

Friday, October 23

Math 221 02
Fall 2020
Prof. Doug Baldwin

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Sunday 6:00 - 7:30. Review this week and help with any problem set questions.

Mid-Semester Feedback

One theme in feedback surveys from this course and others was students asking for video recordings of classes, or classes that students could attend via Zoom when they can’t come in person.

Unfortunately, there are lots of technical and possibly legal/ethical challenges in doing that routinely. I might be able to make short focused videos outside of class for some things (e.g., Mathematica demonstrations where interaction is important) though.

A related question came up Wednesday: could we hold classes via Zoom after Thanksgiving, when everything is online? My experience with such classes isn’t good — there’s just enough distance between people to cut way down on interaction, technical difficulties can shut some people out, and it’s very hard for me to judge how well students are understanding something. But if you strongly favor Zoom classes for a short period, I might be persuaded. Please let me know if you have a preference either for or against.

Related Rates

From section 4.2 in the textbook and the related rates discussion.

Moving Shadow

A light that swivels around a post casts a shadow of that post onto a screen 2 feet away. If the light rotates at 0.1 radian per second, how fast is the shadow moving when the light has turned 30 degrees (i.e., π/6 radians)?

Light pivots around post and casts shadow on screen

Like other related rates problems, start by finding an equation that relates the things you’re interested in. In this case it looks like there should be a trigonometric function that relates the shadow’s position on the screen, x, to the light’s angle. And indeed there is, namely tangent:

x over 2 equals tangent of theta

Now take derivatives of both sides of the equation to get the derivative we want:

1 half d x d t equals secant squared of theta times d theta d t

And finally, plug in known numbers and rearrange to get a value for the derivative. In doing this we needed sec2 π/6. We can work this out knowing that secant is the reciprocal of cosine, knowing the values that sine and cosine have at principal angles and the shape of the cosine graph (but not necessarily memorizing exactly what value goes with every angle):

d x d t equals 2 times secant squared of pi over 6 times point 1 which equals 4 fifteenth of a foot per second

Extension Ladder

(A variation on the stereotypical ladder-sliding-down-wall problem). Someone is setting up an extension ladder against a wall, meaning that they hold the bottom of the ladder a fixed distance of 6 feet from the wall, while extending the ladder and thereby pushing its top up the wall. How fast is the top of the ladder rising when the ladder is extended a total distance of 10 feet?

10 foot ladder extending up wall 6 feet away

The Pythagorean Theorem is a good way to relate the ladder’s total length to the height of its top:

h equals square root of L squared minus x squared

Differentiating this equation and simplifying gives us dh/dt. Notice when doing this that x is a constant, so its derivative will be 0.

d h d t equals L over square root of L squared minus x squared all times d L d t

Finally, plug in numbers from the problem and simplify:

d h d t equals 10 over square root of 100 minus 36 all times 1 equals 5 fourth of a foot per second

Next

Using derivatives to approximate curved lines as straight: linear approximation (and differentials).

Please read “Linear Approximation of a Function at a Point” and “Differentials” in section 4.2 of the textbook by class time Monday.

Please also participate in this discussion of differentials by class time Monday.

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