SUNY Geneseo Department of Mathematics
Math 221 02
Fall 2020
Prof. Doug Baldwin
This discussion introduces you to working with the chain rule. See section 3.6 of our textbook for explanations and examples of the chain rule.
Please post one or more responses to one or more of the following by class time on Wednesday, October 7. Please keep posting, especially about problems not covered in Wednesday’s class, until class time on Friday, October 9.
Use the chain rule to find the following derivatives:
\[\frac{d}{dx} \left( x^4 - 3x^3 + 6x \right)^7\] \[\frac{d}{dt} \cos^4t\] \[\frac{d}{dx} \left( \sin^2x + \cos^2x\right)\](If you think for a minute, you can probably predict ahead of time what the last of these derivatives will be. What is your prediction, and how does the chain rule get the value?)
I mentioned in the October 5 class that the units for the ferris wheel and silo problems in the trigonometric derivatives discussion were artificial. Now that you know about the chain rule, you can do those problems in more natural units.
In particular, suppose the ferris wheel is turning at a rate of 1 revolution every 2 minutes. Then a rider’s height (in feet) t minutes after the ride starts is given by
\[h(t) = 12 - 12\cos \left( \pi t \right)\]Find the rider’s vertical speed and acceleration, in feet/minute and feet/minute2, respectively.
For the silo, the sun moves up or down in the sky at a speed of about about 0.004 radians per minute. Thus the sun’s angle t minutes after dawn is 0.004t radians, and so the silo’s shadow is
\[L = \frac{30}{\tan \left( 0.004 t \right)}\]feet long. Find the speed at which the shadow is growing or shrinking, in feet per minute, t minutes after dawn.
The other place where we have recently encountered a puzzle that the chain rule can solve was in our “reciprocal rule” for derivatives. This is the rule we derived in class, that says that
\[\frac{d}{dx} \left( \frac{1}{f(x)} \right) = \frac{-f^\prime(x)}{f(x)^2}\]We derived this rule by using the quotient rule. When we tried by using the power rule on f(x)-1 we got a different result that we couldn’t really explain. Does using the chain rule with the power rule and f(x)-1 agree with our version of the reciprocal rule?
Here are some examples of the chain rule in more complicated settings than the ones above.
Here are a couple of examples that use the chain rule, albeit in functions that I picked more for their pedagogical value than because they have any real-world meaning. Find the derivatives of…
\[h(t) = \frac{\sin\left(t^2\right)}{\cos^2t}\] \[g(x) = \tan^5\left(x^3-3x^2\right)\]The mechanics of calculating and simplifying some of these are messy, so feel free to post about how you would approach them, and particularly where the chain rule fits in, rather than posting a complete solution.
Back around the end of March, 2014, the orbits of Earth and Mars brought the two planets close to each other (relatively speaking, of course). In that time period, the distance between the planets was approximated by the formula
\[d = \sqrt{ \left( 1.64\cos \left(\frac{2\pi}{687}t+0.19\right) - \cos\left(\frac{2\pi}{365}t\right) \right)^2 + \left( 1.64 \sin\left( \frac{2\pi}{687}t + 0.19\right) - \sin\left(\frac{2\pi}{365}t \right)\right)^2}\]where t is time in days since March 19, 2014, and d is the distance between the planets, measured in astronomical units (1 AU is the average distance from Earth to the sun). While this formula isn’t terribly accurate over a long period of time, it wasn’t bad over a period of a few days. If you’re wondering why I ever cared about this formula, I used it in a computer programming class where I wanted students to work with a messy but “real world” calculation that would be better done with a computer than by hand.
How fast was the distance between the planets changing in late March, 2014? Because this is a messy calculation, I’m more interested in you commenting on how you would go about doing it than I am in you actually doing it. But if someone wants to carry it all out, you’re certainly welcome to.
You don’t officially know it yet, because we don’t quite know enough calculus to work out formulas for derivatives of roots, but for this problem you can assume that the derivative of the square root function is
\[\frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \frac{du}{dx}\](Notice that this formula acknowledges that u might be a function of x, and therefore incorporates a “du/dx” term that comes from the chain rule.)