SUNY Geneseo Department of Mathematics
Wednesday, October 7
Math 221 02
Fall 2020
Prof. Doug Baldwin
The ferris wheel problem from the chain rule discussion?
The heart of this problem is differentiating the height function. Verify that it (or at least the “12 cos(πt)” part, since the rest has a derivative of 0) fits the form to which the chain rule applies. It does, because cosine is the outer function, f, and multiplication by π is the inner function, g:
Now finding the speed, i.e., the derivative of height, is just a matter of plugging f and g into the formula from the chain rule:
(i.e., a talk by another mathematician sharing something they work on)
Next Tuesday (October 13), from 4:00 to 5:00 via Zoom.
“The Column-Row Factorization A = CR”
Prof. Gilbert Strang, MIT
About teaching linear algebra, but it’s advertised as accessible to everyone.
Prof. Strang is one of the main authors of your textbook.
The next session will be Sunday, and will review what we covered this week plus provide a chance to talk about the latest problem set and/or other questions you have.
Based on “Deriving the Chain Rule” in section 3.6 in our textbook, and the chain rule discussion.
What we did with the ferris wheel problem above illustrates the general strategy for differentiating with the chain rule:
Use the chain rule to find derivatives of…
cos4t
This has (as of shortly before class) a nearly correct solution in the discussion, apart from a typo in applying the power rule.
The key idea is, like in the ferris wheel problem, to recognize that the function matches the form for the chain rule, identify f and g, and then plug them into the chain rule formula:
sin2x + cos2x
(You might be able to predict ahead of time what this derivative will be — but how does the chain rule come up with that answer?)
From the trigonometric identity sin2x + cos2x = 1, you would expect this derivative to be 0. And it is, because applying the chain rule to sin2x and to cos2x produces two terms that are negatives of each other:
Another problem from the trigonometric derivatives discussion that we can now do in more natural units. 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 has length (in feet)
L = 30 / tan( 0.004t )
Find the speed at which the shadow is growing or shrinking, in feet per minute, t minutes after dawn.
As we did originally, it’s easiest to rewrite the equation for the length of the shadow as L = 30 cot(0.004t) and then use the chain rule with cotangent as f and multiplication by 0.004 as g:
Another way to rewrite the equation is as 30 times tan( 0.004t )-1. This is interesting because it involves using the chain rule twice, first to apply the power rule to tan(0.004t) raised to the -1 power, and then again in finding the derivative of tan( 0.004t ).
One way to motivate or understand the chain rule is that it’s calculus’s form of unit conversion. For instance, in the silo problem, the outer function, tangent, gives you the shadow’s length in terms of the sun’s angle in radians, and the derivative gives you speed in the awkward units of feet per radian. To convert feet per radian to feet per minute, you’d have to multiply by a radians per minute conversion factor, which is exactly what the derivative of the inner function gives you.
More about the chain rule, particularly in more sophisticated settings.
Please read the rest of section 3.6 (“The Chain and Power Rules Combined,” “Combining the Chain Rule with Other Rules,” “Composites of Three or More Functions,” and “The Chain Rule Using Leibniz’s Notation”) by class time Thursday.
Also contribute to the ongoing discussion of the chain rule by class time Thursday.