Day 24, Mat128: Calculus A

• I hope that you've had a restful and enjoyable spring break!
• Your quiz Friday will cover section 2.2. There is a new assignment related to section 2.2.
• I'll be zooming this meeting, and recording it, for a student who has a legitimate issue requiring it.

• We had an exam, which was mostly painful -- for you and for me.
  • The exam
  • Median score: 63%
  • Key -- will be posted on Friday
  • You can get 1/3 of the points you missed by fixing the problems up.

    Return your exam, and the corrections, by Friday.

    You can print off a copy of problems to work from the link above, to work on problems anew. Do not write on your original exam, which you should submit attached to the repairs.

• Some observations from the exam:
  • A reminder that if you see "GYE" on a problem, it means that you acted properly "Given Your Error" on a previous part.

  • I apologize for referring to "problem 1" when I meant "problem 2", which happened in problem 3. I switched problems around on the exam at the last minute -- not always a good idea.

  • I wanted to start off with an "easy one", however, and I thought that drawing tangent lines and estimating slopes would be easier than it was.

    Remember to keep working to become a blackbelt by practicing this exercise.

    Let's take a look at that graph, and think about how to choose the four tangents to roughly approximate the second derivative. I've printed off that page, so you can get a start on revising, if you need to.

  • One of the difficulties of problem 2 was the units -- it seems funny to think of time as the dependent variable, and temperature as the independent. Remember your magic power as a mathematician: you can rename things!

    Otherwise it was very similar to one of your recent quizzes.

  • Problem 3 was the trickiest -- in part because it relied on solving problem 2.

    It was also tricky because it tested whether you know the difference between exponential functions and power functions. Several of you tried to use the power rule on this function, but it's not a power! You have to use the exponential rule: \[ \left(a^x\right)' = \ln(a) a^x \] Several of you thought that the derivative should be \(a^x\) -- the function itself -- but that only works for base \(e\); all others have derivatives which are proportional to themselves, with constant of proportionality \(\ln(a)\).

    I think that another problem is that the constant was so ugly, and the base was ugly, and the variable (Temp) was ugly. Don't be put off by notation (but that's easy to say, and harder to do)!

  • Then problem 4 was supposed to be easy as well, since we'd gone over the product rule derivation several times in class, and so it was either hit or miss.

    Then you were supposed to use it!:) Although some of you expanded out that first one, and did it as a polynomial -- which was fine. You needed to use it for the last one, however, since it was a product of the exponential and polynomial functions.

(The derivative of cosine can be obtained by simply shifting the derivative of sine, since the two are just shifts of each other.)

• Theorem (trigonometric derivatives):
  Let $F(x)=\sin(x)$.

  \[ \frac{d}{dx}[F(x)]=\cos(x) \]

  We need the limit definition and a couple of limits to establish this one, as well as one trig identity: \[ \sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a) \]

  There are exactly three trig identities which you ought to know (all the others can be derived from those):

  1. the one just mentioned, \[ \sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a) \]
  2. the symmetric version of that:

     \[ \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b); \] and

  3. the Pythagorean identity: \[ \sin(x)^2+\cos(x)^2=1. \]
  Here are the picture we'll need for the limits:

  So the result is this:

• Corollary:
  Let $F(x)=\cos(x)$. \[ \frac{d}{dx}[F(x)]=-\sin(x) \]

  This is a consequence of the derivative of sine, by symmetry and periodicity:

  • \(\cos(x)=\sin(x+\frac{\pi}{2})\)
  • It should be clear that shifting sine by \(\pi\) just shifts its derivative by the same amount: \[ \sin'(x+\frac{\pi}{2})=cos'(x)=\cos(x+\frac{\pi}{2}) \]
  • And using our second of the identities above, we obtain \[ cos'(x)=\cos(x+\frac{\pi}{2})=\cos(x)\cos(\frac{\pi}{2})-\sin(x)\sin(\frac{\pi}{2}) \] and simplifying, we obtain \[ cos'(x)=-\sin(x) \]

  Further consequences: \[ \frac{d^2}{dx^2}[\sin(x)]=-\sin(x) \] \[ \frac{d^3}{dx^3}[\sin(x)]=-\cos(x) \] \[ \frac{d^4}{dx^4}[\sin(x)]=\sin(x) \]

  The sine (and cosine) functions are their own fourth derivatives.

• Perhaps the most important of the relationships above is this one: \[ \frac{d^2}{dx^2}[\sin(x)]=-\sin(x) \]

  Consider \(y(x)=\sin(x)\), and think of it as the position of a particle. If we recall that the second derivative is the acceleration, then we can write \[ \frac{d^2}{dx^2}[y(x)]=-y(x) \]

  Thus Newton's second law (\(F=ma\) as famously expressed) in this case becomes \[ F=ma=-my(x) \]

  and so we can say that the force on the particle is opposed to the displacement of the particle; this is the differential equation for a frictionless spring.

• Derivatives of other trig functions are now easily obtained as well, using the quotient rule:
  • Derivative of tangent: \[ \tan'(x) = \left(\frac{\sin(x)}{\cos(x)}\right)' = \frac{1}{\cos(x)^2} = \sec(x)^2 \]
  • Derivative of cotangent (this is also a shift -- the identify is \(\cot(x)=-\tan(x+\pi/2)\)): \[ \cot'(x) = \frac{-1}{\sin(x)^2} = -\csc(x)^2 \]