Day 13, Mat129: Calculus I

• Your 2.2 homework is returned (graded #14 and 22)

• 2.3 homework is due today. Hopefully you're enjoying all these new rules!

• Your first exam is this Thursday, 2/8, and will cover up through section 2.2: that's five sections -- 1.5, 1.6, 1.8, 2.1, and 2.2.

  You should review those five sections, review your notes, and look over the examples that we've done in class, your quizzes, your homework, etc. Things on the exam should look familiar.

  I've told you flat out some of the things that you're going to see on this, if you've been paying attention.

  I don't do study guides, because students have weaponized them: if I ask something that's not explicitly on the study guide, students will complain that I've done something unfair. So why would I tie my hands?

• So I'll start with any questions about the material to be covered on the exam.

• Please put your rectangles away.

Today we'll start on Section 2.4: Derivatives of trig functions. (This is good practice for your exam!)

• The derivatives of trig functions rely on some basic understanding of the circle:
  • Radian measure: there is a linear relationship between radians and degrees: \[ Radians=\frac{2\pi}{360}Degrees \] In calculus, we almost always use radians. You should try to start thinking in terms of radians, too.

  • Length of a sector, related to angle $\theta$ subtending and radius $r$:

    Arc Length: $L = \theta r$

    Again, this is an example of a linear relationship, but it's called "bi-linear", because it's linear in two different ways: $L$ is proportional to each of $\theta$ and $r$. So

    • if you double the radius, you double the arc length;
    • if you double the angle, you double the arc length.

    Now you might wonder if there should be, or could be, an extra constant in there, but you can see that the formula works out for the circumference formula for a circle: \[ \theta=2\pi, \textrm{ so } L=2 \pi r \]

    Yep! That's the formula for the circumference of a circle.

    

    We can use this diagram (and the arc length formula) to prove a really important limit: \[ \lim\limits_{\theta \to 0}{\frac{\sin(\theta)}{\theta}}=1 \]

    (See page 142, which I'll walk you through now. The proof uses some elementary trigonometry, plus the squeeze theorem, which is one of the really cool things about this proof!)

  • Definitions of the trig functions, as functions of $x$, in radians (which is how we think about them in calculus):
  

  We will derive the derivative of the sine function from the limit definition of the derivative (although we can see the derivative graphically above).

  There are exactly three important trig identities one needs to know (all the others can be derived from these three):

  • The Pythagorean theorem:
    $\sin^2(x)+\cos^2(x)=1$

  • The sine of a sum:
    $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$

  • The cosine of a sum:
    $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

  So let's see how to derive the derivative of the sine from the MIDIC -- the Most Important Definition In Calculus -- the limit definition of the derivative -- using the second of these identities, plus one more important limit (which can be proven with conjugates, and the first limit we derived): \[ \lim\limits_{\theta \to 0}{\frac{\cos(\theta)-1}{\theta}}=0 \]

  Then the derivative of the cosine can be derived by simply shifting the sine function, and using the second trig identity above.

• Examples:
  • Derivative of tangent (use the quotient rule, of course!)
  • Fourth derivatives of sine, cosine
  • #7, p. 146
  • #26
  • #35
  • #49