## Last Quarter’s Final

1. Let $f(x) = {1\over{x^2}}$. Compute $f'(x)$ using the definition (the limit of a difference quotient).

2. Differentiate (find $D_x(f)$):
a. $f(x) = 5^{3x}$

b. $f(x) = \log_2(x^3)$

c. $f(x) = \arctan({1\over x})$

3. Find formulas for $y'$ using a. logarithmic differentiation and b. the “Quotient Rule”, where $= {{(2x+1)^2}\over x}$.

4. Evaluate the limits (where defined; “DNE” for any that are not):
a. $\lim_{x\rightarrow 3}{{x-3}\over{|x-3|}}$
b. $\lim_{x\rightarrow \infty} {{3x^2 + 5x - 11}\over{2x^2 + 7}}$
c. $\lim_{x\rightarrow 0} {{e^x-1}\over{e^{2x}-1}}$

5. Find the equation of the line tangent to g(x) = ln(x) at x = 1.

6. Use implicit differentiation to find ${{dy}\over{dx}}$ given that $y = e^{xy} + x^2$.

7. Find an algebraic formula for cos(arctan(x)) using a sketch of a right triangle.

8. Compute the second derivative, ${{d^2 y}\over{dx^2}}$, where $y = \sin(x) + xe^x$.

9. A balloon rises at a rate of 10 feet per second from a point on the ground 100 feet from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 100 feet off the ground.

10. Find (both co-ordinates of) the local extrema and inflection point for the cubic function $y = x^3 - 9x - 27$. Sketch the curve.

11. Estimate $\root3\of{123}$ using a tangent line approximation (hint: $\root3\of{125} = 5$).

12. Find the antiderivatives.
a. $\int {1\over{x^2 +1}}dx$
b. $\int x^3 dx$
c. $\int dx$

Marry, sir. Find all the typos.
Moreover. Convince a skeptic that problem 1 or one much like it is worth learning how to do (and belongs on a Calc I final) even if they know $f'(x) = {{-2}\over{x^3}}$. Hint: even a god-damn computer knows that; this is a course for college credit.
Secondarily. Explain why one does not go on, “Harvard Calculus” style, to require discussions about graphs (without calculations)… I thought this was a course for college credit. For extra credit, justify the lack of an “epsilon-delta” proof on this test.
Sixth and lastly. Were notations for differentiation (${{dy}\over{dx}}, f'(x), D_x (f)$) just tossed around at random to confuse everyone? Or what? Justify your conclusions.
Thirdly. How in the name of the nameless name is it even remotely possible that after using TeX for seventeen freaking years, Vlorbik doesn’t know how to typeset those limits properly? For that matter, why didn’t he just scan last year’s test and upload the S.O.B.? How can one survive without being able to work the most basic tools?
Explain why, despite having been present on two or more occasions when Vlorbik explained how to do the tangent line problem on the calculator with no calculus whatsoever, students will blank out on problem 5. No, that’s easy. Explain why they think we made up a whole new symbol for the natural log when we really meant the reciprocal function… now that’s a mystery. And why should it be hard to recognize a constant function, for hecksake? What’s up with these people who haven’t learned calculus yet? Why don’t they get it?

And to conclude. Any questions?

## News From The Gulag

$\bullet$“Math deprivation as punishment” at Coffee and Graph Paper.
$\bullet$Teach For Awhile: Sarah Cannon.

## What Can You Do With This?

$\bullet$Piaget roundly debunked at The Number Warrior.
$\bullet$Things R. Talbot used to think, at Casting Out Nines.
$\bullet$The students are onto Mr. Meyer.
$\bullet$

## Make It New

Last year’s posts about Arc Length On The Cardiod, Vector Products, and integral-secant-cubed are relevant again. The cardiod was a recent in-class example; the “vector” note corrects a mistake I didn’t make until yesterday; the “surprisingly useful integral” is one I had students to look up in a table on the recent exam (and work with… nobody had this one all the way right but me though in principle it’s an “easy” problem when the table is available…).

What the heck. Here’s the problem: The polar equation $r = \theta$, for $\theta \in [0, \infty)$, determines an “Archimedean Spiral”, S. Determine the arc length along S for $\theta$ between 0 and $\pi\over6$ (exactly; a messy “formula”… check the work numerically). And the answer: ${{\pi\sqrt{36+ \pi^2}}\over{72}} + {1\over2} ln({{\pi +\sqrt{36+\pi^2}}\over6})$. This evaluates to about .5466… as does $\int_0^{\pi\over6} \sqrt{\theta^2 + 1}d\theta$… so we can be reasonably sure this is right.

In my own work messy expressions like this are almost never right the first time… on the copies of the handwritten solution I distributed yesterday, quite a bit of erased work is clearly visible. Rooting out every last little mistake until things are just right is of course a vital part of the process for a lot of problems. One student was real close.

$\bullet$ Carnival of Mathematics #52.
$\bullet$Planar Curves at Numericana.
$\bullet$Sumidiot on Wolfram Alpha and calculus reform.