## Mathochism: Annoyance and dismay (Update)

One woman’s attempt to revisit the math that plagued her in school. But can determination make up for 25 years of math neglect? My initial faith in the calculus professor has disappeared. We took our second test today, and two of the nine questions were in bad faith.

One involved linear approximation and implicit differentiation. That was not the issue — I could do that. But the y value was zero. And the x value, 2, when plugged into 3x², yielded 12. Which would have worked if it hadn’t been for the fact that the constant in the polynomial was -12. So the f(a) went to zero. The derivative was undefined, since Y prime became 0/0. So it was 0 + undefined times an increment of x. So it didn’t work out.

Was it a good way to show we could do implicit differentiation? No. I wound up just doing it, then leaving it as f(a) + f'(a) (increment), just to show I know how to handle the process. But what a waste of fucking time.

Then, we had to differentiate (sin2x-cos2x)½, with x at π/4, and find the equations of the tangent and normal lines. Sine and cosine at π/4 is √2/2. So that meant it was 2√2/2 – 2√2/2, which was the square root of 0. So, y was 0. When I differentiated, the result was an equation with zero in the denominator, so that didn’t work. I left it to show I could differentiate, but again, a waste of time.

That wasted time cost me on the related rates problem, which I rushed through and probably didn’t get right. So, that’s 3 out of 6 wrong. I made a genuine mistake on a trig limit problem, so that’s four.

Another failing grade! I’m not dropping this time, though, because I am learning calculus, in spite of the tests failing to show it. This format is clearly not working for me, and sadly, the teacher has let me down.

Update: I fear I may have been unfair on the second problem. It occurred to me that sin 2π/4 is actually π/2, as is cos 2π/4. Which means the problem was solvable, and y=1. Live and learn. I’m still pissed about the linear approximation thing. And the format still doesn’t work for me.

Live and learn.

All text copyrighted by A.K. Whitney, and cannot be used without permission.

• I’m trying to reconstruct the original first problem from your description. Could it have been solving 3x^2+y^2-12 = 0 by implicit differentiation, and extrapolating from the point (2, 0)?

In that case, you have the equation of an ellipse, and (2,0) is one of the x-intercepts. Implicit differentiation says dy/dx = -3x/y, which is -6/0, not 0/0, at the point in question, and the tangent is the vertical line x=2. If you want to approximate from there, you should probably switch the variables and express x as a function of y, since x isn’t going to be changing much in the immediate neighborhood of (2,0). But it’s possible that the real question was something entirely different.

• akwhitney Maybe. But we have not been switching variables in this context, in fact, all the problems we’ve discussed, or been given, have been either implicit differentiation or linear approximation, not a combo. So even if the point was to switch the x for the y, it was a strategy we had not been offered either in the book or the lectures.

• akwhitney Oh, and the increment we were supposed to approximate was 1.79.