# Parametrization, Parabolas, Calculus, OH MY!

Okay, so a second problem in a row! This one is a straight up calculus one, from the 2008 AP Calculus BC exam — multiple choice section. The teacher of that class asked me if I could work this problem — and I admit I struggled. She showed me her solution, and then I left thinking “it couldn’t be that hard…”

When trying to fall asleep today, I started thinking of it and I was able to solve it in a different way.

Without any more preamble, if you care to try your hand at this:

A particle is moving along the curve $y=x^2-x$ at a constant speed of $2\sqrt{10}$. When it reaches the point $(2,2)$, you know $\frac{dx}{dt}>0$. Find the value of $\frac{dy}{dt}$ at that point.

As usual, feel free to throw your thoughts, solutions, etc. in the comments below, if you want. I bet for many of you this will be super easy, but for the few of you who struggle through it (sigh) like me, you might find it actually frustratingly enjoyable.

Oh, and also throw down there if you get stuck and care to see my solution… It’ll motivate me to actually type it up in a timely fashion.

1. CalcDave says:

Since dy/dx = (dy/dt) / (dx/dt) and dy/dx = 2x – 1, then (2x – 1)*(dx/dt) = dy/dt and at x = 2, 3*(dx/dt) = dy/dt. For a constant speed, (dx/dt)^2 + (dy/dt)^2 = 40. Now substitute the first piece into the second to get (dx/dt)^2 + 9*(dx/dt)^2 = 40 and find that (dx/dt) = 2 (also using that it’s positive). So, (dy/dt) = 6.

2. samjshah says:

Nice. That’s almost how the AP Calc teacher eventually solved it.

3. @thescamdog says:

That’s basically what I did, other than differentiating wrt t right off the bat to get dy/dt=(2x-1)dx/dt. Do you have a simpler solution, Sam?

4. Sue VanHattum says:

I didn’t write it out so nicely, and wasn’t sure I could adequately explain my first step, but I did it pretty much the same way. Procedurally, it seemed easy.

5. jd2718 says:

Eeks, I haven’t done this stuff in far too long.

Because the slope (dy/dx) is 3, I make a teensy right triangle: horizontal leg is dx/dt, vertical is 3dx/dt, hypotenuse is sqr(40). Then pythagoras.

(and then pythagoras again, since I solved for the wrong component).

Actually, first I thought it over and “saw” the parabola: (0,0), (1,0), oops (.5,-.25), (2,2), (3,6)… “the steeper it gets” I thought “the more Y-ish the speed, and the less X-ish. I could break it into components”

Jonathan