Implicit derivative and related rates?

Hi!

I'm having a bit of trouble with this assignment. I had to miss a couple class, and I'm afraid the teacher talked about this. I have the notes, but they are not of much help and don't talk about related rates at all. I tried looking around, I found lots of PDF from different universities, but I still don't get how to solve this problem. If I could get some pointers, that'd be great :D

Thanks

Here's the problem (translated from French):

Quote:

A particle oscillates on the x-axis. It's position $\displaystyle x(t)$ (in meters) at the time $\displaystyle t$ (in seconds) is related to its speed $\displaystyle v(t)$ by the equation:

$\displaystyle x^2+x \cdot v=v^2=13$

Give the acceleration of this particle when its position is $\displaystyle x = -3 meters$, knowing that at that moment, the speed is positive.

Hint: This is a related rates problem in which you need to use the implicit derivative. Use the fact that $\displaystyle a(t) = \frac{dv}{dt}$ and $\displaystyle v(t) = \frac{dx}{dt}$

Re: Implicit derivative and related rates?

Hey Kevindqc.

Hint: Try differentiating both sides with respect to x and collect all the dv/dt terms and get that in terms of the rest of the expression. (Note the chain rule d/dx u(g(x)) = dg/dx u'(g(x)))

Re: Implicit derivative and related rates?

I'm not sure I get what you're saying :(

is v' the dv/dt terms you are talking about? it's not dv/dx? I'm confused :(

like:

$\displaystyle x^2+x \cdot v + v^2 = 13$

$\displaystyle (x^2)' + (x \cdot v)' + (v^2)' = (13)'$

$\displaystyle 2x + (1 \cdot v + v' \cdot x) + 2 \cdot v \cdot v' = 0$

$\displaystyle v' \cdot x + 2 \cdot v \cdot v' = -2x - v$

$\displaystyle v' ( x + 2 \cdot v) = -2x - v$

$\displaystyle v' = \frac{-2x-v}{x+2v}$

What am I supposed to do with $\displaystyle v'$ :/

I don't understand what "get that in terms of the rest of the expression" means :x

I guess what really bugs me is the t