# Implicit derivative and related rates

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):

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

$$x^2 + x \cdot s + s^2 = 13$$

Give the acceleration of this particle when its position is $x = 3$ 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 $a(t) = {dv \over dt}$ and $s(t) = {dx \over dt}$

I did this, but I'm not sure if it's the right thing to do, or what to do with my result :/

$$x^2+x \cdot s + s^2 = 13$$ $$(x^2)' + (x \cdot s)' + (s^2)' = (13)'$$ $$2x + (1 \cdot s + s' \cdot x) + 2 \cdot s \cdot s' = 0$$ $$s' \cdot x + 2 \cdot s \cdot s' = -2x - s$$ $$s' ( x + 2 \cdot s) = -2x - s$$ $$s' = \frac{-2x-s}{x+2s}$$

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It looks correct. Do you wanna evaluate $a(t)$? – Babak S. Feb 7 at 17:21
@BabakSorouh ... but it isn't. Note that $\cdot'$ stands for $\frac d{dt}$, not $\frac d{dx}$ here. – Hagen von Eitzen Feb 7 at 17:25

## 1 Answer

Taking the derivative (with respect to time $t$, not $x$!) of $$x^2+xs+s^2=13,$$ you should obtain $$2xx'+(x's+xs') +2ss'=0$$ or after inserting $s,a$ for $x',s'$ $$2xs+(s^2+xa) +2sa=0$$ and hence $$a=-\frac{(2x+s)s}{x+2s}.$$ Remark: You fell for interpreting the prime $'$ as taking the derivative with respect to $x$. It is very usual in physics to denote the derivative with respect to time by a dot instead (among others, to avoid such confusion). So we aould then rather write $2x\dot x+\dot xs+x\dot s+2s\dot s=0$ and insert $\dot x=s, \dot s=a$.

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Honestly, when I saw your approach, I found out that. Yes Hagen. I am losing my information about Classical Mechanics...... Thanks +1 – Babak S. Feb 7 at 17:31
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