Basic implicit differentiation question

[influx]

So it has been quite a few years since I learned about implicit differentiation so the content is a bit rusty in my mind.

x=rcos(θ)

How do you find dx/dt?

I know the answer but I am trying to figure out why. I mean dx/dt can be written as (dx/dθ)*(dθ/dt) so why is the answer not just (-rsinθ)*(dθ/dt)?

 

[blue_leaf77]

You have to know which variables in the RHS which depends on ##t##. By the way, you don't need implicit differentiation in this case since what you want to find is ##dx/dt## and ##x## in your equation has been expressed explicitly in terms of the other variables.

so why is the answer not just (-rsinθ)*(dθ/dt)?

So you know the right answer? It will be helpful to post it as well.

 

[Svein]

x=rcos(θ)

How do you find dx/dt?

Since you have not specified otherwise, I assume that you really have [itex]x(t)=r(t)\cdot \cos(\theta (t)) [/itex]. Then [itex]\frac{dx(t)}{dt}=\frac{dr(t)}{dt}\cdot \cos(\theta(t))+r(t)\cdot\frac{dcos(\theta(t))}{dt}=\frac{dr(t)}{dt}\cdot \cos(\theta(t))+r(t)\cdot(-\sin(\theta(t)))\frac{d\theta(t)}{dt} [/itex]

 

[HallsofIvy]

For x a function of two variables, r and [itex]\theta[/itex], where r and [itex]\theta[/itex] are functions of t, the "chain rule" is
[tex]\frac{dx}{dt}= \frac{\partial x}{\partial r}\frac{dr}{dt}+ \frac{\partial x}{\partial \theta}\frac{d\theta}{dt}[/tex]
That gives Svein's answer.