Basic implicit differentiation question

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The discussion revolves around finding dx/dt for the equation x = rcos(θ) using implicit differentiation. The initial confusion stems from the assumption that dx/dt could simply be calculated as (-rsinθ)(dθ/dt), but this overlooks the dependence of both r and θ on time t. The correct approach involves applying the chain rule, leading to the formula dx/dt = (dr/dt)cos(θ) - r(sin(θ))(dθ/dt). This highlights the necessity of considering all variables that depend on t when differentiating. The conversation emphasizes the importance of understanding the relationships between the variables in the context of differentiation.
influx
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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)?
 
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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.
influx said:
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.
 
influx said:
x=rcos(θ)

How do you find dx/dt?
Since you have not specified otherwise, I assume that you really have x(t)=r(t)\cdot \cos(\theta (t)). Then \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}
 
For x a function of two variables, r and \theta, where r and \theta are functions of t, the "chain rule" is
\frac{dx}{dt}= \frac{\partial x}{\partial r}\frac{dr}{dt}+ \frac{\partial x}{\partial \theta}\frac{d\theta}{dt}
That gives Svein's answer.
 
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