Particle movement on a polar curve

[artsydragon]

Hello,

I am not sure how to approach this problem (from AP Calculus BC, semester 2)

In the video and notes we did not go in depth into how to solve anything beyond the basic derivatives of polar curves (dy/dx = (dy/d(theta))/(dx/d(theta) as well as how to calculate the second derivative). Because there is time in the problem as well, I am not sure how to approach it.

Either a hint to start the problem, or the full explained steps, would be helpful.

Thank you

 

[Deleted member 4993]

View attachment 24903
Hello,

I am not sure how to approach this problem (from AP Calculus BC, semester 2)

In the video and notes we did not go in depth into how to solve anything beyond the basic derivatives of polar curves (dy/dx = (dy/d(theta))/(dx/d(theta) as well as how to calculate the second derivative). Because there is time in the problem as well, I am not sure how to approach it.

Either a hint to start the problem, or the full explained steps, would be helpful.

Thank you

The question asks you to calculate

\(\displaystyle \frac{dr}{dt}\)

from r = 5 + 2*sin(3*Θ)

Have you done that?

Exactly where are you stuck?

 

[artsydragon]

The question asks you to calculate \(\displaystyle \frac{dr}{d\theta}\).

Have you done that? Exactly where are you stuck?

It's asking for the derivative of r in respect to t, which I'm not sure how to solve. If it were asking for the derivative in respect to theta, it would be simpler.

 

[Deleted member 4993]

It's asking for the derivative of r in respect to t, which I'm not sure how to solve. If it were asking for the derivative in respect to theta, it would be simpler.

You are correct. I had a typo there - fixed it

Do you know that:

\(\displaystyle \frac{df(x,y)}{dt} = \left[ \frac{\partial f}{\partial x} * \frac {dx}{dt}\right] + \left[\frac{\partial f}{\partial y} * \frac {dy}{dt}\right] \).......slightly edited for clarity

It is almost like chain-rule.