Derivative of polar coordinate

amelemühendis

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could anyone help me to find out dQ/dt and d^2Q/dt^2 please.

r = 5+ 2cos (Q)

I need to find "angular velocity" and "angular acceleration"
respect of time
 

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amelemühendis said:
could anyone help me to find out dQ/dt and d^2Q/dt^2 please.

r = 5+ 2cos (Q)

I need to find "angular velocity" and "angular acceleration"
respect of time

Do you know how to differentiate

f(x) = cos(x)
 
Thank you very much Subhotosh Khan for your interest.
ofcourse i know it is f'(x) = -sin(x) but i couldn't solve my problem with this.

in fact my real problem is how to establish equation which is equal to Q = ........
 
amelemühendis said:
Thank you very much Subhotosh Khan for your interest.
ofcourse i know it is f'(x) = -sin(x) but i couldn't solve my problem with this.

in fact my real problem is how to establish equation which is equal to Q = ........

\(\displaystyle r \ \ = \ \ 5 \ \ + \ \ 2cos(\theta)\)

\(\displaystyle \theta \ \ = \ \ cos^{-1}\left ( \frac{r-5}{2}\right )\)

However, if you are trying to find " expression for d?/dt", it will be lot easier (and neater) if you perform implicit differentiation.
 
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