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Chapter 2: Problem 32

Use the transformations relating polar and Cartesian coordinates to prove that $$ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left[x \frac{d y}{d t}-y \frac{d x}{d t}\right] $$

Short Answer

Expert verified
The provided identity is indeed correct. It's proven by leveraging the relationships between Cartesian and polar coordinates and applying the chain rule of differentiation.

Step by step solution

01

Recall relationship between cartesian and polar coordinates

Polar and Cartesian coordinates can be related through the following equations: \( x = r cos(\theta) \), \( y = r sin(\theta) \)
02

Calculate dx/dt and dy/dt

By taking the derivative of \( x = r cos(\theta) \) with respect to \( t \), we get: \(\frac{dx}{dt} = \frac{dr}{dt} cos(\theta) - r sin(\theta) \frac{d\theta}{dt}\). Similarly, taking the derivative of \( y = r sin(\theta) \) with respect to t, we get: \(\frac{dy}{dt} = \frac{dr}{dt} sin(\theta) + r cos(\theta) \frac{d\theta}{dt}.\
03

Substitute dx/dt and dy/dt in given expression

Substitute the expressions for dx/dt and dy/dt from step 2 into the given expression \( x \frac{dy}{dt} - y \frac{dx}{dt} \). This becomes \( r cos(\theta) * [ \frac{dr}{dt} sin(\theta) + r cos(\theta) \frac{d\theta}{dt} ] - r sin(\theta) * [ \frac{dr}{dt} cos(\theta) - r sin(\theta) \frac{d\theta}{dt} ] \).
04

Simplify expression

On simplifying the above expression, \(r^2 \frac{d\theta}{dt}\) terms cancel out and we get \(r^2 \frac{d\theta}{dt}\)
05

Divide by r^2

Finally, divide the whole expression by \(r^2\), bears out the required result, \(\frac{d\theta}{dt} = \frac{1}{r^{2}} [ x \frac{dy}{dt} - y \frac{dx}{dt} ]\)

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Most popular questions from this chapter

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