Question

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

Short answer

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

Recall relationship between cartesian and polar coordinates

Polar and Cartesian coordinates can be related through the following equations: $x=rcos(\theta )$, $y=rsin(\theta )$

Calculate dx/dt and dy/dt

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

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 $rcos(\theta )\ast [\frac{dr}{dt}sin(\theta )+rcos(\theta )\frac{d\theta }{dt}]-rsin(\theta )\ast [\frac{dr}{dt}cos(\theta )-rsin(\theta )\frac{d\theta }{dt}]$.

Simplify expression

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

Divide by r^2

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

Key concepts

Differentiation in Polar Coordinates

Differentiation in polar coordinates requires us to understand how functions change with respect to their variables in this coordinate system, which is based on radius and angle, unlike the traditional Cartesian system.
While Cartesian coordinates use x and y, polar coordinates involve an angle $\theta$ and radius $r$.
When differentiating in polar coordinates, we focus on $\frac{dr}{dt}$ and $\frac{d\theta }{dt}$, which represent the rates of change of the radial distance and angle with respect to time.

• For $x=r\cos (\theta )$, the derivative with respect to time is $\frac{dx}{dt}=\frac{dr}{dt}\cos (\theta )-r\sin (\theta )\frac{d\theta }{dt}$.


• For $y=r\sin (\theta )$, it is $\frac{dy}{dt}=\frac{dr}{dt}\sin (\theta )+r\cos (\theta )\frac{d\theta }{dt}$.

These derivatives reflect how changes in $r$ and $\theta$ translate into changes in x and y.
They are crucial for solving problems involving motion or transformations from polar to Cartesian coordinates.

Cartesian Coordinate Derivatives

In Cartesian coordinates, derivatives focus on changes in x and y with respect to another variable, often time $t$.
It's essential to grasp how the velocities in Cartesian coordinates $\frac{dx}{dt}$ and $\frac{dy}{dt}$ link back to polar forms.

• The derivative $\frac{dx}{dt}$ is derived from $r\cos (\theta )$, factoring in how both r and $\theta$ change over time.


• Similarly, $\frac{dy}{dt}$ comes from $r\sin (\theta )$ and includes both $\frac{dr}{dt}$ and $\frac{d\theta }{dt}$.

These equations tie back to how motion in a polar system translates to a Cartesian plane.
This helps in solving problems where the position is defined in polar terms, but needs to be understood or evaluated in Cartesian ones.

Polar Coordinates Calculus

Polar coordinates calculus explores how we can compute derivatives and integrals when our variables are radius $r$ and angle $\theta$.
This requires understanding and leveraging the relationships between polar and Cartesian forms.
When integrating or differentiating in polar coordinates, we often start by converting our equations to Cartesian or utilize transformations.
For instance:

• The expression $x\frac{dy}{dt}-y\frac{dx}{dt}$ relates polar velocities through Cartesian expressions.


• After substitutions and simplifications in such calculus problems, we often end up simplifying in a way that leverages these interrelationships.

The calculation shown in the exercise demonstrates how operations in polar coordinates can be proven equivalently, using their Cartesian counterparts.
Resulting expressions, like $\frac{d\theta }{dt}=\frac{1}{r^2}[x\frac{dy}{dt}-y\frac{dx}{dt}]$, showcase the powerful insights calculus can provide when transitioning between these two systems of coordinates.