Polar differential problem with my workings...

Hi All,

I'm having problems with this question and need someone to look at my workings and give a gentle push in the right direction. If anyone can help I'd be most grateful.

Calculate$\displaystyle \frac{dy}{dx}$for the curve whose equation in polar coordinates is $\displaystyle r=1+sin^2\theta$

So my understanding is that polar coords are given as $\displaystyle r,\theta$ so not quite sure what to do with this. Would I rearrange to find $\displaystyle \theta$ such that $\displaystyle \theta = sin^{-1}(\pm\sqrt{r-1})$ and keep r as it is????

Or is it using r and $\displaystyle \theta$ with $\displaystyle x = rcos \theta $ & $\displaystyle y=rsin \theta$

Then find $\displaystyle \frac{dx}{d\theta}$ and hence $\displaystyle \frac{d\theta}{dx}$ and $\displaystyle \frac{dy}{d\theta}$ and finally get the result $\displaystyle \frac{dy}{dx}$$\displaystyle =-cot\theta$

Best regards, Felix

Re: Polar differential problem with my workings...

Quote:

Originally Posted by

**FelixHelix** Hi All,

I'm having problems with this question and need someone to look at my workings and give a gentle push in the right direction. If anyone can help I'd be most grateful.

Calculate$\displaystyle \frac{dy}{dx}$for the curve whose equation in polar coordinates is $\displaystyle r=1+sin^2\theta$

So my understanding is that polar coords are given as $\displaystyle r,\theta$ so not quite sure what to do with this. Would I rearrange to find $\displaystyle \theta$ such that $\displaystyle \theta = sin^{-1}(\pm\sqrt{r-1})$ and keep r as it is????

Or is it using r and $\displaystyle \theta$ with $\displaystyle x = rcos \theta $ & $\displaystyle y=rsin \theta$

Then find $\displaystyle \frac{dx}{d\theta}$ and hence $\displaystyle \frac{d\theta}{dx}$ and $\displaystyle \frac{dy}{d\theta}$ and finally get the result $\displaystyle \frac{dy}{dx}$$\displaystyle -cot\theta$

Best regards, Felix

$\displaystyle \frac{dx}{d\theta}=-r\sin \theta+\cos \theta\frac{dr}{d\theta}$

$\displaystyle \frac{dy}{d\theta}=r\cos \theta+\sin \theta\frac{dr}{d\theta}$

Note that $\displaystyle \frac{dr}{d\theta}\neq 0$ in this case.

Re: Polar differential problem with my workings...

I'm not sure what you have done here. Can you expalin? Also I'm guessing that that is a typo on your second line and should read $\displaystyle \frac{dy}{d\theta}$...

Re: Polar differential problem with my workings...

Quote:

Originally Posted by

**FelixHelix** I'm not sure what you have done here. Can you expalin? Also I'm guessing that that is a typo on your second line and should read $\displaystyle \frac{dy}{d\theta}$...

Sorry about the typo.

I have merely used the product rule to differentiate $\displaystyle x=r\cos \theta$ and $\displaystyle y=r\sin \theta$.

Re: Polar differential problem with my workings...

Quote:

Originally Posted by

**FelixHelix** Hi All,

I'm having problems with this question and need someone to look at my workings and give a gentle push in the right direction. If anyone can help I'd be most grateful.

Calculate$\displaystyle \frac{dy}{dx}$for the curve whose equation in polar coordinates is $\displaystyle r=1+sin^2\theta$

So my understanding is that polar coords are given as $\displaystyle r,\theta$ so not quite sure what to do with this. Would I rearrange to find $\displaystyle \theta$ such that $\displaystyle \theta = sin^{-1}(\pm\sqrt{r-1})$ and keep r as it is????

Or is it using r and $\displaystyle \theta$ with $\displaystyle x = rcos \theta $ & $\displaystyle y=rsin \theta$

Then find $\displaystyle \frac{dx}{d\theta}$ and hence $\displaystyle \frac{d\theta}{dx}$ and $\displaystyle \frac{dy}{d\theta}$ and finally get the result $\displaystyle \frac{dy}{dx}$$\displaystyle =-cot\theta$

Best regards, Felix

You should know that $\displaystyle \displaystyle x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2$.

So

$\displaystyle \displaystyle \begin{align*} r &= 1 + \sin^2{\theta} \\ \sqrt{x^2 + y^2} &= 1 + \left(\frac{y}{r}\right)^2 \\ \sqrt{x^2 + y^2} &= 1 + \frac{y^2}{r^2} \\ \sqrt{x^2 + y^2} &= 1 + \frac{y^2}{x^2 + y^2} \\ \sqrt{x^2 + y^2} &= \frac{x^2 + 2y^2}{x^2 + y^2} \\ x^2 + y^2 &= \left(\frac{x^2 + 2y^2}{x^2 + y^2}\right)^2 \\ x^2 + y^2 &= \frac{\left(x^2 + 2y^2\right)^2}{\left(x^2 + y^2\right)^2} \\ \left(x^2 + y^2\right)^3 &= \left(x^2 + 2y^2\right)^2 \\ \frac{d}{dx}\left[\left(x^2 + y^2\right)^3\right] &= \frac{d}{dx}\left[\left(x^2 + 2y^2\right)^2\right] \\ 3\left(x^2 + y^2\right)^2\left(2x + 2y\,\frac{dy}{dx}\right) &= 2\left(x^2 + 2y^2\right)\left(2x + 4y\,\frac{dy}{dx}\right) \\ 6x\left(x^2 + y^2\right)^2 + 6y\left(x^2 + y^2\right)^2\frac{dy}{dx} &= 4x\left(x^2 + 2y^2\right) + 8y\left(x^2 + 2y^2\right)\frac{dy}{dx} \\ 6y\left(x^2 + y^2\right)^2\frac{dy}{dx} - 8y\left(x^2 + 2y^2\right)\frac{dy}{dx} &= 4x\left(x^2 + 2y^2\right) - 6x\left(x^2 + y^2\right)^2\end{align*}$

$\displaystyle \begin{align*} 2y\left[3\left(x^2 + y^2\right)^2 - 4\left(x^2 + 2y^2\right)\right]\frac{dy}{dx} &= 2x\left[2\left(x^2 + 2y^2\right) - 3\left(x^2 + y^2\right)^2 \right] \\ \frac{dy}{dx} &= \frac{2x\left[2\left(x^2 + 2y^2\right) - 3\left(x^2 + y^2\right)^2\right]}{2y\left[3\left(x^2 + y^2\right)^2 - 4\left(x^2 + 2y^2\right)\right]} \\ \frac{dy}{dx} &= \frac{x\left[2\left(x^2 + 2y^2\right) - 3\left(x^2 + y^2\right)^2\right]}{y\left[3\left(x^2 + y^2\right)^2 - 4\left(x^2 + 2y^2\right)\right]} \end{align*}$