Tangent Line of Polar Curve

Question

i start by changing polar coords into x and y and then find the derivatives to get the slope.

$$x=(3-3\sin\theta)\cos\theta $$

$$x=3\cos\theta -3\cos\theta \sin\theta $$

and took $x'=(-3\sin\theta +3\sin^2\theta -3\cos^2\theta )$

$$= -3(\sin\theta -\sin^2\theta +\cos^2\theta)$$

$$= -3\sin\theta +3$$

$$\:y=(3-3\sin\theta)\sin\theta $$

$$y=3\sin\theta -3\sin^2\theta $$

$$y'=3\cos\theta -3\cos^2\theta $$

Now that i got the (hopefully) correct derivatives i need to find the tangent line for $\theta =\frac{3\pi }{4}$

so i plug in $\theta =\frac{3\pi }{4}$ into $\frac{3cos\theta -3cos^2\theta }{-3sin\theta +3}$ and got $-\frac{\sqrt{2}}{2}$ for the slope.

How do i get the tangent line though?

Answer 1

Hint: $y - y_0 = m(x - x_0)$ is the equation of the tangent line for any curve. You have $m = -\dfrac{1}{\sqrt{2}}$, and $x_0 = (3-3\sin 3\pi/4)\cos 3\pi/4$....

Answer 2

It is advisable to identify the tangent by the angle $\phi$ that it forms with the vector radius $r$. $\phi=\alpha-\theta$,

$\alpha$ is the angle that the tangent forms with the abscis axis $x$:

$\alpha=atan(\frac{dy}{dx})$,

$\theta$ is radial-angle.

Then calculating the tangent:

$tan(\phi)=\frac{\frac{dy}{dx}-tan(\theta)}{1+\frac{dy}{dx}.tan(\theta)}$,

Moving to polar coordinates:

$tan(\phi)=\frac{r}{\frac{dr}{d\theta}}$.

then

$x=(3-3sin(\theta))cos(\theta)$,

$\:y=(3-3\sin\theta)\sin\theta$,

$r=3-3sin\theta$

To calculate the tangent line

$y-y_{0}=(\frac{dy}{dx})_{x_{0}}(x-x_{0})$

You need the coordinates of the tangent point

$x_{0}=(3-3sin(\frac{3\pi}{4}))cos(\frac{2\pi}{4})=\frac{3}{2}-\frac{3\sqrt{2}}{2}$,

$y_{0}=-x_{0}$

And the value of the angular coefficient calculated in $(x_{0},y_{0})$:

$\frac{dy}{dx}=\frac{tan(\phi)+tan(\theta)}{1-tan(\phi)tan(\theta)}$

Substituting

$\theta=\frac{3\pi}{4}$,

we get:

$m=(\frac{dy}{dx})_{x_{0}}=\frac{cos(\theta)(-1+2sin(\theta))}{-1+sin(\theta)+2cos(\theta)^{2}}$

$m=-1+\sqrt{2}$;

The equation of tangent line is:

$y=(1-\sqrt{2})x+\frac{9\sqrt{2}}{2}-6$.