find the equation of the tan line tothe cardioid r=1+cos(th)

[Smily]

find the equation of the tangent line to the cardioid r=1+cos(theta) at the point (theta)=pi/4
what happens to the slope of the cardioid as (theta)--> pi ?

i have no idea how to solve it.can you help me? :roll:

 

[galactus]

$\displaystyle x=rcos({\theta})$

$\displaystyle y=rsin({\theta})$

If we relate polar and rectangular coordinates, we get the parametrics:

$\displaystyle x=f({\theta})cos({\theta})......[1]$ and $\displaystyle y=f({\theta})sin({\theta}).......[2]$

Without going through the derivation, the slope is given by:

\(\displaystyle \L\\\frac{dy}{dx}=\frac{\frac{dy}{d{\theta}}}{\frac{dx}{d{\theta}}}=\frac{rcos({\theta})+sin({\theta})\frac{dr}{d{\theta}}}{-rsin({\theta})+cos({\theta})\frac{dr}{d{\theta}}}\)

In your case, $\displaystyle r=1+cos({\theta})\ and\ \frac{dr}{d{\theta}}=-sin({\theta})$

Now, enter in your $\displaystyle {\theta}=\frac{\pi}{4}$ and find the slope.

Then use the line equation:

y=mx+b, where m is the slope you just found.

From [1] and [2]:

$\displaystyle x=(1+cos({\theta}))cos{\theta})$

$\displaystyle y=(1+cos({\theta}))sin({\theta})$

So, you have:

\(\displaystyle \L\\(1+cos({\theta}))sin({\theta})=m((1+cos({\theta}))(cos({\theta})))+b\)

=\(\displaystyle \L\\\frac{\sqrt{2}}{2}+\frac{1}{2}=m(\frac{\sqrt{2}}{2}+\frac{1}{2})+b\)

Enter in your slope and solve for b.

 

[soroban]

Re: find the equation of the tan line tothe cardioid r=1+cos

Hello, Smily!

Here's part (a) . . .

(a) Find the equation of the tangent line to the cardioid $\displaystyle r\:=\:1\,+\,\cos\theta$ at the point $\displaystyle \theta\,=\,\frac{\pi}{4}$

You are expected to know the Slope Formula for polar functions:

\(\displaystyle \L\;\;\frac{dy}{dx}\;=\;\frac{r\cdot\cos\theta\,+\,r'\cdot\sin\theta}{-r\cdot\sin\theta\,+\,r'\cdot\cos\theta}\)


We have: $\displaystyle \,r\:=\:1\,+\,\cos\theta,\;\;r'\:=\:-\sin\theta$

Then: \(\displaystyle \L\,\frac{dy}{dx}\;=\;\frac{(1\,+\,\cos\theta)\cos\theta\,+\,(-\sin\theta)\sin\theta}{-(1\,+\,\cos\theta)\sin\theta\,+\,(-\sin\theta)\cos\theta} \;=\;\frac{\cos\theta\,+\,\cos^2\theta\,-\,\sin^2\theta}{-\sin\theta\,-\,2sin\theta\cos\theta}\)

$\displaystyle \;\;$Hence: \(\displaystyle \L\,\frac{dy}{dx}\;=\;-\frac{\cos\theta\,+\,\cos2\theta}{\sin\theta\,+\,\sin2\theta}\)


At $\displaystyle \theta\,=\,\frac{\pi}{4}$, we have:
\(\displaystyle \L\;\;\;\frac{dy}{dx} \;= \;-\frac{\cos\frac{\pi}{4}\,+\,\cos\frac{\pi}{2}}{\sin\frac{\pi}{4} \,+ \,\sin\frac{\pi}{2}} \;=\; -\frac{\frac{1}{\sqrt{2}}\,+\,0}{\frac{1}{\sqrt{2}}\,+\,1}\;=\;-\frac{1}{\sqrt{2}\,+\,1}\)

Rationalize: \(\displaystyle \L\,-\frac{1}{\sqrt{2}\,+\,1}\cdot\frac{\sqrt{2}\,-\,1}{\sqrt{2}\,-\,1} \;= \;-\frac{\sqrt{2}\,-\,1}{2\,-\,1}\;=\;-(\sqrt{2}\,-\,1)\;=\;1\,-\,\sqrt{2}\)

And we have the slope of the tangent line . . . can you finish it now?

 

[Smily]

thank you