find the equation of the tan line tothe cardioid r=1+cos(th)
- Thread starter Smily
- Start date
\(\displaystyle x=rcos({\theta})\)
\(\displaystyle y=rsin({\theta})\)
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.
\(\displaystyle y=rsin({\theta})\)
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.
Re: find the equation of the tan line tothe cardioid r=1+cos
Hello, Smily!
Here's part (a) . . .
\(\displaystyle \L\;\;\frac{dy}{dx}\;=\;\frac{r\cdot\cos\theta\,+\,r'\cdot\sin\theta}{-r\cdot\sin\theta\,+\,r'\cdot\cos\theta}\)
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?
Hello, Smily!
Here's part (a) . . .
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?
