Tangent Line for Polar Coordinates

[KoryME]

I am having trouble figuring out how to start this problem.

"Find the equation of the tangent line in cartesian coordinates of the curve given in polar coordinates by: r=10cos(theta) at (theta)=pi/6."

I think I might need to convert to rectangular form first but I'm not sure... Thanks in advance!

 

[Deleted member 4993]

I am having trouble figuring out how to start this problem.

"Find the equation of the tangent line in cartesian coordinates of the curve given in polar coordinates by: r=10cos(theta) at (theta)=pi/6."

I think I might need to convert to rectangular form first but I'm not sure... Thanks in advance!

For this problem, that would be probably easiest way to attack.

The expression for \(\displaystyle \frac{dy}{dx}\) related to a polar graph is given at:

http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx

 

[mmm4444bot]

I think I might need to ...

Did you try yet what you're thinking?

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[galactus]

\(\displaystyle r=10cos\theta\)

Converting to cartesian:

\(\displaystyle \sqrt{x^{2}+y^{2}}=\frac{10x}{\sqrt{x^{2}+y^{2}}}\)

\(\displaystyle y=\pm\sqrt{10x-x^{2}}\)

Since \(\displaystyle x=rcos\theta, \;\ y=rsin\theta, \;\ r=10cos\theta\), you can use \(\displaystyle \theta=\frac{\pi}{6}\) to find the point of tangency for the line, its slope, and thus the equation.

 

[soroban]

Hello, KoryME!

Find the equation of the tangent line in cartesian coordinates of the curve
given in polar coordinates by: .\(\displaystyle r\:=\:10\cos\theta\,\text{ at }\theta =\frac{\pi}{6}\)

We can work it without converting to rectangular, but it will take some work.

\(\displaystyle x \:=\:r\cos\theta \:=\: (10\cos\theta)\cos\theta \:=\:10\cos^2\theta\)
\(\displaystyle y \:=\:r\sin\theta \:=\: (10\cos\theta)\sin\theta \:=\:10\sin\theta\cos\theta \)

\(\displaystyle \text{At }\theta = \frac{\pi}{6}\!:\; x\:=\:10\cos^2\!\left(\frac{\pi}{6}\right)^2 \:=\:10\left(\frac{\sqrt{3}}{2}\right)^2 \:=\:10\left(\frac{3}{4}\right) \quad\Rightarrow\quad x \:=\:\frac{15}{2}\)

\(\displaystyle \text{At }\theta \,=\,\frac{\pi}{6}\!:\,y \:=\:10\sin\left(\frac{\pi}{6}\right)\cos\left( \frac{\pi}{2}\right) \;=\; 10\cdot\frac{1}{2}\cdot\frac{\sqrt{3}}{2} \quad\Rightarrow\quad y\;=\;\frac{5\sqrt{3}}{2} \)

We have the point: .\(\displaystyle \left(\frac{15}{2},\:\frac{5\sqrt{3}}{2}\right)\)


We can determine the slope like this: .\(\displaystyle \begin{Bmatrix}r &=& 10\cos\theta \\ r' &=& \text{-}10\sin\theta \end{Bmatrix}\)

\(\displaystyle \dfrac{dy}{d\theta} \:=\:r\cos\theta + r'\sin\theta \:=\: (10\cos\theta)\cos\theta + (\text{-}10\sin\theta)\sin\theta \:=\:10(\cos^2\theta-\sin^2\theta)\)

\(\displaystyle \dfrac{dx}{d\theta} \:=\: \text{-}r\sin\theta + r'\cos\theta \:=\: \text{-}(10\cos\theta)\sin\theta + (\text{-}10\sin\theta)\cos\theta \:=\: \text{-}20\sin\theta\cos\theta\)

\(\displaystyle \text{Then: }\:\dfrac{dy}{dx} \:=\:\dfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \:=\:\dfrac{10(\cos^2\theta - \sin^2\theta)}{-20\sin\theta\cos\theta} \:=\:\dfrac{10\cos2\theta}{-10\sin2\theta} \:=\:-\cot2\theta \)

. . \(\displaystyle \text{At }x = \frac{\pi}{6}\!:\;\dfrac{dy}{dx} \:=\:-\cot\frac{\pi}{3} \:=\:-\sqrt{3}\)


\(\displaystyle \text{We have: }\:\begin{Bmatrix}\text{Point:} & \left(\frac{15}{2},\,\frac{5\sqrt{3}}{2}\right) \\ \text{Slope:} & -\sqrt{3} \end{Bmatrix}\)

\(\displaystyle \text{Write the equation of the tangent line . . .}\)