What is the equation of the tangent line of #r=theta-sin(-theta+(pi)/3) # at #theta=(2pi)/3#?

Question

Emma Aldridge · Accepted Answer

#y = -0.48(x - sqrt(3)/4 + (2pi)/6) + (sqrt(3)pi)/3 - 3/4#

We need the Cartesian coordinates of the point of tangency:

#r((2pi)/3) = (2pi)/3 -sin(-(2pi)/3 + pi/3) = (2pi)/3 - sqrt(3)/2#

#x = rcos(theta) and y = rsin(theta)#

#x = ((2pi)/3 - sqrt(3)/2)cos((2pi)/3) and y = ((2pi)/3 - sqrt(3)/2)sin((2pi)/3)#

#x = sqrt(3)/4 - (2pi)/6 and y = (sqrt(3)pi)/3 - 3/4#

For the slope of the line we need #dy/dx# in polar form evaluated at #theta = (2pi)/3#. This reference Tangent Lines to polar equations gives us the equation:

#dy/dx = ((dr)/(d""theta)sin(theta) + rcos(theta))/((dr)/(d""theta)cos(theta) - rsin(theta)#

#(dr)/(d""theta) = 1 + cos( -theta + pi/3)#

Evaluated at #theta = (2pi)/3#

#(dr)/(d""theta) = 3/2 #

The slope, m, of the tangent line is:

#m = ((3/2)sin((2pi)/3) + ((2pi)/3 - sqrt(3)/2)cos((2pi)/3))/((3/2)cos((2pi)/3) - ((2pi)/3 - sqrt(3)/2)sin((2pi)/3)#

#m = ((3/2)(sqrt3/2) + ((2pi)/3 - sqrt(3)/2)(-1/2))/((3/2)(-1/2) - ((2pi)/3 - sqrt(3)/2)(sqrt3/2)#

#m = -(9sqrt(3) + sqrt(3) - 2pi)/(9 + (2pi - sqrt(3))(sqrt3)#

#m ~~ - 0.48#

Use the point-slope form of the equation of a line:

#y = m(x - x_0) + y_0#

#y = -0.48(x - sqrt(3)/4 + (2pi)/6) + (sqrt(3)pi)/3 - 3/4#

Charles Anctil · Answer

undefined To find the equation of the tangent line of $r = \theta - \sin\left(-\theta + \frac{\pi}{3}\right)$ at $\theta = \frac{2\pi}{3}$, we need to find the slope of the tangent line at that point and the point where it touches the curve.

First, differentiate $r$ with respect to $\theta$ to find the slope of the tangent line:
$$\frac{dr}{d\theta} = 1 - (-\cos(-\theta + \frac{\pi}{3}))$$
$$\frac{dr}{d\theta} = 1 + \cos(\theta - \frac{\pi}{3})$$

Evaluate $\frac{dr}{d\theta}$ at $\theta = \frac{2\pi}{3}$:
$$\frac{dr}{d\theta} \Bigg|_{\theta = \frac{2\pi}{3}} = 1 + \cos\left(\frac{2\pi}{3} - \frac{\pi}{3}\right) = 1 + \cos(\frac{\pi}{3}) = 1 + \frac{1}{2} = \frac{3}{2}$$

This is the slope of the tangent line at $\theta = \frac{2\pi}{3}$.

Next, plug $\theta = \frac{2\pi}{3}$ into the equation $r = \theta - \sin\left(-\theta + \frac{\pi}{3}\right)$ to find the corresponding $r$ value:
$$r \Bigg|_{\theta = \frac{2\pi}{3}} = \frac{2\pi}{3} - \sin\left(-\frac{2\pi}{3} + \frac{\pi}{3}\right) = \frac{2\pi}{3} - \sin(-\pi) = \frac{2\pi}{3}$$

So, at $\theta = \frac{2\pi}{3}$, $r = \frac{2\pi}{3}$.

Now, we have the slope ($m = \frac{3}{2}$) and a point ($\frac{2\pi}{3}, \frac{2\pi}{3}$) on the tangent line. Using the point-slope form of the equation of a line, we can write the equation of the tangent line:
$$y - y_1 = m(x - x_1)$$
$$y - \frac{2\pi}{3} = \frac{3}{2}\left(x - \frac{2\pi}{3}\right)$$

This equation can be simplified further if desired.