What is the slope of the tangent line of #r=theta/3+sin((3theta)/8-(5pi)/3)# at #theta=(7pi)/6#?

Question

Aurora Alvarado · Accepted Answer

#-1.847#

First, let's go ahead and find the values of #r# and #(dr)/(d theta)# at #theta = (7pi)/6#, just to make things easier later.

#r = theta/3 + sin((3theta)/8-(5pi)/3)#

#= (7pi)/18 + sin((21pi)/48 - (5pi)/3)#

#~~ 1.881#

#(dr)/(d theta) = 1/3 + 3/8 cos((3theta)/8 - (5pi)/3)#

#= 1/3 + 3/8cos((21pi)/48 - (5pi)/3)#

#~~ 0.0514#

Now, the slope of the tangent line at any point is #dy/dx#, but the problem is that we don't have #y# and #x#, we have #r# and #theta#.

Luckily, we can apply a version of the chain rule which states that

#dy/dx = (dy"/"d theta)/(dx"/"d theta)#

We will also have to use the rectangular --> polar coordinate formulas:

#x = rcostheta# #y = rsintheta#

Since we have expressions for #x# and #y#, we can use the product rule to derive them with respect to #theta#.

#dy/(d theta) = d/(d theta) rsintheta = (dr)/(d theta)sintheta + rcostheta#

#dx/(d theta) = d/(d theta) rcostheta = (dr)/(d theta)costheta -rsintheta#

With these formulas, we can find the slope of the line at #theta = (7pi)/6#.

#dy/dx = (dy"/"d theta)/(dx"/"d theta) = ((dr)/(d theta)sintheta + rcostheta)/((dr)/(d theta)costheta -rsintheta)#

We already know that at #theta = (7pi)/6#, #r = 1.881# and #(dr)/(d theta) = 0.0514#. All we need to do now is find the values of #sintheta# and #costheta#, and then plug everything in.

#sin((7pi)/6)=-0.5# #cos((7pi)/6) = -0.866#

Now we can find our slope:

#((dr)/(d theta)sintheta + rcostheta)/((dr)/(d theta)costheta -rsintheta) = (0.0514 * (-0.5) + 1.881 * (-0.866))/(0.0514 * (-0.866) - 1.881 * (-0.5))#

#= -1.847#

Final Answer

Isabella Ackerman · Answer

undefined To find the slope of the tangent line of $ r = \frac{\theta}{3} + \sin\left(\frac{3\theta}{8} - \frac{5\pi}{3}\right) $ at $ \theta = \frac{7\pi}{6} $, you need to find the derivative $ \frac{dr}{d\theta} $ and then evaluate it at $ \theta = \frac{7\pi}{6} $.

First, differentiate $ r $ with respect to $ \theta $:

$$ \frac{dr}{d\theta} = \frac{1}{3} + \frac{3}{8}\cos\left(\frac{3\theta}{8} - \frac{5\pi}{3}\right) $$

Now, evaluate $ \frac{dr}{d\theta} $ at $ \theta = \frac{7\pi}{6} $:

$$ \frac{dr}{d\theta}\Bigg|_{\theta = \frac{7\pi}{6}} = \frac{1}{3} + \frac{3}{8}\cos\left(\frac{3}{8}\left(\frac{7\pi}{6}\right) - \frac{5\pi}{3}\right) $$

$$ = \frac{1}{3} + \frac{3}{8}\cos\left(\frac{7\pi}{8} - \frac{5\pi}{3}\right) $$

$$ = \frac{1}{3} + \frac{3}{8}\cos\left(\frac{21\pi}{24} - \frac{40\pi}{24}\right) $$

$$ = \frac{1}{3} + \frac{3}{8}\cos\left(\frac{-19\pi}{24}\right) $$

$$ = \frac{1}{3} + \frac{3}{8}\cos\left(-\frac{19\pi}{24}\right) $$

$$ = \frac{1}{3} + \frac{3}{8}\cos\left(\frac{5\pi}{24}\right) $$

Now, compute the cosine of $ \frac{5\pi}{24} $, then substitute it into the expression:

$$ \cos\left(\frac{5\pi}{24}\right) \approx 0.9659 $$

$$ \frac{dr}{d\theta}\Bigg|_{\theta = \frac{7\pi}{6}} \approx \frac{1}{3} + \frac{3}{8} \times 0.9659 $$

$$ \approx \frac{1}{3} + 0.3629 $$

$$ \approx 0.6959 $$

Therefore, the slope of the tangent line at $ \theta = \frac{7\pi}{6} $ is approximately $ 0.6959 $.