# What is the slope of the tangent line of #r=5theta+cos(-theta/3-(pi)/2)# at #theta=(-5pi)/6#?

slope would be

Formula for slope in polar coordinates is

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To find the slope of the tangent line at a given point on a polar curve, you can first find the derivative of the polar equation with respect to θ. Then, evaluate the derivative at the given value of θ to find the slope of the tangent line at that point.

The derivative of the polar equation ( r = 5\theta + \cos\left(-\frac{\theta}{3} - \frac{\pi}{2}\right) ) with respect to ( \theta ) is:

( \frac{dr}{d\theta} = 5 - \frac{1}{3}\sin\left(-\frac{\theta}{3} - \frac{\pi}{2}\right) )

Evaluate ( \frac{dr}{d\theta} ) at ( \theta = -\frac{5\pi}{6} ) to find the slope of the tangent line:

( \frac{dr}{d\theta}\bigg|_{\theta = -\frac{5\pi}{6}} = 5 - \frac{1}{3}\sin\left(-\frac{-5\pi}{6 \times 3} - \frac{\pi}{2}\right) )

( = 5 - \frac{1}{3}\sin\left(\frac{5\pi}{18} - \frac{\pi}{2}\right) )

( = 5 - \frac{1}{3}\sin\left(\frac{5\pi}{18} - \frac{9\pi}{18}\right) )

( = 5 - \frac{1}{3}\sin\left(\frac{-4\pi}{18}\right) )

( = 5 - \frac{1}{3}\sin\left(-\frac{2\pi}{9}\right) )

Since ( \sin(-\theta) = -\sin(\theta) ), we have:

( = 5 + \frac{1}{3}\sin\left(\frac{2\pi}{9}\right) )

Now, calculate the value of ( \sin\left(\frac{2\pi}{9}\right) ), which is approximately ( \sin\left(40^\circ\right) ).

Substitute the value into the equation:

( \frac{dr}{d\theta}\bigg|_{\theta = -\frac{5\pi}{6}} \approx 5 + \frac{1}{3}\sin\left(40^\circ\right) )

Calculate the value, and that will be the slope of the tangent line at ( \theta = -\frac{5\pi}{6} ).

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