Home > Calculus > Determining the Slope and Tangent Lines for a Polar Curve

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

Answer 1

Christopher Allers

Starting function: #r(\theta) = 5\theta + cos(-\theta/3 - \pi/2)#

Derivative: #r'(\theta) = (dr(\theta))/(d\theta) = 5 +1/3 sin(-\theta/3-\pi/2)#

Assessing at #\theta = -pi/6#: #r'(-pi/6) = 5+1/3sin(-\pi/18-\pi/2)# #r'(-pi/6) = 5+1/3sin(-(5\pi)/(9)) \approx 4.67#

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Answer 2

Emilia Aguilar

The slope of the tangent line of ( r = 5\theta + \cos\left(-\frac{\theta}{3} - \frac{\pi}{2}\right) ) at ( \theta = -\frac{\pi}{6} ) is ( \frac{dr}{d\theta} ) evaluated at ( \theta = -\frac{\pi}{6} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.