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What is the slope of the polar curve #f(theta) = -4theta - thetacos^2theta # at #theta = (15pi)/8#?

Answer 1

Samantha Adkison

Slope of the polar curve at # theta = ((15 pi)/8 )# is # -9.02#

#f(theta) = -4 theta - theta cos^2 theta#

# theta = (15 pi)/8 =337.5^0#

#f^'(theta) = -4 - (-theta* 2 cos theta sin theta+cos^2 theta) #

#f^'(theta) = -4 +theta* 2 cos theta sin theta- cos^2 theta #

#f^'((15 pi)/8)= -4 +theta* sin 2 theta- cos^2 theta #

#f^'((15 pi)/8)= -4 +(15 pi)/8* sin ((15 pi)/4)- cos^2 ((15 pi)/8) #

#f^'((15 pi)/8)= -9.02(2 dp)#

Slope of the polar curve at # theta = ((15 pi)/8 )# is # -9.02# [Ans]

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Answer 2

Isaiah Allen

To find the slope of the polar curve (f(\theta) = -4\theta - \theta \cos^2(\theta)) at (\theta = \frac{15\pi}{8}), we first find the derivative (f'(\theta)) and then evaluate it at the given angle.

The derivative of (f(\theta)) is (f'(\theta) = -4 - \cos^2(\theta) + 2\theta \cos(\theta) \sin(\theta)).

Now, plug in (\theta = \frac{15\pi}{8}) into (f'(\theta)) to find the slope at that angle.

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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.