What is the slope of #r=tantheta-theta# at #theta=pi/8#?

Answer 1

Derivative with Polar Coordinates is # (\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)#

#(\partial r) / (\partial theta) = 1/cos^2(theta)+1# #r(theta = pi/8)=0.0 22# # (\partial x) / (\partial y) = (tan(theta)/cos(theta)+sin(theta)+rcos(theta))/(1/cos(theta)+cos(theta)-rsin(theta))# # (\partial x) / (\partial y)(theta=pi/8) =1.57 #
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Answer 2

To find the slope of ( r = \tan(\theta) - \theta ) at ( \theta = \frac{\pi}{8} ), we need to differentiate ( r ) with respect to ( \theta ) and then evaluate the result at ( \theta = \frac{\pi}{8} ). The derivative of ( r ) with respect to ( \theta ) is ( \frac{d}{d\theta}(\tan(\theta) - \theta) ). The derivative of ( \tan(\theta) ) is ( \sec^2(\theta) ), and the derivative of ( -\theta ) is ( -1 ). So, the derivative of ( r ) is ( \sec^2(\theta) - 1 ). Evaluating this at ( \theta = \frac{\pi}{8} ) gives ( \sec^2(\frac{\pi}{8}) - 1 ), which can be simplified to ( 2 - \sqrt{2} ). So, the slope of ( r = \tan(\theta) - \theta ) at ( \theta = \frac{\pi}{8} ) is ( 2 - \sqrt{2} ).

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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.

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.

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.

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.

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