What is the slope of the polar curve #f(theta) = theta - sec^2theta+costheta # at #theta = (7pi)/12#?
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To find the slope of the polar curve ( f(\theta) = \theta - \sec^2(\theta) + \cos(\theta) ) at ( \theta = \frac{7\pi}{12} ), we need to compute the derivative ( \frac{df}{d\theta} ) and then evaluate it at ( \theta = \frac{7\pi}{12} ).
Given ( f(\theta) = \theta - \sec^2(\theta) + \cos(\theta) ), we differentiate with respect to ( \theta ):
[ \frac{df}{d\theta} = \frac{d}{d\theta}(\theta) - \frac{d}{d\theta}(\sec^2(\theta)) + \frac{d}{d\theta}(\cos(\theta)) ]
[ = 1 - \frac{d}{d\theta}(\sec^2(\theta)) - \sin(\theta) ]
Using the chain rule and the derivative of ( \sec^2(\theta) ), which is ( 2\sec(\theta)\tan(\theta) ), we get:
[ = 1 - 2\sec(\theta)\tan(\theta) - \sin(\theta) ]
Now, we evaluate this expression at ( \theta = \frac{7\pi}{12} ):
[ = 1 - 2\sec\left(\frac{7\pi}{12}\right)\tan\left(\frac{7\pi}{12}\right) - \sin\left(\frac{7\pi}{12}\right) ]
After evaluating, we have the slope of the polar curve at ( \theta = \frac{7\pi}{12} ).
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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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