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

Derivative with Polar Coordinates is

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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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