What is the slope of the polar curve #f(theta) = theta - sec^2theta+costheta # at #theta = (7pi)/12#?

Answer 1

#dy/dx = -6.968 if theta = (7pi)/12#

The slope of the polar curve at #theta = (7pi)/12# is #dy/dx# evaluated in terms of #theta# at #theta = (7pi)/12#.
To find #dy/dx#, we can first find #(dy)/(d theta)# and #(dx)/(d theta)# and use the chain rule for derivatives to determine that:
#dy/dx = (dy)/(d theta) times (d theta)/(dx) = (dy)/(d theta) div (dx)/(d theta)#
#y=r sintheta = (theta - sec^2theta+costheta)(sintheta)# #(dy)/(d theta) = (1 - 2sec^2(theta)tan(theta) - sintheta)(sintheta) + (theta-sec^2theta+costheta)(costheta)#
Evaluating this at #theta = (7pi)/12# gives that #(dy)/(d theta) = 111.12 if theta = (7pi)/12#
#x=r costheta = (theta - sec^2theta + costheta)(costheta)# #(dx)/(d theta) = (1 - 2sec^2(theta)tan(theta) - sintheta)(costheta) + (theta-sec^2theta+costheta)(-sintheta)#
Evaluating this at #theta = (7pi)/12# gives that #(dx)/(d theta) = -15.948 if theta = (7pi)/12#
#(dy)/(d theta) div (dx)/(d theta) = 111.12/(-15.948) = -6.968#
Therefore, the slope of the line tangent to #f(theta)# at the point where #theta = (7pi)/12# is #-6.968#.
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Answer 2

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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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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