What is the slope of the line normal to the tangent line of #f(x) = sec^2x-xcos(x-pi/4) # at # x= (7pi)/4 #?

Answer 1

The reqd. slope = #4/((16+7pi)#.

#f(x)=sec^2x-xcos(x-pi/4)# #:.f'(x)=2*secx*secx*tanx-{x*(-sin(x-pi/4))+cos(x-pi/4)} = 2sec^2x*tanx+x*sin(x-pi/4)-cos(x-pi/4)#
But #f'(x)# is the slope tangent (tgt.) line at #(x, f(x))# Hence, slope of the tangent line at #x=(7pi)/4# is #f'((7pi)/4)=2sec^2((7pi)/4)tan((7pi)/4)+((7pi)/4)sin((7pi)/4-pi/4)-cos((7pi)/4-pi/4)=2*2*(-1)+((7pi)/4)*(-1)-0=-4-(7pi)/4=-(16+7pi)/4# As normal is perpendicular to tgt. line, the reqd. slope = #4/((16+7pi)#.
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Answer 2

To find the slope of the line normal to the tangent line of ( f(x) = \sec^2(x) - x \cos(x - \frac{\pi}{4}) ) at ( x = \frac{7\pi}{4} ), we need to first find the derivative of the function and evaluate it at ( x = \frac{7\pi}{4} ) to get the slope of the tangent line. Then, we'll find the negative reciprocal of this slope to get the slope of the normal line.

  1. Find the derivative of ( f(x) ): [ f'(x) = 2\sec(x)\tan(x) + \cos(x - \frac{\pi}{4}) + x\sin(x - \frac{\pi}{4}) ]

  2. Evaluate ( f'(x) ) at ( x = \frac{7\pi}{4} ): [ f'\left(\frac{7\pi}{4}\right) = 2\sec\left(\frac{7\pi}{4}\right)\tan\left(\frac{7\pi}{4}\right) + \cos\left(\frac{7\pi}{4} - \frac{\pi}{4}\right) + \frac{7\pi}{4}\sin\left(\frac{7\pi}{4} - \frac{\pi}{4}\right) ]

  3. Simplify ( f'\left(\frac{7\pi}{4}\right) ): [ f'\left(\frac{7\pi}{4}\right) = 2\sec\left(\frac{7\pi}{4}\right)\tan\left(\frac{7\pi}{4}\right) + \cos\left(\frac{3\pi}{2}\right) + \frac{7\pi}{4}\sin\left(\frac{3\pi}{2}\right) ] [ f'\left(\frac{7\pi}{4}\right) = 2(-1)(1) + 0 + \frac{7\pi}{4}(-1) ] [ f'\left(\frac{7\pi}{4}\right) = -2 - \frac{7\pi}{4} ]

  4. This is the slope of the tangent line. Now, find the negative reciprocal to get the slope of the normal line: [ \text{Slope of normal line} = -\frac{1}{f'\left(\frac{7\pi}{4}\right)} = -\frac{1}{-2 - \frac{7\pi}{4}} = \frac{1}{2 + \frac{7\pi}{4}} = \frac{1}{2} - \frac{\pi}{8} ]

So, the slope of the line normal to the tangent line of ( f(x) ) at ( x = \frac{7\pi}{4} ) is ( \frac{1}{2} - \frac{\pi}{8} ).

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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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