Find k such that "function" is normal to the curve

[oleaha]

Hello!

I'm having problems figuring out how to solve this question. There is two main questions here for me, what does the question ask for and how to derivate absolute values.

Question:
Find k such that y-16x=k is normal to the curve y = 1/abs(x-1)

 

[Bob Brown MSEE]

Hello!

how to derivate absolute values.

Question:
Find k such that y-16x=k is normal to the curve y = 1/abs(x-1)

Break into 2 problems:

y(x)= 1/(x-1) for x>1
y(x)= 1/(1-x) for x<1

 

[DrPhil]

Question:
Find k such that y-16x=k is normal to the curve y = 1/abs(x-1)

Break into 2 problems:

y(x)= 1/(x-1) for x>1
y(x)= 1/(1-x) for x<1

Something is wrong with the question: the slope of the line is 16, and no possible value of \(\displaystyle k\) will change that. The best you could do is find a value of \(\displaystyle x\) such that the slope of the curve is -1/16.

Do you have the question copied correctly?

EDIT: I suppose you could choose \(\displaystyle k\) so that the line passes through that point.

 

[HallsofIvy]

Hello!

I'm having problems figuring out how to solve this question. There is two main questions here for me, what does the question ask for and how to derivate absolute values.

Question:
Find k such that y-16x=k is normal to the curve y = 1/abs(x-1)

As Bob Brown and DrPhil have said, do this as two separate problems: for x> 1 and x< 1. If x>1 y= 1/(x- 1)= (x- 1)^{-1} and y'= -(x- 1)^{-2} while if x< 1, y= 1/(-(x-1))= 1/(1- x)= (1- x)^{-1} and y'= (1- x)^{-2}.

y= 16x- k has slope 16 and will be normal to this curve, for x> 1, when y'= -(x- 1)^{-2}= -1/16 which implies (x- 1)^2= 16. x- 1= 4. x= 5. (We don't need to consider x- 1= -4 because this is "for x> 1"). When x= 5, y= 1/(5-1)= 1/4 so we must have y= 16(5)- k= 80- k= 1/4. Solve that for k.

For x< 1, we must have y'= (1- x)^{-2}= -1/16 which implies (1- x)^2= -16 which is impossible if x is a real number.

That is, y= 16x- k is normal to 1/|x-1|^2 if and only if 80- k= 1/4.