Calculus Question

Question

Find the point on the curve y = sqrt 2x-1 where the tangent line is parallel to the line x-3y=16.

Help!?

Answer 1

Put the line into the y = mx+b format where m is the slope:

 

      x - 3y = 16

     -3y = -x + 16

     y = (1/3)X - 16/3

 

The slope of this line is 1/3.  A tangent line to y = (2x-1)^1/2 will be parallel if it has the same slope (1/3).  The slope of the tangent line is equal to the derivative of y = (2x-1)^1/2.  Use the chain  rule and the power rule to find the derivative.  Let u = 2x-1:

 

     dy/dx = dy/du · du/dx = du^1/2/du · d(2x-1)/dx = (1/2)u^-1/2 · 2 = (2x-1)^-1/2

 

We want:

     1/3 = 1/(2x-1)^1/2

     (2x-1)^1/2 = 3

     2x-1 = 9

     2x = 10

     x = 5

 

     y = (2·5-1)^1/2 = 9^1/2 = 3

 

The point is (5,3).

Answer 2

y = √(2x-1) is assumed (I need parentheses--is this what you intended?

y' = (2x-1)^-½ = 0 is always positive on its domain x > ½.

m = 1/3 is desired.

1/3 = 1/√(2x-1) 

squaring & equating denominators gives 9 = 2x-1 so x = 5. Note: find the y coordinate by substituting into the curve's equation. At one time, I got off the curve, mistaking the linear equation for that of the tangent line. Earlier error removed by editing. KS