# Thread: tanget line parallel to curve positino

1. ## tanget line parallel to curve positino

Hello, I am trying to solve the following problem:
At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5.

My attempt is:
rewrite 3x-y=5 as y=3x-5
differentiated to 3

Differentiate y=2(x-cosx) = 2(1+sinx) = 2 + 2sinx
so
3 = 2+2sinx
=
1 = 2sinx2
=
1/2 = sinx
=
sin^-1(1/2) = x
= 0.524
plugging back into original curve formula gives a y value of -0.684
so my final answer is "the tangent is parallel to the line 3x-y=5 at the point .524(x), -0.684(y)."

I am unsure if I am correct. Could someone please cast their eyes over this and let me know if I am on the right track ?

2. ## Re: tanget line parallel to curve positino

Looks correct to me.

3. ## Re: tanget line parallel to curve positino

actually I think I got this one wrong.

To find the points on the curve (2(x-cosx) where the tangent is parallel to the line y=3x-5.

differentiating 2(x-cosx) = 3x-5:
2+2sinx = 3

Then solving for x to find where the tangent is parallel to the curve:
2sinx=1
sin^-1 (0.5) = x

At this stage I am not sure what to do next. This trig equation now gives me 30 degrees and 150 degrees ?, but I am trying to find an x and a y position.