Looks correct to me.
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
3 = 2+2sinx
1 = 2sinx2
1/2 = sinx
sin^-1(1/2) = x
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 ?
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:
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.
Can someone help please.
You must work in radians in this question. You were correct with your first solution but you only gave one answer. Did they not ask for answers in a given range. If sinx=1/2 another answer would be x=2.618 (that is 150 degrees in radians)