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March 31st, 2012, 12:52 PM   #1
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points on curve parallel to line

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

1. rewrite 3x-y=5 as y-3x-5
2. equate 2(x-cosx) = y-3x-5
3. differentiate: 2+2sinx = 3
4. solve for x: sin^-1(,5) = 0.524
5. My answer would be "x=pi/6 +2pi*n" and y=????????

So, as this is a trig equation I would have an infinite number of answers as per my answer above in 5.
My problem is I am not sure how I should express the y values in order to answer the original question "At what point(s) on the curve y=2(x-cosx) is the tangent parallel to the line 3x-y=5".

Any help would be much appreciated.
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March 31st, 2012, 01:19 PM   #2
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Re: points on curve parallel to line

?/6 + 2k?, k ? Z is correct. You also have 5?/6 + 2k?, k ? Z; since sin(5?/6) = 1/2.

Using ?/6, y = 2(?/6 - ?(3)/2) = ?/3 - ?(3), since cos(?/6) = ?(3)/2.
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March 31st, 2012, 02:23 PM   #3
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Re: points on curve parallel to line

thanks for that.
I presume I would also need to provide a y answer for the 5?/6 equation also ?

So my total answer would be

x= ?/6 + 2k?, y= 2(?/6 - ?(3)/2)
or
x= 5?/6 + 2k?, y= 2(5?/6 - cos5?/6)

Would that be correct ?
Thanks kindly
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March 31st, 2012, 05:07 PM   #4
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Re: points on curve parallel to line

Quote:
Originally Posted by fran1942
x= ?/6 + 2k?, y= 2(?/6 - ?(3)/2)
You are expected to substitute the value you have for x into 2(x - cos(x)), as follows,

x = ?/6 + 2k?, y = 2(?/6 + 2k? - ?(3)/2).

Apply the same idea to the remaining equation.
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April 2nd, 2012, 07:07 PM   #5
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Re: points on curve parallel to line

thank you. I understand but I am unclear on why you inserted the repetition factor 2k? only once in the equation for y.
As x = x = ?/6 + 2k?, wouldn't you insert that whole equation wherever x appears in the original function
eg. original function = 2(x-cosx)
therefore y = 2(?/6 + 2k? - ?(3)/2 + 2k?)
instead of y = 2(?/6 + 2k? - ?(3)/2) ?

Quote:
Originally Posted by greg1313
Quote:
Originally Posted by fran1942
x= ?/6 + 2k?, y= 2(?/6 - ?(3)/2)
You are expected to substitute the value you have for x into 2(x - cos(x)), as follows,

x = ?/6 + 2k?, y = 2(?/6 + 2k? - ?(3)/2).

Apply the same idea to the remaining equation.
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April 2nd, 2012, 08:48 PM   #6
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Re: points on curve parallel to line

Since that is what you would use for cos(x) in the expression.
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