My Math Forum points on curve parallel to line

 Calculus Calculus Math Forum

 March 31st, 2012, 12:52 PM #1 Member   Joined: Feb 2012 Posts: 42 Thanks: 0 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.
 March 31st, 2012, 01:19 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond 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.
 March 31st, 2012, 02:23 PM #3 Member   Joined: Feb 2012 Posts: 42 Thanks: 0 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
March 31st, 2012, 05:07 PM   #4
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,968
Thanks: 1152

Math Focus: Elementary mathematics and beyond
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.

April 2nd, 2012, 07:07 PM   #5
Member

Joined: Feb 2012

Posts: 42
Thanks: 0

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.

 April 2nd, 2012, 08:48 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: points on curve parallel to line Since $\cos$$\frac{\pi}{6}+2k\pi$$=\cos$$\frac{\pi}{6}$$= \frac{\sqrt{3}}{2}$ that is what you would use for cos(x) in the expression.

 Tags curve, line, parallel, points

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jaredbeach Calculus 4 May 12th, 2017 09:47 AM Shamieh Calculus 3 October 8th, 2013 08:06 PM tx3 Algebra 4 May 23rd, 2013 11:47 PM tomkoolen Geometry 0 January 24th, 2013 02:42 PM tankertert Calculus 3 May 8th, 2012 12:02 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top