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(xcosx) is the tangent parallel to the line 3xy=5. 1. rewrite 3xy=5 as y3x5 2. equate 2(xcosx) = y3x5 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(xcosx) is the tangent parallel to the line 3xy=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:
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(xcosx) therefore y = 2(?/6 + 2k?  ?(3)/2 + 2k?) instead of y = 2(?/6 + 2k?  ?(3)/2) ? Quote:
 
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 that is what you would use for cos(x) in the expression.


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