October 25th, 2018, 11:38 AM  #11  
Member Joined: Oct 2018 From: Netherlands Posts: 39 Thanks: 3  Quote:
I guess the only escape from that is to assume that the coordinate values are not the same as the meters. Last edited by Arisktotle; October 25th, 2018 at 11:44 AM.  
October 25th, 2018, 01:03 PM  #12 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
The problem is complete as given. y=ax^2 is the shape of the track. The parabola has to be shifted along x axis so that x=100m (W) and y=100m (N) is a point on the track. Then ck the sign is outside the track and find the line from sign tangent to track (equate slope of line through sign to slope of a point on parabola. EDIT: To translate, y=a(xk)^2 and 100,100 is a point on the curve gives k. Last edited by zylo; October 25th, 2018 at 01:38 PM. 
October 25th, 2018, 01:58 PM  #13 
Member Joined: Oct 2018 From: Netherlands Posts: 39 Thanks: 3  
October 25th, 2018, 02:18 PM  #14 
Global Moderator Joined: Dec 2006 Posts: 19,878 Thanks: 1835  
October 26th, 2018, 06:08 AM  #15 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115  
October 26th, 2018, 08:50 AM  #16  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115  Quote:
x'=xcost+ysint y'=xsint+ycost y'=ax'^2 Find t for case y=x and then let x=100. x(sint+cost)=ax^2(costsint)^2 (sint+cost)/(costsint)^2=ax=18 Couldn't solve it.  
October 26th, 2018, 10:10 AM  #17 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
There is a way to find rotation angle. Let R,$\displaystyle \phi$ = (sqrt2)100,135 be the location of 100.100 in polar coordinates. In polar coordinates, x=rcos$\displaystyle \theta$, y=rsin$\displaystyle \theta$ and y=ax^2 becomes sin$\displaystyle \theta$=(ar)cos^2$\displaystyle \theta$=ar(1sin^2$\displaystyle \theta$) substitute a=.18, R=(sqrt2)100 and solve for $\displaystyle \theta$. Then the rotation angle is 135$\displaystyle \theta$ Now you can find equation of rotated track in x,y coordinates, ck sign is outside track, and find tangent to parabola from line through sign. EDIT Better yet. Solve the problem in the rotated coordinate system. Then all you need is the coordinates of the sign in the rotated coordinate system and y'=ax'^2. Last edited by zylo; October 26th, 2018 at 10:33 AM. 
October 26th, 2018, 12:34 PM  #18 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
Solving for $\displaystyle \theta$ gives that a radius of (sqrt2)100 intersects y=ax^2 at 78 and 102deg. To 100,100 the parabola has to be rotated 135102 =33deg ccw. In a coordinate system rotated $\displaystyle \alpha$ = 33deg ccw the coordinates of the sign become: x'=xcos$\displaystyle \alpha$+ysin$\displaystyle \alpha$=5cos33+4sin33=6.37 y'=ycos$\displaystyle \alpha$xsin$\displaystyle \alpha$=4cos335sin33=.631 In the new coordinate system, dropping primes, the problem becomes y=ax^2 with sign at 6.37,.631 At x=6.37 distance to track is .18(6.37)^2= 7.30 and sign is outside track. Final step is to find line through sign tangent to track, and give result in original coordinate system. OP can do that. 
October 30th, 2018, 04:22 PM  #19 
Newbie Joined: Oct 2018 From: Deep Outta Space Posts: 4 Thanks: 0 
The equation is broken because the sign is inside the parabola therefore it's never going to touch.
Last edited by skipjack; October 30th, 2018 at 11:23 PM. 
November 2nd, 2018, 06:08 PM  #20 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 
Equation of curve in rotated coordinate system: y=ax$\displaystyle ^{2}$ Coordinates of signal in rotated coordinate system: (r,s)=(6.37,.631) See my previous post. Slope of curve = dy/dx = 2ax Equation of straight line through signal: y=mx+b s=mr+b $\displaystyle \rightarrow$ b=smr y=mx+smr This line is also tangent to curve at x$\displaystyle _{0}$, y$\displaystyle _{0}$ where m=2ax$\displaystyle _{0}$ and y$\displaystyle _{0}$=ax$\displaystyle _{0}^{2}$ $\displaystyle ax_{0}^{2}= (2ax_{0})x_{0}+s2ax_{0}r$ This gives two tangent points from signal as expected. (solve for x$\displaystyle _{0}$ then y$\displaystyle _{0}$=ax$\displaystyle _{0}^{2}$) 

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