October 17th, 2007, 01:28 PM  #1 
Newbie Joined: Oct 2007 Posts: 1 Thanks: 0  calc help
A car is traveling at night along a highway shaped like a parabola with its vertex at the origin. The car starts at a point 100 m west and 100 m north of the origin and travels inan easterly direction. There is a statue located 100 m east and 50 m north of the origin. At what point on the highway will the car's headlights illuminate the statue? what i have so far: y'(x)=m=(50y)/(100x) delta x=100x delta y=50y delta y=f(x+deltax)f(x) =50y=100x y'=1 y=x50 I dont no how to finish the problem off so and i feel like i am going in circles, so any help would be great! Thanks!!!! 
October 18th, 2007, 05:57 AM  #2 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
If we assume that the car's range of lights is a straight line, then your question is: find the coordinates of the parabola mentioned in your text what, if we put a tangent to the parabola in that coordinates, we will get a line that will have the point of the statue as a solution! So, first, we have to find the function of the car's path. We have: f(x)=a*x^2+b*x+c f(0)=0 > vector is in the origin (1) f'(0)=0 > vector is in the origin (2) f(100)=100 > the car's original point of starting from the first data we get: a*0^2+b*0+c=0 > c=0 Now: f(x)=a*x^2+b*x from the second data we get: f'(x)=2a*x+b f'[0]=b=0 Again: f(x)=a*x^2 and from the third data we get: f8100)=a*100^2=100 > a=1/100 Finally: f(x)=1/100*x^2 Now, the formula for a tangent to a function through a point (x0,y0) is: yy0=f'(x0)(xx0) f'(x0)=x0/50 and since one of the solutions of the tangent has to be (100,50), we get: 50y0=x0/50(100x0) (*) finally, since the tangent has to touch the parabola in x0, we can get: f(x0)=y0=x0^2/100 plugging that in (*) we get: 50x0^2/100=x0/50(100x0) 50x0^2/100=2*x0x0^2/50 x0^2/1002x0+50=0 x0^2200x0+5000=0 solving for x0 we get: x0=50*(2sqrt(2)) and x0=50*(2+sqrt(2)) the reason we get two point is justified, but our solution is only one, and that's: x0=50*(2sqrt(2)) The reason for this is the fact that the car is traveling from the west to the east, and the light are in front of the car. If it would have been the other way, our solution would have been the other one! Now, finally, the point we are looking for is: (x0,f(x0))=(50*(2sqrt(2)),25(2sqrt(2))^2) 
October 18th, 2007, 08:40 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,713 Thanks: 1806 
Does that solution really satisfy the requirement that the car initially travels in an easterly direction?

October 18th, 2007, 08:53 AM  #4 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
Well, I think so.. Would you disagree?

October 20th, 2007, 02:40 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,713 Thanks: 1806 
I would. It starts 100 m west and 100 m north of the origin, yet it reaches the origin. On a straight line route, it would be going southeast. For the parabola considered, the initial direction was much nearer south than east.

October 21st, 2007, 08:00 AM  #6 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0  

Tags 
calc 
Search tags for this page 
use of mathematics in car headlights using parabola,a car is traveling at night along a highway shaped like a parabola
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
calc lim help!  venom21  Calculus  10  October 11th, 2012 10:58 AM 
Calc  henoshaile  Calculus  4  October 6th, 2012 10:59 PM 
Should I skip trig and pre calc and go straight to Calc?  trekster  New Users  5  January 13th, 2012 02:13 PM 
Calc Help  JordanReich  Calculus  14  September 22nd, 2010 08:50 PM 
calc. help  calc. route based on waypoints, angle and speed  miroslav_karpis  Algebra  1  February 16th, 2009 07:00 AM 