My Math Forum calc help

 Calculus Calculus Math Forum

 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=(50-y)/(100-x) delta x=100-x delta y=50-y delta y=f(x+deltax)-f(x) =50-y=100-x y'=1 y=x-50 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: f8-100)=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: y-y0=f'(x0)(x-x0) f'(x0)=x0/50 and since one of the solutions of the tangent has to be (100,50), we get: 50-y0=x0/50(100-x0) (*) 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: 50-x0^2/100=x0/50(100-x0) 50-x0^2/100=2*x0-x0^2/50 x0^2/100-2x0+50=0 x0^2-200x0+5000=0 solving for x0 we get: x0=50*(2-sqrt(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*(2-sqrt(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*(2-sqrt(2)),25(2-sqrt(2))^2)
 October 18th, 2007, 08:40 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,286 Thanks: 1681 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: 19,286 Thanks: 1681 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 south-east. 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 This is how I pictured it! The red arrow represents the original car direction, the green one the solution I find to be true, and the blue one the one I find to be wrong! (The parabola, of course represents the car's path) [/url]

 Tags calc

,

### 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 Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post venom21 Calculus 10 October 11th, 2012 10:58 AM henoshaile Calculus 4 October 6th, 2012 10:59 PM trekster New Users 5 January 13th, 2012 02:13 PM JordanReich Calculus 14 September 22nd, 2010 08:50 PM miroslav_karpis Algebra 1 February 16th, 2009 07:00 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top