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Forum nameThe Pits v3.1
Topic subjectPrecalc/Calc help
Topic URLhttps://forums.2gnt.com/dcboard.php?az=show_topic&forum=i60&topic_id=26379
26379, Precalc/Calc help
Posted by edxmon, Aug-01-07 01:57 PM
I have an extra credit problem that goes like this:

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 meters west and 100 meters north of the origin and travels in an easterly direction. There is a statue located 100 meters east and 50 meters north of the origin. At what point on the highway will the car's headlights illuminate the statue?

Here is what I know:

-I need to find the derivative of the equation for the parabola to find the equation of the line tangent to it, and that goes through point (100,50) the location of the statue.

-The car starts at point (-100,100)

But for the life of me I can not come up with the equation to model this parabola that goes through point (100,100)!

What equation in the form y=ax^2, will give you such point? y=100 when x=-100?! I realize it must be pretty damn wide thus the value of "a" needs to be small. Ive been doing math all day and I am rather tired, I need a little help, I think I'm brain drained. :(
26398, RE: Precalc/Calc help
Posted by Nine5eclipse03, Aug-02-07 02:40 PM
:scratch

Got it. Its easy dude, when you really think about it, the answer is:

Yes.
26450, RE: Precalc/Calc help
Posted by MCubed45, Aug-05-07 03:52 PM
a=1/100

100 = 1/100 * (-100)^2
y=1/100 x^2

finding the point is a little more complicated though. you're gonna use the deriviate (slope) and location to find an equation of a line that intersects the parabola at that point AND the point (100,50)
26451, RE: Precalc/Calc help
Posted by MCubed45, Aug-05-07 04:08 PM
x = 100 - 50^(.5)

i think. i gotta double check but i'm pretty sure that's the solution.
26455, RE: Precalc/Calc help
Posted by edxmon, Aug-05-07 04:54 PM
Once I have the line tangent to the parabola I just make them equal to each other to find the point "x" at which both functions have the same value. Once I have the point "x" I just plug that in to the equation to either one and find the point "y". Thanks for the help Mcubed :thumbsup
26458, RE: Precalc/Calc help
Posted by MCubed45, Aug-05-07 06:15 PM
yup yup. you basically have two points on a line in the form y=mx+b.

the first point is (100,50):

50=m100+b

you know the slope of this line is the derivative of the parabola at that point. therefore y'=x/50 = m

50 = (x/50)100+b = 2x + b

-------------------------------------------------
the second point on your line is the point which lies on the parabola:
(x, (1/100)x^2)

this gives you the second equation:
(1/100)x^2 = mx +b

again we substitute m:
(1/100)x^2 = (x/50)x + b
0 =(1/100)x^+b

-----------------------------------------------
you now have 2 equations with 2 unknowns. get rid of b and solve for x. you end up with a quadratic and 2 possible real values for x. you take the one that lies in the region 0<x<100. the other solution comes from when the TAILights are pointing at the statue. the tangent at this point also intersects the statue but that's not what the problem asked for.
26492, RE: Precalc/Calc help
Posted by edxmon, Aug-07-07 03:25 PM
You rock Mc! Ima post up some more of my math woes when I have any, if thats okay with you I mean, this differentiation/integration justs gets more and more hardcore. God bless :thumbsup