Calc Help!?

Question

A car is traveling at night along a highwa 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!!!!

Answer 1

First find the equation of the parabola. The vertex is the orgin and it passes thru the point P(-100, 100). The equation of the parabola is:

y = x²/100

The slope of the parabola is the slope of the tangent line.

dy/dx = 2x/100 = x/50

Now use the point slope formula to find the line that passes thru the point of the statue Q(100, 50).

y - 50 = m(x - 100)
x²/100 - 50 = (x/50)(x - 100)
x² - 5000 = 2x(x - 100) = 2x² - 200x
0 = x² - 200x + 5000

x = {200 ± √[(-200)² - 4*1*5000]} / (2*1)
x = {200 ± √[40,000 - 20,000]} / 2
x = {200 ± √20,000} / 2
x = {200 ± 100√2} / 2
x = 100 ± 50√2

There are two tangent lines but only one results in the car's headlights, rather than tail lights, hitting the statue. The car is going east and so must be west of the statue to light it.

x = 100 - 50√2 = 50(2 - √2) ≈ 29.289322

y = x²/100 = (100 - 50√2)²/100 = (15,000 - 10,000√2)/100
y = 150 - 100√2 = 50(3 - 2√2) ≈ 8.5786438

The point on the road at which the car's headlights illuminate the statue is:

(x,y) = (100 - 50√2, 150 - 100√2)

Answer 2

Ok so first I found the equation of the parabola.
y=ax^2
100=a*(100)^2
a=1/100
Therefore: y=0.01x^2

The question is when do the car headlights light up the statue. The thing to think about is, how is the direction of the car headlight related to the highway? And the answer is that it is TANGENT to it.
Get out a piece of graph paper and draw the parabola and draw a point where the statue is. Draw a straight line from point (100,50) to the graph, approximately where the straight line is tangent to the graph. This is where the car will be when its lights illuminate the statue. So now you have a rough idea of where the car will be if you drew this accurately. My guess was around x=40 from my sketch. So this is a good starting point.

Basically the main thing to figure out is, the slope of the straight line that you drew must be the same as the slope of the parabola at that point.

Slope of the parabola:
dy/dx (f(x)) = dy/dx (0.01x^2) = 0.02*x

Slope of the line:
rise / run = (50 - Y) / (100 - X)
where (X,Y) will be the coordinates of the car on the parabola.

So we know that these 2 slopes have to be equal. We have

0.02*X = (50-Y) / (100-X)
you just need to relate Y and X now. But it must be a point on the parabola so Y = 0.01*X^2 !!!!

0.02*X = (50 - 0.01X^2) / (100-X)
multiply through:
0.02(100X-X^2) = 50 - 0.01X^2
100X - X^2 = 2500 - 0.5X^2
0.5X^2 - 100X + 2500 = 0

so this is just a quadratic equation you can solve.
the roots turn out to be
X = 29.289, 170.711

The only answer that makes sense is 29.289. So the car will illuminate the statue at the point (28.289, 0.01*28.289^2)

(28.289,8.579)

Answer 3

86.5mph