HELP! ASAP!! A rally car in cross country race is travelling at night.

Oct 2018
4
0
Deep Outta Space
A rally car in cross country race is travelling, at night, along a section of the track that is in the shape of a parabola. The function can be modelled by y = ax^2, where a = 0.14.

The car starts at a point 100m west and 100m north of the apex of the corner; it travels in an easterly direction with its headlights on.

There is a road sign situated at 5m east of the apex of the corner and 4m north of this point.

At what point on the track will the cars headlights illuminate the road sign?

Been stuck for ages with no luck. :(
 
Last edited by a moderator:

skipjack

Forum Staff
Dec 2006
21,482
2,472
It seems a bit odd that the problem doesn't relate the directions mentioned (north, west, etc.) to the direction of the axis of the parabola. Where did you get the problem from? Have you changed any values?
 

SDK

Sep 2016
805
545
USA
The derivative of a quadratic is monotone (increasing in this case) which means that the linearization at every point (which points in the direction of the headlghts) is an underapproximation at every point. So the headlights will NEVER illuminate the sign. If you play a bit more loosely with the physical meanings, you might also say they will illuminate the sign only when the car collides with the sign.
 
Oct 2018
39
3
Netherlands
Another way of stating it is that the shape of the parabola is convex. Therefore the headlight will always shine outside the parabola.

The headlight metaphor is actually a good way to explain the concepts convex and concave!
 
Oct 2018
4
0
Deep Outta Space
The question is from my Maths Assignment and I haven't changed any values.
 
Oct 2018
4
0
Deep Outta Space
I've been told the car lights have to illuminate the sign. :/
 
Oct 2018
39
3
Netherlands
I suppose that "4 m north" is really "4 m south" in our standard coordinate perception (or a = -0.14) and they want you to calculate the point where the tangent through the sign touches the parabola using differentiation.
 
Last edited:

skipjack

Forum Staff
Dec 2006
21,482
2,472
That wouldn't suffice to explain how the car's starting point lies on the parabola.
 
Oct 2018
39
3
Netherlands
That wouldn't suffice to explain how the car's starting point lies on the parabola.
I guess the only escape from that is to assume that the meters do not correspond to the coordinates. It gives us the equation:

-0.14 x^2 = x or x=-1/0.14 as the starting point corresponding to 100 meters.

Hmm, still doesn't seem to work while north and south remain reversed but that may be a glitch in the original challenge.
 
Last edited:
Mar 2015
1,720
126
New Jersey
At first glance, assuming the sign is outside the parabola, the problem is to find where a line from the sign is tangent to the parabola- a nice little calculus problem.

However, The sign is located at x=5 and y=4.
The parabola at x=5 is at y=.14(25)=3.5 so the sign is inside the parabola from where you can't draw a tangent.

It becomes a viable problem if the sign is at x=5 and y=-4.

(Assuming x is East and y is North)
 
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