Thread: Driving a car

1. Driving a car

Hey people! I got a problem here I'm stuck on:

A car is driving at night along a level, curved road. It starts in the origin, the equation of the road is y = x2, and the car's x-coordinate is an increasing function of time. There is a signpost located at (2,3.75).
a) What is the position of the car when its headlight illuminates the signpost? Do you have any implicit physical assumptions in your solution?
b) What is the shortest distance between the signpost and the car?
c) Let dx/dt=vx and dy/dt=vy. The car's velocity is then[vx,vy]. How are vx and vy related?

I got some ideas for how I am to solve it, but I do not know where to begin. Please help me xXxPapaLouxXx

2. Re: Driving a car

I'll help you begin. In order for the headligts to shine on the sign the slope of the road dy/dx must equal the slope of a line conecting the car to the sign. So given that the sign is at coordinates (A,B), the car is at (x,y), and y = x^2, you have:

$\displaystyle \frac {dy}{dx} = 2x = \frac {B-y}{A-x} = \frac {B-x^2}{A-x}$

Now you can solve for x, then y.

As for part (b) the car is at its closest approach when the slope of the line connecting the car to the sign is the negative inverse of the slope of the road:

$\displaystyle \frac {B-y}{A-x} = \frac {-1}{2x}$

For (c) consider the chain rule.

3. Re: Driving a car

Thank you so much for help! I managed to do a) with no problems (I didn't actually need help on that :P ), but I still can't seem to get b) right. And c) I don't have a clue for how I am to solve!

Can you help me a bit more in the right direction? That would be great! 4. Re: Driving a car

For (b), do you understand how I got: $\displaystyle \frac {B-y}{A-x} = \frac {-1}{2x}$? Now replace y with x^2, solve for x, then y=x^2 gives you the position of the car on the road.

For (c) - are you familiar with the chain rule for derivatives? If you have dx/dt and dy/dt, and from the curve of the road you know the value for dy/dx, the chain rule lets you do this:

$\displaystyle \frac {dx}{dt} = \frac {dx}{dy}\ \times \ \frac{dy}{dt} = \frac {dy/dt}{dy/dx}$

Hope this helps.

5. Re: Driving a car

No, I'm not quite sure how you come up with the "= -1/2x"-part :P

But when solving for x, I end up with a third order polynominal, and get three values for x. How can I explain which one to choose?

So the chain rule for derivatives gives the relation between the velocity for x and y?

Sorry for asking so much. Hehe, I just want to understand everything I do! And thanks for your help 6. Re: Driving a car

The slope of the parabola is dy/dx=2x, so the negative inverse of that is -(1/2x), and that's the slope of the line connecting the car to the sign post at the closest point of approach. And yes, you do get a cubic, because there are three points on the car's path when the sign is at right angle's to the car's trajectory. Of course only one of these points is the point of closest approach. In the attached figure note that two of the solutions are with negative values of x, and the third is for positive x and hence the correct answer, though it's a little hard to see because it almost overlaps the car's path.

The chain rule allows you to relate Vx to Vy. Since Vx = dx/dt and Vy = dy/dt, and slope is dy/dx, then Vy divided by the slope of the car's path equals Vx. 7. Re: Driving a car

Thank you so much! Search Tags

car, differentiation, driving, tangent, velocity 