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need help on calc

X

Xylitol

Diamond Member
A ball is thrown from the origin of a coordinate systme. The equation of its path is y = mx - (e^(2m))/1000 * x^2, where m is positive and represents the slope of the path at the origin.

a) For what value of m will the ball strike the horizontal axis at the greatest distance from the origin? Justify your answer.

I just don't know how to solve this problem since I can't isolate the X in order to optimize it and find the maximum value of it. Is there something that I'm missing?

Thanks

edit: got m = 1/2
sounds right - anyone wanta double check?
 
At the maximum distance, y=0. So the y disappears and you have an equation x(m) that you can optimize.
 
Yes you can. mx - ( (x^2) * e^(2m) )/ 1000 = 0.

Get x in terms of m. x = 1000m/(e^2m) Differentiate that and set it equal to 0.
 
Well, it's a parabola, and upside down one at that. What you want to do is differentiate it, and you can find the max in terms of m. Since it starts at the origin, and parabolas are symmetrical, simply double the x value of the max to find where it hits 0, again. That x value will be in terms of m. You can find the max of the that x value in the same way you found the max of the parabola.
 
Originally posted by: TridenTBoy3555
Originally posted by: Xylitol
got it thanks guys
i got m=1/2

anyone wanta check my answer?
Click to expand...

Nope, I do not check answers of people who say wanta.
Click to expand...

Stop being a douche.

I got 1/sqrt(2) but I haven't done this stuff in a while so could be wrong.

Edit - I just checked again, 1/2 is correct.
 
Originally posted by: TridenTBoy3555
Originally posted by: Xylitol
got it thanks guys
i got m=1/2

anyone wanta check my answer?
Click to expand...

Nope, I do not check answers of people who say wanta.
Click to expand...

Somebody ban this guy already...

Edit: From a physics standpoint, 45 degrees will give you the most horizontal distance with a projectile, so there's some further reinforcement for your answer.
 
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