Math geniuses of letsrun, help with this easy word problem.

12 feet.

Let d be the distance from the goal,

r(d) be the angle from vertical to the right post, from Datsyuk's perspective, and l(d) the angle to the left post. pi/2 > l(d) > r(d) > 0.

r(d) = arctan(6/d) and

l(d) = arctan(12/d)

for d > 0.

let the shooting angle be

a(d) = l(d) - r(d) = arctan(12/d) - arctan(6/d)

then

a'(d) = 1/(1+(12/d)^2)(-12/d^2) - 1/(1+(6/d)^2)(-6/d^2)

= -12/(d^2 + 144) + 6/(d^2 + 36)

equate this result to zero and multiply both sides by the product of the two denominators:

0 = 6(d^2 + 144) - 12(d^2 + 36)

6d^2 = 432

d = +/- sqrt72

Verify that a local max occurs in a' at sqrt72. Finally Datsyuk just has to calculate sqrt72 in his head, which can be done quickly since it's sqrt(2x4x9) = 6sqrt2 ~ 6x1.4 = 8.4 feet.

Take the shot now. The angle only gets smaller as he approaches the net. Providing he can shoot an absolutely straight shot, the answer is "now."

Johnny Applseed wrote:

Take the shot now. The angle only gets smaller as he approaches the net. Providing he can shoot an absolutely straight shot, the answer is "now."

Two of the math geniuses of letsrun have already replied contrary to this.

does he shoot left or right?

Johnny Applseed wrote:

Take the shot now. The angle only gets smaller as he approaches the net. Providing he can shoot an absolutely straight shot, the answer is "now."

err, no. if he is infinitely far away, the angle reduces to 0. as he approaches the line, the shooting angle again is approaching zero. in between (o, inf) the value is obviously greater than zero. so, the question is to determine at what point the shooting angle reaches a maximum.

Infinity. It's obvious, but the math is actually a little complicated.

Ah, crete is correct. It's not obvious. I'm an idiot...after a few beers.

The optimal shot is not the one with the largest angle.

I used the law of cosines, but the algebra was more of a pain in the ass.

Let 'a' be the distance from Pavel to the near post, 'b' be the distance to the far post, and 'd' be the distance to the goal line.

We'll call the shooting angle 'g'.

Law of cosines says 6^2 = a^2 + b^2 - 2*a*b*cos(g)

Substitute: a = sqrt(d^2 + 6^2), b = sqrt(d^2 + 12^2)

Rearrange to: cos(g) = (d^2 + 72)/sqrt(l^2 + 6^2)/sqrt(l^2 = 12^2)

As the others did, take the derivative with respect to d, set it equal to zero, solve for d.

After shaking out the algebra mistakes, you get d = sqrt(72).

The angle of the dangle (x) is necessary to find y (the heat of the meat) provided tht the maxis of the axis (graphing in y=mx+b) and the gravity of the cavity (solving inequalities) remain constant.

Working time wrote:

Johnny Applseed wrote:

Take the shot now. The angle only gets smaller as he approaches the net. Providing he can shoot an absolutely straight shot, the answer is "now."

Two of the math geniuses of letsrun have already replied contrary to this.

No offense but if you think that anyone posting above is a 'math genius' then you have absolutely no idea what you are talking about.