LP Geniuses - Page 3

On June 12 2014 17:15 TalentedTom wrote: Show nested quote + On June 12 2014 12:56 devon06atX wrote: I mean, the optimal point to shoot from is the very middle of the net. Even if you travelled along the goalline, you would need to travel exactly 6ft to get to the optimal spot to shoot. However, since you're travelling up and over, the distance would increase significantly. That's why I say 5.196 feet has got to be wrong. By quite a bit. Someone please explain to me why I'm wrong. Cuz I just don't see it. this was almost exactly my original thought process, it makes a lot of logical sence to shoot from exactly the middle, but apparently math says otherwise, kinda akward optimized cosine law dont lie,

This is incorrect. If you were exactly in the middle of the two posts, the optimal distance will be when you and the two posts form an equilateral triangle, not when your 6ft away. Surprisingly, the height of a 6ft equilateral triangle is also 5.196ft.

On June 12 2014 17:19 TalentedTom wrote: Show nested quote + On June 12 2014 10:18 Rhaegar99 wrote: not too sure how everyone got the first question but i got the same answer as everyone else theta = atan(9/x) - atan(3/x) dtheta/dx = 3/(x^2+9) - 9/(x^2+81) find maxima by setting dtheta/dx to 0 1/(x^2+9) = 3/(x^2+81) x^2+81 = 3x^2+27 2x^2 = 54 x^2 = 27 x = 5.196 i am looking for a more simplified compressed solution, something that would not require technological (other then a calculator) intervention, will check this out later when I get home. Gonna go vote on provincial ellections

I believe this is as simple of a solution that you will get. This is a basic optimization question which you learn in early calculus.

Firstly you want to look for the function of theta (let us use t) with respect to x (our distance); that is, the angle S of triangle SGP minus the angle S of triangle SGO.
Both of these are right angle triangles so we can use the tan trigonometry function tan(t) = opposite/adjacent for each triangle
To get t by itself we inverse tan (aka atan), the equation to get t = atan(opposite/adjacent)
Therefore our t function is: t = atan(9/x) - atan(3/x)

On our next step, we need to derive the function.
The derivative of our function is: dt/dx = 3/(x^2+9) - 9/(x^2+81)

We calculate the derivative of the function (dt/dx) so that we get the rate at which t changes at any point with respect to x.
A positive dt/dx indicates that t is increasing when x increases, and a negative dt/dx indicates that t is decreasing when x increases.
Therefore to get the the maximum value of t, it is at the point where dt/dx is neither increase nor decreasing, which is when dt/dx = 0
This is also called the maxima of the function.
Therefore to get the maximum distance, we solve for d when dt/dx = 0
The rest is just basic algebra

0 = 3/(x^2+9) - 9/(x^2+81)
1/(x^2+9) = 3/(x^2+81)
x^2+81 = 3x^2+27
2x^2 = 54
x^2 = 27
x = 5.196

nah, trax, you're right. I (stupidly) thought crosby was skating at an angle towards the net. aka - the right angle would be at the middle of the net, not at the goal-line 3 ft outside the post.

Even thought it clearly states in the question that he's skating perpendicular, I stupidly thought he was skating at 45 degrees... and the diagram just fucked me all up haha.

My idiotic mistake.

I hope that clears up why I had an impossible time accepting the correct answer. I was sorta embarassed once I figured out my own error heh.



edit - sorry for being '*facepalmy' yesterday heh

man everyone got super hung up on these two classic calculus problems :O

On June 12 2014 17:19 TalentedTom wrote: Show nested quote + On June 12 2014 10:18 Rhaegar99 wrote: not too sure how everyone got the first question but i got the same answer as everyone else theta = atan(9/x) - atan(3/x) dtheta/dx = 3/(x^2+9) - 9/(x^2+81) find maxima by setting dtheta/dx to 0 1/(x^2+9) = 3/(x^2+81) x^2+81 = 3x^2+27 2x^2 = 54 x^2 = 27 x = 5.196 i am looking for a more simplified compressed solution, something that would not require technological (other then a calculator) intervention, will check this out later when I get home. Gonna go vote on provincial ellections

I think this solution is slightly simpler than the one i had originally (i am gitpush lol), since if you remember what the derivative of arctan(x) is you can do this by hand. (my solution based on the cosine law requires computer to solve the derivative and optimizations)

i remember i had to memorize derivative of arctan(x) when i learned calculus.. but most of the time we use Maple / Mathematica (Wolfram Alpha) for most of our school work.

On June 11 2014 22:07 ggplz wrote: really? daut is exactly who i expected to solve this

Np all good