anyone can you help me? please..
ho to simplify this,
[tanθ + (d/x)]/( 1+ (d/x)·tanθ)= (d + 5.6)/x
can you show me step by step?
(tanθ+(d/X))/(1+tanθ(d/X) )=(d+5.6)/X
Simplify the equation,
tanθ+(d/X)=(d+5.6)(1+tanθ(d/X))/X
tanθ=d/X+d^2/X^2 tanθ+5.6/x+5.6d/X^2 tanθ-d/X
Factorize tanθ and get,
tanθ(1-5.6d/X^2 -d^2/X^2 )=5.6/X
tanθ=-5.6X/(-X^2+d^2+5.6d)
is it correct?
$\dfrac{x\tan{t}+d}{x+d \tan{t}}=\dfrac{d+5.6}{x}$
$x^2\tan{t}+dx = dx + 5.6x + d^2\tan{t} + 5.6d \tan{t}$
$x^2\tan{t} - d^2\tan{t} - 5.6d \tan{t} = 5.6x$
$\tan{t}\bigg(x^2 -d^2 - 5.6d \bigg) = 5.6x$
$\tan{t} = \dfrac{5.6x}{x^2-d^2-5.6d}$
ok cool. thanks for the answer. but then I want to derive it and find the critical point.
sec^2 θ dθ/dX=5.6((-d^2-5.6d-X^2)/(X^2-d^2-5.6d)^2 )
-d^2-5.6d-X^2=0
X^2=-d^2-5.6d
X=√(-d^2-5.6d)
I don't think this is correct. can you help me?
im sorry i didn't mean to trouble anybody its just that i confused.. im really sorry
this is the real problem by daewoo_lowrider
Hi, can someone please help me with this problem? It states:
Suppose a try is scored at T, then the kicker can kick to the posts from anywhere along the line TF perpendicular to the try line PQ. (Refer to Figure 1) Where is the best place to kick from?
Many kickers seem to think that the further they go back (within their kicking limit) the better. But this is not necessarily the case. The best position for the kick to be taken is clearly where angle ACB subtended by the posts along TF is a maximum.
Here is the diagram:
http://img.photobucket.com/albums/v3...ce/figure1.jpg
Basically, I think I have to use trig, and a circle subtending the goal posts, but I'm stuck at that, I have hardly any measurements and angles (the distance between the goal posts is 5.6m.
Thanks for any help,
Daniel
i found it in mhf but i dont understand the solution
let $\theta = \angle{ACB}$, $\phi = \angle{TCA}$, $|TA| = d$ and $|TC| = x$. Note $d$ is a constant and $x$ is variable.Suppose a try is scored at T, then the kicker can kick to the posts from anywhere along the line TF perpendicular to the try line PQ. (Refer to Figure 1) Where is the best place to kick from? ... the distance between the goal posts is 5.6m.
let the distance between the posts, $|AB| = 5.6 = r$, also a constant.
$\phi = \text{arccot}\left(\dfrac{x}{d}\right)$
$\phi + \theta = \text{arccot}\left(\dfrac{x}{d+r}\right)$
$\theta = \text{arccot}\left(\dfrac{x}{d+r}\right) - \text{arccot}\left(\dfrac{x}{d}\right)$
note the derivative of $\text{arccot}(u) = -\dfrac{1}{1+u^2} \cdot \dfrac{du}{dx}$ ...
$\dfrac{d}{dx} \bigg[\theta = \text{arccot}\left(\dfrac{x}{d+r}\right) - \text{arccot}\left(\dfrac{x}{d}\right) \bigg]$
$\dfrac{d \theta}{dx} = - \dfrac{1}{1+\left(\frac{x}{d+r}\right)^2} \cdot \dfrac{1}{d+r} + \dfrac{1}{1+\left(\frac{x}{d} \right)^2} \cdot \dfrac{1}{d} = \dfrac{d}{d^2+x^2} - \dfrac{d+r}{(d+r)^2+x^2}$
$\dfrac{d}{d^2+x^2} - \dfrac{d+r}{(d+r)^2+x^2} = 0$
$\dfrac{d}{d^2+x^2} = \dfrac{d+r}{(d+r)^2+x^2}$
solving for x yields ...
$x = \sqrt{d(d+r)}$