Calculus problem with right triangles In a soccer field ABCD a player kicks the ball from a point P towards the goalpost situated on AB. The ball rolls on the ground without rising. The trajectory of the ball is parallel to AD and it's 40 m from AD. A rangefinder situated 10 m above A (let's call this position T) notes that the distance PT is 50 m and this distance is decreasing with velocity 9 m/s. What's the velocity of the ball in this istant?
Attempted solution:
Let's call R the point in which the trajectory of the ball encounters AB, and x(t) the vector which represents the movement of the ball. We want to know $\frac{dx}{dt}$.  
With the Pythagorean theorem we find out that $AP = \sqrt{50^2-10^2}  = 20\sqrt{6} $ m
Now we need to find the rate of change of AP (I did this with a proportion, but I'm really unsure)
$$v_{TP} : TP = v_{AP} : AP$$
$$ v_{AP} = \frac{AP * v_{TP}}{TP} = \frac {18*\sqrt{6}}{5}$$
In an imaginary right triangle which has x(t) and AR as legs we have that the hypothenuse y(t) varies in function of x.
$$y (x) = \sqrt{AR^2+x^2} = \sqrt{40^2+x^2}$$
so:
$$\frac{dy}{dx} = \frac{x}{\sqrt{40^2+x^2}}$$
with the chain rule we find that:
$$\frac{dx}{dt} = \frac{dy}{dt} * \frac{dx}{dy}$$
we know that $\frac{dy}{dt} = v_{AP} =  \frac {18*\sqrt{6}}{5}$
and we know that in the istant we want to analyze $y = \sqrt{2400}$ so  $x = \sqrt{800}$
Therefore:
$$\frac{dx}{dt} = v_{AP} \frac{\sqrt{1600+800}}{\sqrt{800}} = \frac {18*\sqrt{6}}{5} * \sqrt{3}$$
I'm unsure on how I found $v_{AP}$.
 A: Let consider the right triangle PRT, at the kick we have


*

*$PT^2=50^2=2500$

*$TR^2=AR^2+AT^2=1600+100=1700$

*$PR^2=x_0^2=PT^2-TR^2=800$


for a generic instant we have


*

*$PR=x$

*$PT^2=PR^2+TR^2=x^2+1700\implies PT=\sqrt{x^2+1700}$
ans thus
$$\frac{dPT}{dt}=\frac{dPT}{dx}\frac{dx}{dt}=\frac{x}{\sqrt{x^2+1700}} \frac{dx}{dt}$$
and at the kick
$$\left(\frac{dPT}{dt}\right)_{x=x_0}=\frac{\sqrt{800}}{50}\frac{dx}{dt}=9 m/s\implies\frac{dx}{dt}=\frac{450}{\sqrt{800}}\approx 15.9 m/s$$
A: Some tips:
-You can combine several right triangles with the Pythagorean Theorem. So PT2 = PR2+RA2+AT2 (The problem is poorly worded in that it first presents P as a fixed point, and then treats it a variable point denoting the current position of the ball; this equation is following the latter interpretation).
-In these sorts of problems, taking implicit differentiation is usually better. So PT d(PT) = PR d(PR) + RA d(RA) + AT d(AT). Since RA and AT are constant, PT d(PT) = PR d(PR) and d(PR) = PT*d(PT)/PR.
d(PT) is given as 9 m/s (Don't forget units! You can ignore them in your calculations, but you have to add them back in in the end.) PT is given as 50m,  RA as 40m, and AT as 10m, so plugging that into the equation PT2 = PR2+RA2+AT2 we get 502 = PR2+402+102 and thus PR2 = 2500-1600-100 = 800.
This then gives d(PR) = 9*50/sqrt(800) = 9*5/sqrt(8) = 45*sqrt(2)/4 m/s
A: Let’s take the mirror image of your diagram, so that $A$ is the origin, $P$ is in the first quadrant of the $x$-$y$ plane, and the ball is traveling along the line $y=40$, $z=0$ toward the $y$-axis (i.e., $x$ is decreasing). We then have $T=(0,0,10)$ and for any point $Q=(x,40,0)$ along the ball’s path, $QT=\sqrt{(x-0)^2+(40-0)^2+(0-10)^2} = \sqrt{x^2+1700}$. You’re given that $PT=50$, so can solve for $x$ to find that $P=(20\sqrt2,40,0)$.  
Setting $r=QT=\sqrt{x^2+1700}$, the range to the ball, differentiate with respect to $t$ (applying the chain rule as necessary) to get $$r' = {x x' \over \sqrt{x^2+1700}} = \frac xr x',$$ therefore $x' = \frac rx r'$. Plug in the given values of $r=50$ and $r'=-9$ and the value of $x$ computed above, and you’re done.
