Related rates problem with tower and plane I am currently stuck on the following related rates problem:

An aircraft flying at $4000$ meters and at a ground speed of $450$ km/h passes over the control tower and heads east. At the beacon checkpoint $20$ minutes later, it turns to head northeast. At what rate is it now receding from the control tower? 

I have been stuck on this problem for a while, and I can't seem to get the answer the textbook got ($318$ km/h). Here is what I understand/tried so far:


*

*I feel like it is unnecessary to use a non-right angle triangle since we want the instantaneous rate of change right when the aircraft begins to move northeast. 

*In this case, we use the Pythagorean theorem to obtain $$x^2 + 4^2 = z^2$$ where $x$ represents the base of the triangle and $z$ is the separation between the aircraft and the control tower.

*Differentiating and substituting gives us $$\frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt} = \frac{x}{\sqrt[]{x^2 + 4^2}} \frac{dx}{dt} = \frac{x}{\sqrt[]{x^2 + 16}} \frac{dx}{dt}$$

*If $x=150km$ (the horizontal distance travelled in $20$ minutes), this gives me $449.84$ km/h.


I would appreciate any insight or comments on my approach. I feel like I am close. But, I need another head or two to help me out. 
 A: If we let $z$ be up and $x$ be east the vector from the tower to the plane is $(150,0,4)$ and the velocity of the plane is $(\frac {450}{\sqrt 2},\frac {450}{\sqrt 2},0)$  You are supposed to project the velocity vector on the vector from the tower to the plane.  If the altitude were zero, the projection would just be $\frac {450}{\sqrt 2} \approx 318.198$.  Your result should be just a bit less.
A: Let $(0,0)$ be the Cartesian coordinates of the tower. Let $t = 0$ hours be the moment when the plane turns $45^\circ$ portwards. Hence, the coordinates (in kilometers) of the airplane at $t$ are given by
$$\begin{array}{rl} x (t) &= 150 + 450 \left(\frac{\sqrt 2}{2}\right) t\\ y (t) &= 450 \left(\frac{\sqrt 2}{2}\right) t\end{array}$$
In cylindrical coordinates,
$$r (t) := \sqrt{x^2(t) + y^2(t)} = \cdots = 150 \sqrt{9 t^2 + 3 \sqrt 2 t + 1}$$
and, thus,
$$\dot r (0) = 450 \left(\frac{\sqrt 2}{2}\right) \approx 318 \,\rm{km} \cdot\rm{h}^{-1}$$

Addendum
In spherical coordinates,
>>> from sympy import *
>>> t = Symbol('t')
>>> x = 150 + 450*sqrt(0.5)*t
>>> y = 450*sqrt(0.5)*t
>>> z = 4
>>> rho = sqrt(x**2 + y**2 + z**2)
>>> diff(rho,t)
(202500.0*t + 47729.707730092)/sqrt(101250.0*t**2 + (318.198051533946*t + 150)**2 + 16)
>>> diff(rho,t).subs(t,0)
4.23962584207603*sqrt(5629)
>>> float(4.23962584207603*sqrt(5629))
318.0849747530005

A: HINT
$$ 450 [ (1 + 1/\sqrt 2) i  + (1/\sqrt 2) j ]$$ 
when integrating add initial value altitude $z$ constant value along $k$ vector.
