Moving point and it's distance relative to a fixed point

[Hercuflea]

Not sure whether this is an intro physics or intro calculus/related rates problem.

1. Homework Statement
Suppose a point P lies at (x,y)=(0,1) meters.

A car is traveling at 30 meters/second along the x-axis towards +∞.

Define r to be the distance between P and the car at any time t.

I need to know the rate of change of the distance between P and the car with respect to time (dr/dt) ?

Homework Equations

Pythagorean theorem...I attached a picture

Attached my attempted solution...tried 2 methods. I do not know what θ(t) or dθ/dt is so that's why I haven't been able to get it.

The Attempt at a Solution

Attached[/B]

 

[Doc Al]

Forget the angle. Just think of r as a function of x. (You know what dx/dt is.)

 

[Hercuflea]

Forget the angle. Just think of r as a function of x. (You know what dx/dt is.)

Thanks. I got it now. Sometimes I just get one track minded and forget to think about other ways of looking at it.
Solution:

Spoiler

v = the velocity
x = horizontal component of r

$$ r^2 = x^2+1^2 $$
$$ r = \sqrt(x^2 +1) $$
$$\frac {dr} {dt} =\frac {1} {2} \frac {1} {\sqrt(x^2 +1)}\frac {d} {dt} (x^2 +1) $$
$$ =\frac {1} {2} \frac {1} {\sqrt(x^2 +1)} 2x \frac {dx} {dt} $$
$$ =\frac {xv} {\sqrt(x^2 +1)} $$

Since $$ x(t)=x_0+vt$$
we have $$\frac {dr} {dt} = \frac {v(x_0+vt)} {\sqrt((x_0+vt)^2 +1)} $$

and also $$r(t) = \sqrt(x^2 +1)= \sqrt((x_0+vt)^2+1)$$