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

1. Aug 26, 2016

### Hercuflea

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

1. The problem statement, all variables and given/known data

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

A car is travelling 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) ?

2. Relevant 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 havent been able to get it.

3. The attempt at a solution
Attached

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• ###### RelatedRatesAttempt.png
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2. Aug 26, 2016

### Staff: Mentor

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

3. Aug 26, 2016

### Hercuflea

Thanks. I got it now. Sometimes I just get one track minded and forget to think about other ways of looking at it.
Solution:
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 $$r(t) = \sqrt(x^2 +1)= \sqrt((x_0+vt)^2+1)$$

Last edited: Aug 26, 2016