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

  1. Aug 26, 2016 #1
    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
     

    Attached Files:

  2. jcsd
  3. Aug 26, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Forget the angle. Just think of r as a function of x. (You know what dx/dt is.)
     
  4. Aug 26, 2016 #3
    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
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