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Find the rate that the distance between two projectiles is changing

  • #1

Homework Statement


The parameteric equations for the paths of two projectiles are given below. At what rate is the distance between the two objects changing at t = pi/2?

Homework Equations


x1=12cos(2t) y1 = 6sin(2t)
x2=6cos(t) y2 = 7sin(t)

The Attempt at a Solution


I am completely stuck. So I was thinking that I should use the distance formula

sqrt( (6cos(t) - 12cos(2t))2 + ( 7sin(t) - 6sin(2t))2))

then take the derivative of that then plug in pi/2 but that got really ugly and I think I am overcomplicating it.

I am looking for a hint here that is all. completely stuck
 
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Answers and Replies

  • #2
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I am completely stuck. So I was thinking that I should use the distance formula

sqrt( (6cos(t) - 12cos(2t))2 + ( 7sin(t) - 6sin(2t))2))

then take the derivative of that then plug in pi/2 but that got really ugly and I think I am overcomplicating it.
Why do you believe that is over complicating it?

Also, if you know the chain rule very well the derivative isn’t that bad.

Edit: ignore the deleted part (if you saw it) about the algebra trick. I tried it and it was a bad idea. Pretty beautiful, but not less complicated.
 
  • #3
vela
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I am looking for a hint here that is all. completely stuck
Your method is fine, but there is a way that's a bit less tedious.

Let ##\vec{r}_i = x_i \hat i + y_i \hat j## and ##\vec{r} = \vec{r}_1 - \vec{r}_2##. The distance ##r## between the two particles is ##\| \vec{r} \|##, so you have
$$r^2 = \vec{r}\cdot\vec{r} = (\vec{r}_1 - \vec{r}_2)\cdot(\vec{r}_1 - \vec{r}_2) = \vec{r}_1\cdot \vec{r}_1 - 2 \vec{r}_1 \cdot \vec{r}_2 + \vec{r}_2\cdot\vec{r}_2.$$ Keeping everything in terms of vectors, calculate ##d(r^2)/dt## and then evaluate the resulting expression for ##t=\pi/2##. Finally, relate ##dr/dt## to ##d(r^2)/dt##.

You might find it helpful to sketch the paths of the two particles and their locations at ##t=\pi/2##.
 
  • #4
Delta2
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@vela methods looks nice and it might save you the trouble of calculating a single complex derivative at the cost of calculating dot products and simpler derivatives instead.

I tend to agree with @Isaac0427 the derivative looks scary but it isn't so hard if you know the chain rule well.
 
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