Find the rate that the distance between two projectiles is changing

[isukatphysics69]

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

 

[Isaac0427]

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.

 

[vela]

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 $t=\pi/2$.

 

[Delta2]

@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.