# Related Rates of perpendicular motion

Two objects A and B are connected by a rigid rod that has a length L. The objects slide along perpendicular guide rails. If A slides to the left with a constant speed v, what is the velocity of B when a=60$^\circ$?

I know the answer is .566 v, but I am completely unsure why.

I have tried saying that $$y=L*sin(a), dy/da=d/da L*sina,dy/da=L cosa,dy/da=Lcos60=L/2$$ but that obviously is the wrong aproach, since I need $dy/dt$, not $dy/da$.

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What is $a$ supposed to be? – Alan Guo Oct 22 '12 at 0:05
a is the angle formed by the rod and the x-axis – segiddins Oct 22 '12 at 0:07
Okay, but you also haven't stated which object goes on which axis, though since A slides to the left I'm guessing A is on the x-axis? – Alan Guo Oct 22 '12 at 0:19
Correct. There's a diagram, but I thought it was extraneous. Sorry, a is on the x, b is on the y. – segiddins Oct 22 '12 at 1:04
Updated question – segiddins Oct 22 '12 at 2:26

You can write:

$x^2+y^2 = z^2$

where $x$ is the location of A with respect to where the rails meet, and $y$ is the location of B with respect to the same spot. And $z^2$ is just L, the length of the rod, which is constant.

Differentiate this wrt to $t$, and you get

$2x \frac{dx}{dt}+ 2y \frac{dy}{dt} = 0$

We know the angle is 60 degrees, so we can determine what $x$ and $y$ should be using $y=Lsin(\theta)$ and $x=Lcos(\theta)$. We also know $\frac{dx}{dt}=v$.

Plug in, solve for $\frac{dy}{dt}$

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Thanks so much! Forgot that we know both x and y, and also a bit of how to do partial differentiation. – segiddins Oct 22 '12 at 3:11