Speed of moving particle

1. Sep 26, 2007

kuahji

This is a related rates problem

The coordinates of a particle moving in the metric xy-plane are differentiable functions of time t with dx/dt = 10m/sec & dy/dt = 5m/sec. How fast is the particle moving away from the origin as it passes through the point (3,-4).

First used the Pythagorean theorem & found D (distance) from the origin, which was 5. Then I implicitly differentiated the problem d^2=x^2+y^2

D dD/dt = x dx/dt + y dy/dt (divided out all the 2s). Then I plugged in the rates of change as outlined in the problem.
D dD/dt = 10x + 5y.

But at this step I must be misunderstanding something. I tried to plug (3,-4) in for x & y, & 5 in for D. However, according to the book thats incorrect. So what am I misunderstanding? The book shows 5 dD/dt = (5)(10)+(12)(5) but where are the second five in the equation & 12 coming from?

2. Sep 26, 2007

Dick

I don't know, because I agree with your solution. The 'book' isn't always right.

3. Sep 26, 2007

kuahji

Ok, thank you for taking a look at the problem.

So it would probably look

5 dD/dt = (10)(3)+(-4)(5)

Then just solve? Or because it is distance, would I have to use positive 4 instead of negative? I'm thinking I'd have to use positive 4, but not so sure since I'm just learning all of this.

4. Sep 26, 2007

turdferguson

No, the negative is correct. In a related rates problem, negatives are very important. If dD/dt wound up negative, the distance would be decreasing

The only thing I can think of is that your book may want the magnitude of the particle's velocity. dx/dt is the horizontal component and dy/dt is the vertical component. The vector sum will give you a single velocity vector, and the magnitude is the speed. In this case, the point (3,-4) is just extra information and the last part of the question isnt very clear