Expert: Socrates - 10/23/2007

Question
Hi
I need help with the following problem:
A particle P is moving in an xy plane. At any instant t, the x coordinate of P is increasing at the rate of .2 units/second, and the y coordinate decreasing at the rate .3 units/second. At what time rate is the distance of P from the origin changing at the instant when P is at (7,24)?

Thanks
Dragos

Answer
My first answer was nonsense, here is the right way:

Distance to the origin is D(t)=(x^2 + y^2)^1/2
D'(t) = (1/2)(x^2 + y^2)^-1/2(2xx' + 2yy')
x=7 , y=24 , x'=.2 , y'= -.3
substitute these values into the expression for D' and get

(1/2)(49 + 576 )^-1/2 ((2)(7)(.2) + (2)(24)(-.3)) =
(1/2)(625)^-1/2 (2.8 - 14.4) =
(1/2)(1/25)(-11.6) =
-11.6/50 =
-116/500=
-29/125

The answer is -29/125