Determining the speed of a moving particle

• March 2nd 2013, 01:40 PM
Lambin
Determining the speed of a moving particle
PROBLEM

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

ATTEMPT

The distance between two points can be written as,

(1) $r=\sqrt{x^2+y^2}$

(In this problem, the two points would be the origin and the particle in question.)

When we take the derivative of equation (1), we find the rate of change of r, or the speed of the particle moving away from the origin:

(2) $\frac{dr}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}} {\sqrt{x^2+y^2}}$

Because we are interested in a particular location in the particle's motion, I substituted the values $x=3$, $y=-4$, $\frac{dx}{dt}=10$, $\frac{dy}{dt}=5$.

From this,

(3) $\frac{dr}{dt}=2\frac{m}{sec}$

However, the textbook answer is $22\frac{m}{sec}$.

Any thoughts would be appreciated. Thanks for taking the time to read.
• March 2nd 2013, 03:48 PM
topsquark
Re: Determining the speed of a moving particle
Quote:

Originally Posted by Lambin
PROBLEM

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

ATTEMPT

The distance between two points can be written as,

(1) $r=\sqrt{x^2+y^2}$

(In this problem, the two points would be the origin and the particle in question.)

When we take the derivative of equation (1), we find the rate of change of r, or the speed of the particle moving away from the origin:

(2) $\frac{dr}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}} {\sqrt{x^2+y^2}}$

Because we are interested in a particular location in the particle's motion, I substituted the values $x=3$, $y=-4$, $\frac{dx}{dt}=10$, $\frac{dy}{dt}=5$.

From this,

(3) $\frac{dr}{dt}=2\frac{m}{sec}$

However, the textbook answer is $22\frac{m}{sec}$.

Any thoughts would be appreciated. Thanks for taking the time to read.

I can find no fault in your work. It may just be a typo: 22 instead of 2.

-Dan