Line Integral, Work in physics

Question

Hi there all: I have a problem!

I need to find the work done on a particle that moves from $(0,0)$ to a point $(1,1)$ by a strait line $y=x$. The force acting upon the particle is $F = (y , 2x$).

Now how i have attempted the problem is I paramaterized the line $x = y$ to be $r = < t, t >$ by letting $x = t$ which means that $y = t$ too. (I'm hoping that this is how you do it, it's been a while since ive done that maths course)

I know the formula for work is: The integral of $F\cdot dr$ - thus meaning I have to differentiate $r$ to get $dr = < 1, 1>$ then finding the force acting on the particle using $F$ and the particles position of $(x,y)$ to be $(1,1)$ (i.e. i substituted this point into $F$) so $F= (1,2)$.

Then taking the dot product between the two i get $3$. Then taking the integral between $1$ and $0$ I get the answer to be $3$. Now the answer in the textbook says the answer is $1.5$ :( bummer.

Where have I gone wrong? Do i need to paramiterise F differently? or.. i dno ??? :S

Answer 1

Given are $$F = (x,2y) \ , \ r = (x,y) \ , \ dr = (dx,dy) $$

The path is $(t,t) , t \in [0,1]$. Put $x=t=y$ and then $$F = (t,2t) \ , \ dr = (dt, dt) $$ and then the dot product is $$F \cdot dr = tdt + 2tdt = 3tdt $$

Note that $\int_0^1 t dt = \frac{t^2}{2} |_0^1 = \frac{1}{2}$. This is probably where you made the mistake.

Answer 2

In a different notation (see below) and without making the parametrization using $x$ as the parameter.

The work $W$ carried out by the force $\overrightarrow{F}= y\overrightarrow{i}+2x\overrightarrow{j}$ acting on the particle, when it moves from $P(0,0)$ to $Q(1,1)$ along the line $\gamma :y=x$ (see picture), is given by the integral

$$\begin{eqnarray*} W &=&\int_{\gamma }\overrightarrow{F}\cdot \overrightarrow{dr}\qquad \gamma :y=x,\text{ from }P(0,0)\text{ to}\;Q(1,1) \\ &=&\int_{\gamma } \left( y\overrightarrow{i}+2x\overrightarrow{j} \right) \cdot \left( dx\overrightarrow{i}+dy\overrightarrow{j}\right) \\ &=&\int_{\gamma }y\text{ }dx+2x\text{ }dy,\qquad y=x,\; \; dy=dx \\ &=&\int_{0}^{1}3x\text{ }dx \\ &=&3\int_{0}^{1}x\text{ }dx=3\times \left. \frac{x^{2}}{2}\right\vert _{0}^{1}=3\times \frac{1}{2}=1.5. \end{eqnarray*}$$

Notation:

1. $\overrightarrow{u}\cdot\overrightarrow{v}$ is the dot product of the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$. Above it was used in $$ \left( y\overrightarrow{i}+2x\overrightarrow{j}\right) \cdot \left(dx\overrightarrow{i}+dy\overrightarrow{j}\right) =y\;dx+2x\;dy.$$
2. $\overrightarrow{i}$ and $\overrightarrow{j}$ are vectors of unit length in the direction of the positive $x$ and $y$ axes, respectively.

Answer 3

The logic is that the force acting on the object, changes with position.

Your parameterization of $dr$ is correct; it leads to $dr=<1,1>dt$, but $F$ is not equal to $<1,2>$ along the path, except in $(x,y)=(1,1)$. It is equal to $F=<t,2t>$. Hence $$ \int F\cdot dr=\int_0^1 <t,2t>\cdot <1,1> dt=\int_0^1 3t dt=1.5, $$ which is the correct answer.