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

  • $\begingroup$ First take dot product, then evaluate $F\cdot dr$ along the parameterization you got, last integrate. Also, try to write formulas correctly as it is hard to read this post. $\endgroup$ – Giuseppe Negro Jun 9 '12 at 10:43

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

enter image description here

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

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


  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.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.