c curve,work done by force........

**m777** · January 17th 2007, 04:42 AM

Hello,
plz try do these questions.

**ThePerfectHacker** · January 17th 2007, 05:30 AM

1)The path is from (0,6) to (7,0).
This is the equation,
$y=-\frac{7}{6}x+6$
Thus, if $x=t$
And it goes from $0\leq t\leq 7$ .
Thus,
$\b{C}=t\b{i}-(\frac{7}{6}t+6)\b{j}$
Thus, the line integral is,
$\int_0^7 (3t)(t)' dt+\int_0^7 2\left(-\frac{7}{6}t+6\right)\left(-\frac{7}{6}t+6\right)' dt$
Now you can compute it.

**ThePerfectHacker** · January 17th 2007, 05:34 AM

2)First we find,
$x'=1$
$y'=t^{1/2}$
Thus,
$\sqrt{(x')^2+(y')^2}=\sqrt{1+t}$
Thus, the line integral is,
$\int_0^3 \frac{1}{1+t} \cdot \sqrt{1+t} dt=\int_0^3 (1+t)^{-1/2} dt$
Use subsitution, $u=1+t$ to get answer.

**ThePerfectHacker** · January 17th 2007, 05:46 AM

3)
Are you sure there is supposed to be a four? It seems it is asking for path independence, but the cross pratials test does not work if there is not four.

$M(x,y)=\frac{1}{x^2+y^2}$
$N(x,y)=\frac{4}{x^2+y^2}$
To have path indepence (what it seems)
It is necessary and suffienct for,
$\frac{\partial M}{ \partial y}=\frac{\partial N}{\partial x}$
But that is not work here.

Thus, what is the path? A straight line?

**m777** · January 17th 2007, 06:41 AM

Hello,
yes, i am sure it is four.path is not straight line it is like a curve shape.

**m777** · January 17th 2007, 11:22 AM

Hello,
the correct question no 3 is attached.

**ThePerfectHacker** · January 17th 2007, 11:33 AM

Originally Posted by m777

Hello,
the correct question no 3 is attached.

I am unable to load the document perhaps it is defective.

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