Hello,
plz try do these questions.
1)The path is from (0,6) to (7,0).
This is the equation,
$\displaystyle y=-\frac{7}{6}x+6$
Thus, if $\displaystyle x=t$
And it goes from $\displaystyle 0\leq t\leq 7$.
Thus,
$\displaystyle \b{C}=t\b{i}-(\frac{7}{6}t+6)\b{j}$
Thus, the line integral is,
$\displaystyle \int_0^7 (3t)(t)' dt+\int_0^7 2\left(-\frac{7}{6}t+6\right)\left(-\frac{7}{6}t+6\right)' dt$
Now you can compute it.
2)First we find,
$\displaystyle x'=1$
$\displaystyle y'=t^{1/2}$
Thus,
$\displaystyle \sqrt{(x')^2+(y')^2}=\sqrt{1+t}$
Thus, the line integral is,
$\displaystyle \int_0^3 \frac{1}{1+t} \cdot \sqrt{1+t} dt=\int_0^3 (1+t)^{-1/2} dt$
Use subsitution, $\displaystyle u=1+t$ to get answer.
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.
$\displaystyle M(x,y)=\frac{1}{x^2+y^2}$
$\displaystyle N(x,y)=\frac{4}{x^2+y^2}$
To have path indepence (what it seems)
It is necessary and suffienct for,
$\displaystyle \frac{\partial M}{ \partial y}=\frac{\partial N}{\partial x}$
But that is not work here.
Thus, what is the path? A straight line?