How would you calculate the work done

[stunner5000pt]

if the force depended on the DIRECTION of the VELOCITY
that is \vec{F} = - F \hat{v}.
so suppose it was dragged along a path - say from (0,0) to (1,1) i na straight line what is the work done?
It going in a striaght line so we could parametrize the curve into r(t) = (t,t)
and r'(t) = (1,1) right
and thus v hat = (1,1) as well then?
now the problem is the work depends on the direction of the velocity

so then the work done is W = -F \int_{0}^{1} (1,1) (1,1) dt is that right?

ok so let's say it was dragged along the curve y = x^2
then r(t) = (t,t^2)
and r'(t) = (1,2t)
and \hat{v} = \frac{(1,2t)}{\sqrt{1+4t^2}}
ok but what the dirrefetial quantity fo the function here? Is it (t,t^2) dt?
so then is
W = -F \int_{0}^{1} \frac{(1,2t))}{\sqrt{1+4t^2}} (t,t^2) dt
is that right?

Please help!

 

[StatusX]

You should use:

W=\int \vec F \cdot d\vec x = \int \vec F \cdot \vec v \mbox{ } dt

If \vec F = -|F| \hat v, then:

W= \int -|F|\hat v \cdot \vec v \mbox{ } dt = -|F|\int |v|\hat v \cdot \hat v \mbox{ } dt

= -|F| \int |v|dt = -|F| \int ds

So you don't have to do most of that work you did.

 

[stunner5000pt]

oh ok that seems much simpler
but was i did correct by teh way?

doing it the simpler way though...
for the straight line from (0,0) to (1,1)
the integrand would be
\int_{0}^{1} \int_{0}^{1} (1,1) \bullet (1,1) dx dy ?
and for hte line from (0,0) to (1,1) along y = x^2
\int_{0}^{1} \ int_{y=0}^{y=x^2} (1,1) \bullet (1,1) dy dx

are thos correct?

 

[StatusX]

No, there aren't any double integrals in this. 'ds' means the differential arclength. Note that I substituted it in for |v|dt. The integral of ds over a segment of a curve is the length of that segment of the curve. For a curve y(x) in 2D, ds²=dx²+dy²=dx²(1+(dy/dx)²). For example, for the straight line from (0,0) to (1,1), \int ds =\sqrt{2}.

 

[stunner5000pt]

ooooo i see i understnad now
there was another part of this problem that involved proving what you proved in your first post. I totally disregaraded the meaning of dS. SO the integral of dS represents the arc length.

so for the y=x^2 function the arc legnth would be for r(t) = (t,t^2) and r'(t) = (1,2t)
\int_{0}^{1} \sqrt{1 + 4t^2} dt yes?

 

[stunner5000pt]

can i get a confirmation on my answer? Is it right? I m not sure on how to integrate it though. I tried using mathematica and got an answer that involved sinh which i haven ot come across as of yet.