Calculate the work done by a particle under the influence of a force #y^2 hat(i) - x^2hat(j)# along the curve #y=4x^2# from #(0,0)# to #(1,4)#?

#C=(x,4x^2)# and #F=((4x^2)^2,-x^2)# but #dC=(1,8x)# so
#dW=<< F, dC >>dx = 16x^4-8x^3# then

#W=int_0^1 << F, dC >>dx = 16/5-8/4=1.2#

The work done in moving a particle from the endpoints #A# to #B# along a curve #C# is.

# int_C \ vec(F) * d vec(r) \ \ # where # \ \ {: (vec(F),=F_1 hat(i) + F_2 hat(j)),(d vec(r),=dx hat(i) +dy hat(j)) :} #

The integral is known as a line integral.

So we have:

# vec(F) = y^2hat(i) -x^2hat(j) #

and #C# is the arc of #y=4x^2# from #(0,0)# to #(1,4)#

To evaluate the line integral we convert it to a standard integral by choosing an appropriate integration variable, In this case integrating wrt #x# would seem to make sense.

On #C#, the variable #x# varies from #x=0# to #x=1#. Differentiating the equation for #C# wrt #x# gives:

#dy/dx = 8x#

And so we can express our vector fields in terms of #x# alone:

# vec(F) = y^2hat(i) -x^2hat(j) #
# \ \ \ \ = (4x^2)^2hat(i) -x^2hat(j) \ \ \ \ # (from the equation of #C#)
# \ \ \ \ = 16x^4 hat(i) -x^2 hat(j) #

And:

# d vec(r)=dx hat(i) + dy hat(j) #
# \ \ \ \ \ =dx hat(i) + 8x \ dx hat(j) \ \ \ \ # (from derivative of eqn for #C#)

Hence,

# int_C \ vec(F) * d vec(r) = int_C \ ( 16x^4 hat(i) -x^2hat(j) ) * (dx hat(i) + 8x \ dx hat(j))#
# " "= int_0^1 \ 16x^4 \ dx -x^2(8x)dx #
# " "= int_0^1 \ 16x^4 -8x^3 \ dx #
# " "= [ 16/5 x^5 -2x^4 ]_0^1 #
# " "= (16/5 -2) - 0#
# " "= 6/5#
# " "= 1.2#