Calculate the work done by a particle under the influence of a force y^2 hat(i) - x^2hat(j)y2ˆix2ˆj along the curve y=4x^2y=4x2 from (0,0)(0,0) to (1,4)(1,4)?

2 Answers
Nov 21, 2016

1.21.2

Explanation:

C=(x,4x^2)C=(x,4x2) and F=((4x^2)^2,-x^2)F=((4x2)2,x2) but dC=(1,8x)dC=(1,8x) so
dW=<< F, dC >>dx = 16x^4-8x^3dW=F,dCdx=16x48x3 then

W=int_0^1 << F, dC >>dx = 16/5-8/4=1.2W=10F,dCdx=16584=1.2

Feb 18, 2017

int_C vec(F) * d vec(r) = 1.2CFdr=1.2

Explanation:

The work done in moving a particle from the endpoints AA to BB along a curve CC 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