Calculate the work done by a particle under the influence of a force y^2 hat(i) - x^2hat(j)y2ˆi−x2ˆj along the curve y=4x^2y=4x2 from (0,0)(0,0) to (1,4)(1,4)?
2 Answers
Explanation:
int_C vec(F) * d vec(r) = 1.2∫C→F⋅d→r=1.2
Explanation:
The work done in moving a particle from the endpoints
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
To evaluate the line integral we convert it to a standard integral by choosing an appropriate integration variable, In this case integrating wrt
On
dy/dx = 8x
And so we can express our vector fields in terms of
vec(F) = y^2hat(i) -x^2hat(j)
\ \ \ \ = (4x^2)^2hat(i) -x^2hat(j) \ \ \ \ (from the equation ofC )
\ \ \ \ = 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 forC )
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