Calculate the work done by a particle under the influence of a force x^2 hat(i) + y^2hat(j) + z^2hat(k) along the curve x=cost, y=sint and z=e^(2t) from (1,0,1) to (1,0,4)?
2 Answers
Explanation:
int_C \ vec(F) * d vec(r) = 1/3e^(6pi)-1 ~~ 18829655
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)+F_3 hat(k)),(d vec(r),=dx hat(i) +dy hat(j)+dz hat(k)) :}
The integral is known as a line integral.
So we have:
vec(F) = x^2 hat(i) + y^2hat(j) + z^2hat(k)
and
x=cost\ \ ,y=sint\ \ and\ \ z=e^(2t)
from
To evaluate the line integral we convert it to a standard integral by choosing an appropriate integration variable, In this case integrating wrt the parameter variable
On
dx/dx = -sint \ \ ;dy/dt=cost \ \ and\ \ dz/dt = 2e^(2t)
And so we can express our vector fields in terms of
vec(F) = x^2 hat(i) + y^2hat(j) + z^2hat(k)
\ \ \ \ = (cost)^2 hat(i) + (sint)hat(j) + (e^(2t))^2hat(k) \ \ \ \ (from the equation ofC )
\ \ \ \ = cos^2t hat(i) + sin^2t hat(j) + e^(4t) hat(k)
And:
d vec(r)=dx hat(i) + dy hat(j) + dz hat(j)
\ \ \ \ \ =(-sint \ dt) hat(i) + (cost \ dt) hat(j) + (2e^(2t) \ dt) hat(k)\ \ \ \ (from derivatives)
\ \ \ \ \ =-sint \ dt hat(i) + cost \ dt hat(j) + 2e^(2t) \ dt hat(k)
Hence,
int_C \ vec(F) * d vec(r) = int_C \ (cos^2t hat(i) + sin^2t hat(j) + e^(4t) hat(k)) * (-sint \ dt hat(i) + cost \ dt hat(j) + 2e^(2t) \ dt hat(k))
" "= int_0^pi \ (cos^2t)(-sint)dt+(sin^2tcost)dt+(e^(4t)2e^(2t))dt
" "= int_0^pi \ -sintcos^2t+sin^2tcost+2e^(6t) \ dt
" "= [1/3cos^3t+1/3sin^3t+1/3e^(6t)]_0^pi
" "= 1/3[cos^3t+sin^3t+e^(6t)]_0^pi
" "= 1/3{(cos^3pi+sin^3pi+e^(6pi)) - (cos^3 0+sin^3 0+e^0)}
" "= 1/3{(-1+0+e^(6pi)) - (1+0+1)}
" "= 1/3(e^(6pi)-3)
" "= 1/3e^(6pi)-1
" "= 18829655.02 ...