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
Dec 18, 2016

1/3 (e^(6 pi)-3)

Explanation:

F(r(t))=(cos^2t,sin^2t,e^(4t))

dr = (-sint,cost,2e^(2t))dt

F(r(t))cdotdr=(-sint cos^2t+cost sin^2t+2e^(6t))dt

int_(t=0)^(t=pi)(-sint cos^2t+cost sin^2t+2e^(6t))dt = 1/3 (e^(6 pi)-3)

Feb 18, 2017

int_C \ vec(F) * d vec(r) = 1/3e^(6pi)-1 ~~ 18829655

Explanation:

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)+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 C is the arc of the parametrised curve:

x=cost\ \ , y=sint\ \ and \ \ z=e^(2t)

from (1,0,1) to (1,0,4)

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 t would seem to make sense.

On C, the variable z varies from e^(2t)=1 => t=0 to e^(2t)=e^(2pi)=>t=pi, and these values of t are also consistent with the x and y initial and end values. Differentiating the equations for C wrt t gives:

dx/dx = -sint \ \ ; dy/dt=cost \ \ and \ \ dz/dt = 2e^(2t)

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

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 of C)
\ \ \ \ = 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 ...