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

Answer:

# 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

Answer:

# 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 ... #