   Chapter 16.2, Problem 39E

Chapter
Section
Textbook Problem

Find the work done by the force fieldF(x, y) = x i + (y + 2) jin moving an object along an arch of the cycloidr(t) = (t − sin t) i + (1 − cos t) j 0 ⩽ t ⩽ 2π

To determine

To find: The work done by the force field F(x,y)=xi+(y+2)j .

Explanation

Given data:

The force field is F(x,y)=xi+(y+2)j .

The vector function of arch of the cycloid is r(t)=(tsint)i+(1cost)j,0t2π .

Formula used:

Write the expression to find the work done by the force field F(x,y) along the arch of the cycloid r(t) .

CFdr=abF(r(t))r(t)dt (1)

Here,

r(t) is the vector function of the arch of the cycloid,

a is the lower limit of curve, and

b is the upper limit of the curve.

Write the vector function as follows.

r(t)=(tsint)i+(1cost)j

Write the point (x,y) from the vector function as follows.

(x,y)=(tsint,1cost)

Write the force field as follows.

F(x,y)=xi+(y+2)j (2)

Calculation of F(r(t)) :

Substitute (tsint) for x and (1cost) for y in equation (2),

F(r(t))=(tsint)i+[(1cost)+2]j=(tsint)i+(3cost)j

Calculation of r(t) :

To find the derivative of the vector function, differentiate each component of the vector function.

Differentiate each component of the vector function r(t)=(tsint)i+(1cost)j as follows.

ddt[r(t)]=ddt[(tsint)i+(1cost)j]

Rewrite the expression as follows.

r(t)=ddt(tsint)i+ddt(1cost)j=(1cost)i+[0(sint)]j=(1cost)i+(sint)j

Calculation of CFdr :

Substitute [(tsint)i+(3cost)j] for F(r(t)) , [(1cost)i+(sint)j] for r(t) , 0 for a , and 2π for b in equation (1),

CFdr=02π[(tsint)i+(3cost)j][(1cost)i+(sint)j]dt=02π[(tsint)i(1cost)i+(tsint)i(sint)j+(3cost)j(1cost)i+(3cost)j(sint)j]dt=02π[(tsint)(1cost)+0+0+(3cost)(sint)]dt=02π[ttcostsint+sintcost+3sintsintcost]dt

Simplify the expression as follows

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