Solution Quiz 17 Calculus III Summer 2016 PDF

Title Solution Quiz 17 Calculus III Summer 2016
Course Calculus III
Institution Florida Atlantic University
Pages 2
File Size 88.3 KB
File Type PDF
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Summary

Solution Quiz 17 Calculus III Summer 2016...


Description

Department of Mathematical Sciences Instructor: Daiva Pucinskaite Calculus III July 27, 2016

Quiz 17 Given the force field F: F(x, y) = h −y, x i. Find the work done by the force field F(x, y) = hf (x, y), g(x, y)i on a particle that moves along the given oriented curve C : Recall: Let F be a force field in a region of R2 , and r(t) = hx(t), y(t)i for a ≤ t ≤ b, be a curve. The work done in moving a particle along C in the positive direction is    Z b Z b     F(r(t)) · r′ (t) dt = W = f x(t), y(t) , g x(t), y(t) • x′ (t), y′ (t) dt a

a

(1) C is the upper half of the unit circle centered at the origin oriented counterclockwise. A parametric description of the curve C is * + r(t) =

cos sin t , for |{z} π 0 ≤ t ≤ |{z} |{z}t, |{z} x(t)

a

y(t)

at the time t = t the particle is at position



b

 cos(t), sin(t) .

t= π 2 1 t= π 4

t = 3π 2

t=π

t=0 -1

0

1

-1

The work done by the force field F(x, y) =

*

−y , x |{z} |{z}

+

on a particle that moves along C

g(x,y)

f (x,y)

   Z π • − − cos t W = | {zx} dt | sin {z x}, cos | sin {z }t , |{z} 0 f (x(t),y(t)) g(x(t),y(t))

=

Z

π

0

2 + sin2 }t dt = |sin t {z

= t|0π = π.

1

x′ (t)

Z

0

π

1dt

y ′ (t)

(2) C is the upper half of the unit circle centered at the origin oriented clockwise. A parametric description of the curve C is * + r(t) =

− sin t , for |{z} π 0 ≤ t ≤ |{z} | cos {z }t, |{z} x(t)

a

y(t)

at the time t = t the particle is at position



b

 − cos(t), sin(t) .

t= π 2 1 t= π 4

t = 3π 2

t=0

t=π -1

0

1

-1

The work done by the force field F(x, y) =

W =

Z

π

Z

π

0

=

0



*

−y , x |{z} |{z} f (x,y)

+

on a particle that moves along C

g(x,y)

   x , |cos sin }t , − {zx} dt |{z} |− {z | cos {z }t • sin x′ (t)

f (x(t),y(t)) g(x(t),y(t))

− sin2 t − sin2 t dt = − | {z } −1

= −t|0π = −π.

Z

0

π

1dt

y ′ (t)...


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