Engineering Mathematics: Finding Work Done by a Force in Vector Fields and Line Integrals, Lecture notes of Engineering Mathematics

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MAT 247 Engineering Mathematics

Hakkı Ula¸s Unal¨ Dept. of Electrical-Electronics Eng. Eski¸sehir Technical University, Turkey

October 10, 2018

Today

1 Conservative and Non-Conservative fields

(^2) Double integral

(^3) Surfaces and their integrals

(^4) Triple Integrals

Work done by the Force

Let F be a constant force (along the direction of motion) applied to move an object along a straight segment d. Then, W = F · d corresponds to work done by the force F. However, the above formula only holds whenever the angle between the direction of the force and the direction of the motion is constant.

Work done by the Force

Let F be a constant force (along the direction of motion) applied to move an object along a straight segment d. Then, W = F · d corresponds to work done by the force F. However, the above formula only holds whenever the angle between the direction of the force and the direction of the motion is constant. What about to move a electrical charge in some electromagnetic field?

10 20 -10^0

A

B

0

10

20

0

5

10

15

20

Work done in a vector field

Let us consider to find the work done by the force F to move a electrical charge along the curve described by a vector function r in some electromagnetic field.

10 20 -10^0

A

B

0

10

20

0

5

10

15

20

Let us parametrize the curve by the vector function r = [x(t), y(t)], where t 0 ≤ t ≤ tf such that r(t 0 ) = A and r(tf ) = B.

Work done in a vector field: Line Integral

Let us consider to find the work done by the force F to move a electrical charge on an infinitesimal piece of the curve from position r(t) to r(t + ∆t). Then, the distance vector is

d = r(t + ∆t) − r(t) = r′∆t,

and the work done in this segment is

F · d = F(r)(t) · r′∆t,

Since this piece of the curve is a straight line segment, then, the angle between the direction of the force and the direction of the motion is constant.

Path Independence

C

F(r) · dr,

does not depend on how you reach to B from A in a defined domain.

A

B

Fundamental theory of Calculus

Fundamental Theory of Calculus

Let f (t) be a continuous function on the interval [a, b] and F (t) is any antiderivative of f (t), then, ∫ (^) b

a

f (t)dt = F (b) − F (a),

Fundamental theory of Calculus

Fundamental Theory of Calculus

Let f (t) be a continuous function on the interval [a, b] and F (t) is any antiderivative of f (t), then, ∫ (^) b

a

f (t)dt = F (b) − F (a),

One of the interpretations is:

∫ (^) b a f^ ′(t)dt = f (b) − f (a),

Fundamental Theory of Line Integral

Let r(t) represent a curve and let a = r(a) and b = r(b). Then, ∫ (^) b

a

∇f · dr = f (b) − f (a),

C

F(r) · dr,

with some continuous F 1 , F 2 , F 3 in domain D is path independent iff

F = gradf,

holds for some f in D.

A function f in a domain D is called potential of F if F = grad(f ). That is, (^) ∫

C

F(r) · dr,

is path independent iff F is potential of some function in D.

Proof

r(t)

Suppose that (x 1 , y 1 ) and (x 2 , y 2 ) are two points close to the surface, where r(t) = x(t)i + y(t)j. Then, since dr(t) = dx(t)/dti + dy(t)/dtj, ∫

C

F(r) · dr =

C

[0, −mg] · [ dx(t) dt

dy(t) dt ]dt

∫

C

F(r) · dr = −mg

∫ (^) y 2

y 1

dy = −mg(y 2 − y 1 )

Work done by gravity is path-independent

Non-conservative (Dissipative) Systems

Real Physical Systems are non-conservative

J

h

Friction

mg θ

Lemma

C

F · dr,

is path-independent in a domain D iff

F · dr = F 1 dx + F 2 dy + F 3 dz,

is exact in D with cts F 1 , F 2 , and F 3

Lemma

Let F 1 , F 2 , F 3 in (^) ∫

C

F · dr,

be cts and have cts first partial derivatives in a simply connected domain (A domain is simply connected if it does not have a hole) D. Then, the above integral is path-independent iff

curlF = 0.