Arc length function, Examples

[Pages:8]Math 20C Multivariable Calculus '

Lecture 8

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Arc length function, Examples

? Review: Arc length of a curve. ? Arc length function. ? Examples Sec. 13.4.

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Arc length of a curve

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The arc length of a curve in space is a number. It measures the extension of the curve.

Definition 1 The arc length of the curve associated to a vector valued function r(t), for t [a, b] is the number given by

b

ba = |r (t)| dt.

a

Suppose that the curve represents the path traveled by a particle in space. Then, the definition above says that the length of the curve is the integral of the speed, |v(t)|. So we say that the length of the curve is the distance traveled by the particle.

The formula above can be obtained as a limit procedure, adding up

the lengths of a polygonal line that approximates the original curve.

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Math 20C Multivariable Calculus

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Arc length of a curve

In components, one has,

r(t) = r (t) = |r (t)| =

ba =

x(t), y(t), z(t) ,

x (t), y (t), z (t) ,

[x (t)]2 + [y (t)]2 + [z (t)]2,

b

[x (t)]2 + [y (t)]2 + [z (t)]2dt.

a

The arc length of a general curve could be very hard to compute.

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Arc length function

Definition 2 Consider a vector valued function r(t). The arc length function (t) from t = t0 is given by

t

(t) = |r (u)|du.

0

Note: (t) is a scalar function. It satisfies (t0) = 0. Note: The function (t) represents the length up to t of the curve given by r(t).

Our main application: Reparametrization of a given vector valued function r(t) using the arc length function.

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Math 20C Multivariable Calculus

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Arc length function

Reparametrization of a curve using the arc length function: ? With r(t) compute (t), starting at some t = t0. ? Invert the function (t) to find the function t( ). Example: (t) = 3et/2, then t( ) = 2 ln( /3). ? Compute the composition r( ) = r(t( )). That is, replace t by t( ).

The function r( ) is the reparametrization of r(t) using the arc length as the new parameter.

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Examples, Sec. 13.4

? (Probl. 14 Sec. 13.4) Find the velocity, acceleration, and speed of the position function r(t) = t sin(t)i + t cos(t)j + t2k.

? (Probl. 16 Sec. 13.4) Find the velocity and position vectors given the acceleration and initial velocity and position:

a(t) = -10k, v(0) = i + j - k, r(0) = 2i + 3j.

? Problems with projectiles. Given the initial speed |v0| and the initial angle of the projectile with the horizontal, , describe the movement of the projectile.

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Math 20C Multivariable Calculus '

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Scalar functions of 2, 3 variables

? Definition. ? Examples. ? Graph of the functions. ? Level curves and level surfaces.

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Scalar functions of 2 variables

Definition 3 A scalar function f of two variables (x, y) is a rule that assigns to each ordered pair (x, y) D IR2 a unique real number, denoted by f (x, y), that is,

f : D IR2 R IR.

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Comparison: ? Vector valued functions, r : IR IR2 t x(t), y(t)

? Scalar function of two variables, f : IR2 IR

(x, y) f (x, y). &

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Graph and level curves

Definition 4 The graph of a function f (x, y) is the set of all points (x, y, z) in IR3 of the form (x, y, f (x, y)). Definition 5 The level curves of f (x, y) are the curves in in the domain of f , D IR2, solutions of the equation

f (x, y) = k,

for k R, a real constant in the range of f .

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Scalar functions of 3 variables

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Definition 6 A scalar function f of three variables (x, y, z) is a rule that assigns to each ordered triple (x, y, z) D IR3 a unique real number, denoted by f (x, y, z), that is,

f : D IR3 R IR.

Note:

? In order to graph a function f (x, y, z) one needs four space dimensions. So, one cannot do such graph.

? The concept of a level curve can be generalized to functions of more than two variables. In this case they are called level surfaces.

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Limits and continuity

? Limit in 2, 3 space dimensions. ? Continuity. ? Examples.

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Limits

Idea: The function f (x, y) has the number L as limiting value at the point (x0, y0) roughly means that for all points (x, y) near (x0, y0) the value of f (x, y) differs little from L.

Definition 7 Consider the function f (x, y) and a point (x0, y0) IR2. Then,

lim f (x, y) = L

(x,y)(x0 ,y0 )

if for every number > 0 there exists another number > 0 such that |f (x, y) - L| < for every (x, y) D satisfying 0 < (x - x0)2 + (y - y0)2 < .

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Limits

In words: "f (x, y) has a limit L at (x0, y0) if the following holds: for all (x, y) D close enough in distance to (x0, y0) the values of f (x, y) approaches L." A tool to show that a limit does not exist is the following result.

Theorem 1 If f (x, y) L1 along a path C1 as (x, y) (x0, y0), and f (x, y) L2 along a path C2 as (x, y) (x0, y0), with L1 = L2, then

lim f (x, y) does not exist.

(x,y)(x0 ,y0 )

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Limits

Theorem 2 (Squeeze)

Assume f (x, y) g(x, y) h(x, y) for all (x, y) near (x0, y0); Assume

Then

lim f (x, y) = L = lim h(x, y),

(x,y)(x0 ,y0 )

(x,y)(x0 ,y0 )

lim g(x, y) = L.

(x,y)(x0 ,y0 )

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Slide 15

Continuity

Definition 8 A function f (x, y) is continuous at (x0, y0) if lim f (x, y) = f (x0, y0).

(x,y)(x0 ,y0 )

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Continuity

Examples of continuous functions: ? Polynomial functions are continuous in IR2, for example P2(x, y) = a0 + b1x + b2y + c1x2 + c2xy + c3y2.

? Rational functions are continuous on their domain,

f (x, y)

=

Pn(x, y) Qm(x, y)

,

for example,

f (x, y)

=

x2

+

3y - x2y2 x2 - y2

+

y4 ,

x = ?y.

? Composition of continuous functions are continuous, example

f (x, y) = cos(x2 + y2).

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