Comprimento de Arco, Curtavura e Tudo Mais

7/27/2019 Comprimento de Arco, Curtavura e Tudo Mais

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Math 214 Arc Length, Curvature, and All That Solutions

1. Consider the line segment in R2 from the point (0, 0) to (1, 1). Compute and interpret thevector integrals for the following different parametrizations of this segment.

(a)1

0t, t dt = 1

0t dt,

1

0t dt = [t2/2]1

0, [t2/2]1

0 = 1/2, 1/2

(b) /20

cos t, cos t

dt =

/20

cos t dt, /20

cos t dt

=

[sin t]

/20

, [sin t]/20

=

1, 1

(c) 0sin 1

2t

, sin1

2t dt =

0sin(t/2) dt,

0

sin(t/2) dt = [2 cos(t/2)]0

, [2cos(t/2)]0 =

2, 2(d)

1

0t, t2 dt = 1

0t dt,

1

0t2 dt = [t2/2]1

0, [t3/3]1

0 = 1/2, 1/3

(e) Is there an easy geometric representation or any relationship among the integrals of thesevector functions?

Solution: All of these curve join the points (0, 0) and (1, 1), with the first three doingso by the straight line segment while the 4th does so along the parabola y = x2. As thefirst three join these points along the same paths, but at differing speeds, we find thattheir integrals are parallel but not equal (a reflection of the different speeds). However,the last integral is not parallel to the first three since the curve joining (0, 0) to (1, 1) inthe last case is not the same path as that of the first three.

2. Suppose a particle moves along the surface of a sphere. Show that its velocity is alwaysperpendicular to its position vector.

Solution: For simplicity, we may assume that the particle travels along the surface of asphere centered at the origin of radius r, which means that its position vector, r(t), is oflength r for all values of t. That is, |r(t)| =

r(t) r(t) = r for all values of t. Hence

r(t) r(t) = r2 is constant. Differentiating this equation with respect to t, we find thatr(t) r(t) + r(t) r(t) = 0, which is the same as 2r(t) r(t) = 0 since a b = b a for allvectors a, b. Therefore, based on our assumption, we find that r(t) r(t) = 0 for all values oft, which means that the particles position vector, r(t), is always perpendicular to its velocityvector, v(t) = r(t).

3. Find the unit tangent vector of the vector-valued function r(t) = 2sin t, cos t, 1.Solution: We first find r(t) = 2cos t, sin t, 0, which is tangent to the curve and haslength |r(t)| =

(2cos t)2 + ( sin t)2 + (0)2 =

4cos2 t + sin2 t. Thus, the unit tangent

vector of r(t) is

T(t) =1

|r(t)|r(t) =

14cos2 t + sin2 t

2cos t, sin t, 0.

4. A particle starts at the origin with initial velocity v0 = i + 2j + k. Its acceleration at time tis a(t) = t, t2, t3. Find its position vector, r(t).Solution: Since acceleration is the rate of change of velocity, velocity is an antiderivative ofacceleration, so v(t) =

a(t) dt =

t, t2, t3 dt = t2/2, t3/3, t4/4 + C for some constantvector C. Now, given the initial velocity v0 = 1, 2, 1, we find C by computing v(0) =0, 0, 0 + C = C, so C = v0 = 1, 2, 1. Hence v(t) = t2/2, t3/3, t4/4 + 1, 2, 1 = 1 +(t2/2), 2 + (t3/3), 1 + (t4/4). As r(t) is an antiderivative of v(t), we have r(t) = v(t) dt =1 + (t2/2), 2 + (t3/3), 1 + (t4/4) dt = t + (t3/6), 2t + (t4/12), t + (t5/20) + P for someconstant vector P. Yet, as we know r(0) = 0, 0, 0, since the particle starts at the origin,


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and r(0) = 0, 0, 0 + P by the integration we just completed, we see that P = 0 = 0, 0, 0.Hence, the position vector of the particle at time t is

r(t) = t + t3

6, 2t +

4t

12, t +

t5

20.

5. Consider the curve r(t) = sin t,t, cos t, 0 t 2.(a) Find the arc length of r(t).

Solution: The arc length of r(t) is

L =

2

0

|r(t)| dt

=

2

0

|cos t, 1, sin t| dt

=

2

0

(cos t)

2

+ (1)2

+ ( sin t)2

dt

=

2

0

2 dt

= [

2t]20

= 2

2.

(b) Find the arc length of q(t) = sin(t2), t2, cos(t2), 0 t 2.Solution: The arc length of q(t) is

L =

2

0

|q(t)| dt

=20

|2t cos(t2), 2t,2t sin(t2)| dt

=

2

0

(2t cos(t2))2 + (2t)2 + (2t sin(t2))2 dt

=

2

0

4t2 cos2(t2) + 4t2 + 4t2 sin2(t2 dt

=

2

0

8t2 dt

= 2

0

2

2t dt

= [2

2(t2/2)]2

0

= 2

2(2/2) = 2

2.

(c) Find the arc length of p(t) = cos t, t + 2

, sin t, 2 t 3

2.


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Solution: The arc length of p(t) is

L =

3/2

/2|p(t)| dt

=

3/2

/2 | sin t, 1, cos t| dt

=

3/2

/2

( sin t)2 + (1)2 + ( cos t)2 dt

=

3/2

/2

2 dt

= [

2t]3/2/2

= 2

2.

(d) Sketch the graphs of r, q and p. Is there a relationship between r, q and p?

Solution: While I wont sketch these graphs for you, we recognize that the first oneis the graph of the helix on the cylinder x2 + z2 = 1 beginning at the point (0, 0, 1)and ending at the point (0, 2, 1) making exactly one turn around the cylinder frombeginning to end. Then, with the help of a little trig, that all three of these curves are,in fact, the same helix on the cylinder x2 + z2 = 1.

(e) How does this explain your answers to parts (a), (b) and (c)?

Solution: As all three curves are merely different parametrizations of the same curvein space, all three should and do have the same arc length. Hence, our answers to parts(a), (b) and (c) make sense.

6. Consider the curve r(t) = et cos t, et sin t, et, t 0.(a) Graph the curve and prove that the curve exists on the surface z2 = x2 + y2.

Solution: Once again, I wont include the graph here, but I will do the algebra. Forthis curve, its x-component is x = et cos t, the y-component is y = et sin t and thez-component is z = et. Consider x2 + y2 = (et cos t)2 + (et sin t) = e2t cos2 t +e2t sin2 t = e2t(cos2 t + sin2 t) = e2t = z2. Thus this curve lies on the surfacex2 + y2 = z2, which is a circular cone with vertex at the origin in R3.

(b) Find the curvature, (t), of r.

Solution: The curvature of r(t) is

(t) = |dTds|

= |dTdt

dt

ds|

= |T(t)||1/( dsdt

)|= |T(t)|/|r(t)|.


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Thus, we need to find

r(t) =d

dt

etcos t, sin t, 1

= etcos t, sin t, 1 + et sin t, cos t, 0= et

cos t

sin t,

sin t + cos t,

1

,

so

|r(t)| = |et cos t sin t, sin t + cos t,1|= et

( cos t sin t)2 + ( sin t + cos t)2 + (1)2

= et

cos2 t + 2 cos t sin t + sin2 t + sin2 t 2sin t cos t + cos2 t + 1= et

3.

Therefore,

T(t) =1

|r(t)|r(t)

=

1

et3 et

cos t sin t, sin t + cos t,1=

13 cos t sin t, sin t + cos t,1.

Hence,

T(t) =1

3sin t cos t, cos t sin t, 0,

whose length is

|T(t)| = 13

(sin t cos t)2 + ( cos t sin t)2 + (0)2

=1

3sin2 t

2sin t cos t + cos2 t + cos2 t + 2 cos t sin t + sin2 t

=

23

=

2

3.

Thus,

(t) =

2/3

et

3=

et

2

3.

(c) What happens to the curvature as t ?Solution: As t , (t) = 2et , which means that the curve is winding fasterand faster and tighter and tighter as time increases toward infinity.

(d) Prove that as t , the total distance traveled is 3.Solution: The total distance traveled is given by the improper integral

L =

0

|r(t)| dt

= limb

b0

et

3 dt

= limb

[et

3]b0

= limb

[(eb

3) (e0

3)]

=

3.


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7. Using the tangential and normal components of acceleration, describe in words those curvesr for which

(a) a v (i.e., the acceleration is perpendicular to the velocity)Solution: Since a(t) = aTT + aNN, and T is parallel to v, such a curve having a v

must have aT = 0. (Note, 0 =av

= (aTT

+aNN

)v

= aT(Tv

)+aN(Nv

) = aT(ds/dt).Therefore, either the speed, ds/dt, or aT is always zero. Since were moving along thiscurve, it follows that aT = 0.) Thus, for such a curve, all of the acceleration is spent inchanging direction only, and not changing the speed.

For example, the curve r(t) = cos(t), sin(t), 0 has v a at all times, t.(b) a // v (i.e., the acceleration is parallel to the velocity).

Solution: If the acceleration is always parallel to velocity, then aN = 0, so the curvatureis

(t) =|v(t) a(t)||v(t)|3 = 0,

since the cross product of parallel vectors is the zero vector, whose length is zero. Thus,

the only curves where acceleration is parallel to velocity are those with zero curvature,which are straight lines.

Suppose r is a curve of the second type, i.e., its acceleration and velocity are parallel.

i. Assuming initial velocity v0, find the vector function v(t) for the velocity at time t.

Solution: In this case, we never change direction so the unit tangent vector to thiscurve is a constant vector and does not vary with time. Hence a(t) = d

2sdt2 T(t) =

d2sdt2 T, so

v(t) =

a(t) dt =

d2sdt2 T dt =

dsdt T + C for some constant vector C. Then, v is parallel

to a means that 0 = v a = dsdt T + Cd2sdt2 T

=dsdt

d2sdt2

(T T) +

d2sdt2

C T =

d2sdt2

C T, so that C = cT for some scalar c. Thus v(t) = ds

dt + c

T. Giveninitial velocity v(0) = v0, we have v0 =

dsdt |t=0 + c

T, so v0 is parallel to T) and

|v0| = dsdt |t=0 + c, which means that c = |v0| dsdt |t=0. But |v0| is the speed at time t = 0and dsdt is speed as a function of time, so

dsdt |t=0 is also the speed at time t = 0. Hence

c = 0, so

v(t) =ds

dtT

for some constant unit vector T.

ii. Assuming initial position r0, find the vector function r(t) for the position at time t.

Solution: Given that v(t) = dsdt T, we have r(t) =

v(t) dt =

dsdt T dt = s(t)T + C

for some constant vectorC

. Here, s(t) =t0 |

v(t)| dt, so s(0) = 0. Hence

r0=

r(0) =s(0)T + C = C, so

r(t) = r0 + s(t)T.

8. Here is an odd property of space curves:

(a) Give an example of a curve in R2 or in R3 for which the position at time t = 10 is thesame as the position at time t = 0, yet the speed is never zero.

Solution: Consider r(t) = cos(2t), sin(2t). Then r(0) = r(10) = 1, 0. However,the speed is never zero.


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(b) Can you find a differentiable scalar function f(x) for which f(0) = f(10) and f(x) isnever zero for 0 x 10? Why or why not? Are there any theorems to support yourbelief?

Solution: This is impossible (i.e., to have a differentiable function f(x) with f(0) =f(10) but never have f(x) = 0 on the interval 0 x 10) according to the MeanValue Theorem.

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