Lesson 8: Curves, Arc Length, Acceleration

Sections 10.3–4

Curves, Arc Length, and Acceleration

Math 21a

February 22, 2008

Announcements

I Problem Sessions:I Monday, 8:30, SC 103b (Sophie)I Thursday, 7:30, SC 103b (Jeremy)

I Office hours Tuesday, Wednesday 2–4pm SC 323.

Outline

Arc length

Velocity

Pythagorean length of a line segment

Given two points P1(x1, y1) and P2(x2, y2), we can use Pythagorasto find the distance between them:

x

y

P1

P2

x2 − x1

y2 − y1

|P1P2| =√

(x2 − x1)2 + (y2 − y1)2 =√

(∆x)2 + (∆y)2

Length of a curve

Break up the curve into pieces, and approximate the arc lengthwith the sum of the lengths of the pieces:

x

y

Length of a curve

Break up the curve into pieces, and approximate the arc lengthwith the sum of the lengths of the pieces:

x

y

Length of a curve

Break up the curve into pieces, and approximate the arc lengthwith the sum of the lengths of the pieces:

x

y

Length of a curve

Break up the curve into pieces, and approximate the arc lengthwith the sum of the lengths of the pieces:

x

y

Length of a curve

Break up the curve into pieces, and approximate the arc lengthwith the sum of the lengths of the pieces:

x

y

Length of a curve

Break up the curve into pieces, and approximate the arc lengthwith the sum of the lengths of the pieces:

x

y

L ≈n∑

i=1

√(∆xi )2 + (∆yi )2

Sum goes to integralIf 〈x , y〉 is given by a vector-valued function r(t) = 〈f (t), g(t), 〉with domain [a, b], we can approximate:

∆xi = f ′(ti )∆ti ∆xi = g ′(ti )∆ti

So

L ≈n∑

i=1

√(∆xi )2 + (∆yi )2 ≈

n∑i=1

√[f ′(ti )∆ti ]

2 + [g ′(ti )∆ti ]2

=n∑

i=1

√[f ′(ti )]2 + [g ′(ti )]2 ∆ti

As n→∞, this converges to

L =

∫ b

a

√[f ′(t)]2 + [g ′(t)]2 dt

In 3D, r(t) = 〈f (t), g(t), h(t)〉, and

L =

∫ b

a

√[f ′(t)]2 + [g ′(t)]2 + [h′(t)]2 dt

Sum goes to integralIf 〈x , y〉 is given by a vector-valued function r(t) = 〈f (t), g(t), 〉with domain [a, b], we can approximate:

∆xi = f ′(ti )∆ti ∆xi = g ′(ti )∆ti

So

L ≈n∑

i=1

√(∆xi )2 + (∆yi )2 ≈

n∑i=1

√[f ′(ti )∆ti ]

2 + [g ′(ti )∆ti ]2

=n∑

i=1

√[f ′(ti )]2 + [g ′(ti )]2 ∆ti

As n→∞, this converges to

L =

∫ b

a

√[f ′(t)]2 + [g ′(t)]2 dt

In 3D, r(t) = 〈f (t), g(t), h(t)〉, and

L =

∫ b

a

√[f ′(t)]2 + [g ′(t)]2 + [h′(t)]2 dt

Example

Example

Find the length of the parabola y = x2 from x = 0 to x = 1.

SolutionLet r(t) =

⟨t, t2

⟩. Then

L =

∫ 1

0

√1 + (2t)2 =

√5

2+

1

4ln∣∣∣2 +

√5∣∣∣

Example

Example

Find the length of the parabola y = x2 from x = 0 to x = 1.

SolutionLet r(t) =

⟨t, t2

⟩. Then

L =

∫ 1

0

√1 + (2t)2 =

√5

2+

1

4ln∣∣∣2 +

√5∣∣∣

Example

Find the length of the curve

r(t) = 〈2 sin t, 5t, 2 cos t〉

for −10 ≤ t ≤ 10.

AnswerL = 20

√29

Example

Find the length of the curve

r(t) = 〈2 sin t, 5t, 2 cos t〉

for −10 ≤ t ≤ 10.

AnswerL = 20

√29

Outline

Arc length

Velocity

Velocity and Acceleration

DefinitionLet r(t) be a vector-valued function.

I The velocity v(t) is the derivative r′(t)

I The speed is the length of the derivative |r′(t)|I The acceleration is the second derivative r′′(t).

Example

Find the velocity, acceleration, and speed of a particle withposition function

r(t) = 〈2 sin t, 5t, 2 cos t〉

Answer

I r′(t) = 〈2 cos(t), 5,−2 sin(t)〉I∣∣r′(t)

∣∣ =√

29

I r′′(t) = 〈−2 sin(t), 0,−2 cos(t)〉

Example

Find the velocity, acceleration, and speed of a particle withposition function

r(t) = 〈2 sin t, 5t, 2 cos t〉

Answer

I r′(t) = 〈2 cos(t), 5,−2 sin(t)〉I∣∣r′(t)

∣∣ =√

29

I r′′(t) = 〈−2 sin(t), 0,−2 cos(t)〉

Example

The position function of a particle is given by

r(t) =⟨t2, 5t, t2 − 16t

⟩When is the speed a minimum?

SolutionThe square of the speed is

(2t)2 + 52 + (2t − 16)2

which is minimized when

0 = 8t + 4(2t − 16) =⇒ t = 4

Example

The position function of a particle is given by

r(t) =⟨t2, 5t, t2 − 16t

⟩When is the speed a minimum?

SolutionThe square of the speed is

(2t)2 + 52 + (2t − 16)2

which is minimized when

0 = 8t + 4(2t − 16) =⇒ t = 4

Example

A batter hits a baseball 3 ft above the ground towards the GreenMonster in Fenway Park, which is 37 ft high and 310 ft from homeplate (down the left field foul line). The ball leaves with a speed of115 ft/s and at an angle of 50◦ above the horizontal. Is the ball ahome run, a wallball double, or is it caught by the left fielder?

SolutionThe position function is given by

r(t) =⟨115 cos(50◦)t, 3 + 115 sin(50◦)t − 16t2

⟩The question is: what is g(t) when f (t) = 310? The equation

f (t) = 310 gives t∗ =310

115 cos(50◦), so

g(t∗) = 3 + 115 sin(50◦)

(310

115 cos(50◦)

)2

− 16

(310

115 cos(50◦)

)2

≈ 91.051 ft

Home run!

SolutionThe position function is given by

r(t) =⟨115 cos(50◦)t, 3 + 115 sin(50◦)t − 16t2

⟩The question is: what is g(t) when f (t) = 310? The equation

f (t) = 310 gives t∗ =310

115 cos(50◦), so

g(t∗) = 3 + 115 sin(50◦)

(310

115 cos(50◦)

)2

− 16

(310

115 cos(50◦)

)2

≈ 91.051 ft

Home run!