lesson 8: curves, arc length, acceleration
DESCRIPTION
The velocity of a vector function is the absolute value of its tangent vector. The speed of a vector function is the length of its velocity vector, and the arc length (distance traveled) is the integral of speed.TRANSCRIPT
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!