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.

1. Sections 10.3–4
Curves, Arc Length, and Acceleration
Math 21a
February 22, 2008
Announcements
Problem Sessions:
Monday, 8:30, SC 103b (Sophie)
Thursday, 7:30, SC 103b (Jeremy)
Office hours Tuesday, Wednesday 2–4pm SC 323.

2. Outline
Arc length
Velocity

3. Pythagorean length of a line segment
Given two points P1 (x1 , y1 ) and P2 (x2 , y2 ), we can use Pythagoras
to find the distance between them:
P2
y
y2 − y1
P1 x2 − x1
x
|P1 P2 | = (x2 − x1 )2 + (y2 − y1 )2 = (∆x)2 + (∆y )2

4. Length of a curve
Break up the curve into pieces, and approximate the arc length
with the sum of the lengths of the pieces:
y
x

5. Length of a curve
Break up the curve into pieces, and approximate the arc length
with the sum of the lengths of the pieces:
y
x

6. Length of a curve
Break up the curve into pieces, and approximate the arc length
with the sum of the lengths of the pieces:
y
x

7. Length of a curve
Break up the curve into pieces, and approximate the arc length
with the sum of the lengths of the pieces:
y
x

8. Length of a curve
Break up the curve into pieces, and approximate the arc length
with the sum of the lengths of the pieces:
y
x

9. Length of a curve
Break up the curve into pieces, and approximate the arc length
with the sum of the lengths of the pieces:
y
n
L≈ (∆xi )2 + (∆yi )2
i=1
x

10. Sum goes to integral
If 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
n n
L≈ (∆xi )2 + (∆yi )2 ≈ [f (ti )∆ti ]2 + [g (ti )∆ti ]2
i=1 i=1
n
= [f (ti )]2 + [g (ti )]2 ∆ti
i=1
As n → ∞, this converges to
b
L= [f (t)]2 + [g (t)]2 dt
a

11. Sum goes to integral
If 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
n n
L≈ (∆xi )2 + (∆yi )2 ≈ [f (ti )∆ti ]2 + [g (ti )∆ti ]2
i=1 i=1
n
= [f (ti )]2 + [g (ti )]2 ∆ti
i=1
As n → ∞, this converges to
b
L= [f (t)]2 + [g (t)]2 dt
a
In 3D, r(t) = f (t), g (t), h(t) , and
b
L= [f (t)]2 + [g (t)]2 + [h (t)]2 dt
a

12. Example
Example
Find the length of the parabola y = x 2 from x = 0 to x = 1.

13. Example
Example
Find the length of the parabola y = x 2 from x = 0 to x = 1.
Solution
Let r(t) = t, t 2 . Then
√
1
5 1 √
L= 1+ (2t)2 = + ln 2 + 5
0 2 4

14. Example
Find the length of the curve
r(t) = 2 sin t, 5t, 2 cos t
for −10 ≤ t ≤ 10.

15. Example
Find the length of the curve
r(t) = 2 sin t, 5t, 2 cos t
for −10 ≤ t ≤ 10.
Answer
√
L = 20 29

16. Outline
Arc length
Velocity

17. Velocity and Acceleration
Definition
Let r(t) be a vector-valued function.
The velocity v(t) is the derivative r (t)
The speed is the length of the derivative |r (t)|
The acceleration is the second derivative r (t).

18. Example
Find the velocity, acceleration, and speed of a particle with
position function
r(t) = 2 sin t, 5t, 2 cos t

19. Example
Find the velocity, acceleration, and speed of a particle with
position function
r(t) = 2 sin t, 5t, 2 cos t
Answer
r (t) = 2 cos(t), 5, −2 sin(t)
√
r (t) = 29
r (t) = −2 sin(t), 0, −2 cos(t)

20. Example
The position function of a particle is given by
r(t) = t 2 , 5t, t 2 − 16t
When is the speed a minimum?

21. Example
The position function of a particle is given by
r(t) = t 2 , 5t, t 2 − 16t
When is the speed a minimum?
Solution
The square of the speed is
(2t)2 + 52 + (2t − 16)2
which is minimized when
0 = 8t + 4(2t − 16) =⇒ t = 4

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

23. Solution
The position function is given by
r(t) = 115 cos(50◦ )t, 3 + 115 sin(50◦ )t − 16t 2
The question is: what is g (t) when f (t) = 310? The equation
310
f (t) = 310 gives t ∗ = , so
115 cos(50◦ )
2 2
310 310
g (t ∗ ) = 3 + 115 sin(50◦ ) − 16
115 cos(50◦ ) 115 cos(50◦ )
≈ 91.051 ft

24. Solution
The position function is given by
r(t) = 115 cos(50◦ )t, 3 + 115 sin(50◦ )t − 16t 2
The question is: what is g (t) when f (t) = 310? The equation
310
f (t) = 310 gives t ∗ = , so
115 cos(50◦ )
2 2
310 310
g (t ∗ ) = 3 + 115 sin(50◦ ) − 16
115 cos(50◦ ) 115 cos(50◦ )
≈ 91.051 ft
Home run!