MATH 138 Lecture 13 on Section 3.8

1. Applications 3.8 of the Derivative

2. 2
Motion (Physics)

3. 3
Example 1 – Analyzing the Motion of a Particle
The position of a particle is given by the equation
s = f (t) = t
3 – 6t
2 + 9t
where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
(b) What is the velocity after 2 s? After 4 s?
(c) When is the particle at rest?
(d) When is the particle moving forward (that is, in the
positive direction)?
(e) Draw a diagram to represent the motion of the particle.

4. 4
Example 1 – Analyzing the Motion of a Particle
(f) Find the total distance traveled by the particle during the
first five seconds.
(g) Furthest to the right the particle goes in first 2 seconds?
(h) Find the acceleration at time t and after 4 s.
(i) Graph the position, velocity, and acceleration functions
for 0 £ t £ 5.
(j) When is the particle speeding up? When is it slowing
down?
cont’d

5. 5
Example 1 – Solution
Solution:
(a) The velocity function is the derivative of the position
function.
s = f (t) = t
3 – 6t
2 + 9t
v (t) = = 3t
2 – 12t + 9

6. 6
Example 1 – Solution
(b) The velocity after 2 s means the instantaneous velocity
when t = 2 , that is,
v (2) =
= –3 m/s
The velocity after 4 s is
v (4) = 3(4)2 – 12(4) + 9
= 9 m/s
cont’d
= 3(2)2 – 12(2) + 9

7. 7
Example 1 – Solution
(c) The particle is at rest when v (t) = 0, that is,
3t
2 – 12t + 9 = 3(t
2 – 4t + 3)
= 3(t – 1)(t – 3)
= 0
and this is true when t = 1 or t = 3.
Thus the particle is at rest after 1 s and after 3 s.
cont’d

8. Example 1 – Solution
(d) The particle moves in the positive direction when v (t) > 0,
8
that is,
3t
2 – 12t + 9 = 3(t – 1)(t – 3) > 0
This inequality is true when both factors are positive
(t > 3) or when both factors are negative (t < 1).
Thus the particle moves in the positive direction in the
time intervals t < 1 and t > 3.
It moves backward (in the negative direction) when
1 < t < 3.
cont’d

9. Example 1 – Solution
(e) Using the information from part (d) we make a schematic
cont’d
9
sketch in Figure 2 of the motion of the particle back and
forth along a line (the s-axis).
Figure 2

10. Example 1 – Solution
(f) Because of what we learned in parts (d) and (e), we need
10
to calculate the distances traveled during the time
intervals [0, 1], [1, 3], and [3, 5] separately.
The distance traveled in the first second is
| f (1) – f (0) | = | 4 – 0 |
From t = 1 to t = 3 the distance traveled is
| f (3) – f (1) | = | 0 – 4 |
From t = 3 to t = 5 the distance traveled is
| f (5) – f (3) | = | 20 – 0 |
The total distance is 4 + 4 + 20 = 28 m.
cont’d
= 4 m
= 4 m
= 20 m

11. 11
Example 1 – Solution
When at max distance – turn around  velocity = 0!!
v(t) = 0 from part c was t=1, 3, throw out 3 not in interval.
So max distance is s(1) = 4

12. Example 1 – Solution
(h) The acceleration is the derivative of the velocity function:
12
a(t) =
=
= 6t – 12
a(4) = 6(4) – 12
= 12 m/s2
cont’d

13. 13
Example 1 – Solution
(i) Figure 3 shows the graphs of s, v, and a.
cont’d
Figure 3

14. cont’d
14
Example 1 – Solution
(j) The particle speeds up when the velocity is positive and
increasing (v and a are both positive) and also when the
velocity is negative and decreasing (v and a are both
negative).
In other words, the particle speeds up when the velocity
and acceleration have the same sign. (The particle is
pushed in the same direction it is moving.)
From Figure 3 we see that this happens when 1 < t < 2
and when t > 3.

15. 15
Example 1 – Solution
The particle slows down when v and a have opposite
signs, that is, when 0 £ t < 1 and when 2 < t < 3.
Figure 4 summarizes the motion of the particle.
cont’d
Figure 4

16. 16
Exercises

17. 17
Exercises (#10)

18. Exercises (#12)
Sodium Chlorate crystals are easy to grow in the shape of
cubes by allowing a solution of water and sodium chlorate to
evaporate slowly. If V is the volume of such a cube with side
length x, calculate dV/dx when x=3 mm and explain its
meaning.
18

19. 19
Exercises (#4)