lecture 13 applications - section 3.8

Applications of the Derivative3.8

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Motion (Physics)

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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.

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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

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Example 1 – SolutionSolution:

(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

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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

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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

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Example 1 – Solution

(d) The particle moves in the positive direction when v (t) > 0,

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

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Example 1 – Solution

(e) Using the information from part (d) we make a schematic

sketch in Figure 2 of the motion of the particle back and

forth along a line (the s-axis).

cont’d

Figure 2

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Example 1 – Solution(f) Because of what we learned in parts (d) and (e), we need

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

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Example 1 – SolutionWhen 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

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Example 1 – Solution

(h) The acceleration is the derivative of the velocity function:

a(t) =

=

= 6t – 12

a(4) = 6(4) – 12

= 12 m/s2

cont’d

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Example 1 – Solution

(i) Figure 3 shows the graphs of s, v, and a.

cont’d

Figure 3

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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.

cont’d

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Example 1 – SolutionThe 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

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Exercises

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Exercises (#10)

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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.

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Exercises (#4)