M53 Lec2.4 Rates of Change and Rectilinear Motion

Rates of Change and Rectilinear Motion

Mathematics 53

Institute of Mathematics (UP Diliman)

Institute of Mathematics (UP Diliman) Rates of Change and Rectilinear Motion Mathematics 53 1 / 32

For today

1 Rates of Change

2 Rectilinear Motion


Rates of Change

Rates of change occur in many applications.

A microbiologist might be interested in the rate at which the number of bacteriain a culture changes with time.

An engineer might be interested in the rate at which the length of a metal rodchanges with temperature.

An economist might be interested in the rate at which production cost changeswith the quantity of a product that is manufactured.

A medical researcher might be interested in the rate at which the radius of anartery changes with the concentration of alcohol in the bloodstream.


Rates of Change

A hot air balloon started rising vertically from the ground and after 10 seconds, itwas observed to be at an altitude of 40 feet from the ground. 30 seconds after itleft the ground, it was found to be 100 feet from the ground. On average, whatwas its speed between the 10th and the 30th second?

Average speed =100 4030 10 =

6020

= 3 feet/second


Rates of Change

In general, given y = f (x), we can measure how fast y changes with respect tochanges in x.

DefinitionSuppose f is a function and y = f (x).

The average rate of change of y with respect to x on [x0, x] is

f (x) f (x0)x x0 =

fx

.


Rates of Change

Question: In the example earlier, is the velocity at each moment equal to 3feet/second? Not necessarily.

DefinitionSuppose f is a function and y = f (x).

The instantaneous rate of change of y with respect to x at x = x0 is

limxx0

f (x) f (x0)x x0 = limx0

fx

= f (x0)

This gives us a new interpretation for the derivative.Aside from giving the slope of a tangent line, it gives an instantaneous rate ofchange.


Rates of Change

Remarks

Graphically, the average rate of change of y with respect to x on [x0, x] is theslope of the secant line passing through P(x0, f (x0)) and Q(x, f (x)).

The derivative of f at x = x0, f (x0) can be interpreted as the instantaneousrate of change of y with respect at x at x = x0. That is, f (x0) is the rate ofchange of y per unit change in x at the instant when x = x0.


Rates of Change

Remarks

Let y be a function of x.

Ifdydx

> 0 on an interval I, then y increases as x increases, and y decreases as xdecreases.

Ifdydx

< 0 on an interval I, then y decreases as x increases, and y increases as xdecreases.

Ifdydx

= 0 on an interval I, then y does not change with respect to x.


Rates of Change

ExampleA right circular cylinder has a fixed height of 6 units. Find the rate of change of itsvolume with respect to the radius of its base.

Solution.

FinddVdr

.

V = pir2h = 6pir2

dVdr

= 12pir. (Interpretation?)


Rates of Change

ExampleA bactericide was introduced to a nutrient broth in which bacteria were growing.The bacterium population continued to grow for some time but then stoppedgrowing and began to decline. The size of the population at time t (hours) was

P = 106 + 104t 103t2.Determine the growth rates at t = 0, t = 5 and t = 10 hours.

Solution.The growth rate or the rate of change of the bacterium population P withrespect to time t is given by P(t).

= P(t) = 104 2(103)t = 2(103)(5 t). P(0) = 104, P(5) = 0 and P(10) = 104.


Rates of Change

Notice that given

P(t) = 104 2(103)t = 2(103)(5 t)

P(t) > 0 when t [0, 5) and P(t) < 0 when t (5,)Thus, the bacterium population was increasing until t = 5 hours, then itstopped growing and began to decline.


Rates of Change

ExampleA ladder 24 ft. long rests against a vertical wall. Let be the angle between thetop of the ladder and the wall and let x be the distance from the bottom of theladder to the wall. If the bottom of the ladder slides away from the wall, how fast

does x change with respect to when =pi

3?

Solution.

24 ft.

x

From the figure, the equation thatrelates x and is

sin =x24

x = 24 sin .


Rates of Change

We want to finddxd

at =pi

3.

x = 24 sin dxd

= 24 cos .

At =pi

3,

dxd

= pi3

= 24 cos(pi3

)= 12 ft/radian.


Rectilinear Motion

Suppose a particle is moving along a straight line, which we shall refer to as thes-axis.

Suppose the position of the particle at time t is given by the function s(t), calledthe position function of the particle.

The average velocity of the particle on [t0, t] is

vave =s(t) s(t0)

t t0 =st

.


Rectilinear Motion

DefinitionsLet s(t) be the position function of a particle moving along s-axis.

1 The instantaneous velocity of the particle at time t is

v(t) = limt0

st

=dsdt

= f (t).

2 The instantaneous speed of the particle at time t is |v(t)|.3 The instantaneous acceleration of the particle at time t is

a(t) = limt0

vt

=dvdt

= v(t) or a(t) = d2sdt2

= f (t).


Rectilinear Motion

RemarksLet s(t) be the position function of a particle moving along the s-axis. The signs ofv(t) and a(t) give us information about the motion of the particle.1 If v(t) > 0, then the particle is moving in the positive direction of s (usually to

the right or upward) at time t.2 If v(t) < 0, then the particle is moving in the negative direction of s (usually to

the left or downward) at time t.3 If v(t) = 0, either the particle is not moving or is changing direction at time t.


Rectilinear Motion

Remarks1 If a(t) > 0, then the velocity of the particle is increasing at time t. In addition,

if v(t) > 0 then the speed of the particle is increasing at time t (speeding up).if v(t) < 0 then the speed of the particle is decreasing at time t (slowing down).

2 If a(t) < 0, then the velocity of the particle is decreasing at time t. In addition,if v(t) > 0 then the speed of the particle is decreasing at time t.if v(t) < 0 then the speed of the particle is increasing at time t.

3 If a(t) = 0, then the velocity of the particle is constant.(This does not mean that the particle is NOT moving!)


Rectilinear Motion

That is, the particle is

speeding up when v(t) and a(t) have the same sign

slowing down when v(t) and a(t) are opposite in sign


Rectilinear Motion

ExampleA particle moves along a horizontal coordinate line in such a way that its positionat time t is specified by s(t) = t3 12t2 + 36t 30 where s is measured in feetand t in seconds.

1 Find the instantaneous velocity and the instantaneous acceleration in terms oft.

2 Describe the position and motion of the particle in a table that includes theintervals of time when the particle is moving to the left or to the right, when thevelocity is increasing or decreasing, when the speed is increasing ordecreasing, and the particles position with respect to the origin during theseintervals of time.

3 Show the motion of the particle schematically.4 Determine the total distance traveled by the particle during the first 7 seconds.


Rectilinear Motion

ExampleA particle moves along a horizontal coordinate line in such a way that its positionat time t is specified by s(t) = t3 12t2 + 36t 30 where s is measured in feetand t in seconds.1. Find the instantaneous velocity and the instantaneous acceleration in terms of

t.

Solution.s(t) = t3 12t2 + 36t 30

= v(t) = dsdt

= 3t2 24t+ 36

= a(t) = dvdt

= 6t 24


Rectilinear Motion

ExampleA particle moves along a horizontal coordinate line in such a way that its positionat time t is specified by s = t3 12t2 + 36t 30 where s is measured in feet and tin seconds.

2. Describe the position and motion of the particle in a table that includes theintervals of time when the particle is moving to the left or to the right, when thevelocity is increasing or decreasing, when the speed is increasing ordecreasing, and the particles position with respect to the origin during theseintervals of time.

Solution. First, we find t 0 such that v(t) = 0 and a(t) = 0.v(t) = 0

3t2 24t+ 36 = 03(t 6)(t 2) = 0

t = 6, t = 2

a(t) = 06t 24 = 0

t = 4


Rectilinear Motion

s(t) = t3 12t2 + 36t 30, v(t) = 3(t 2)(t 6), a(t) = 6(t 4)s(t) v(t) a(t) Conclusions

t = 0 30 + left of origin, towards right,decreasing velocity, slowing down

0 < t < 2 + towards right, decreasing velocity,slowing down

t = 2 2 0 right of origin, changing direction,decreasing velocity

2 < t < 4 towards left, decreasing velocity,speeding up

t = 4 14 0 left of origin, towards left,constant velocity

4 < t < 6 + towards left, increasing velocity,slowing down

t = 6 30 0 + left of origin, changing direction,increasing velocity

t > 6 + + towards right, increasing velocity,speeding up


Rectilinear Motion

ExampleA particle moves along a horizontal coordinate line in such a way that its positionat time t is specified by s(t) = t3 12t2 + 36t 30 where s is measured in feetand t in seconds.3. Show the motion of the particle schematically.

Solution.Schematically, the particle moved in the following way:

30 2

t = 0t = 2

t = 6


ExampleA particle moves along a horizontal coordinate line in such a way that its positionat time t is specified by s = t3 12t2 + 36t 30 where s is measured in feet and tin seconds.

4. Determine the total distance traveled by the particle during the first 7 seconds.

Solution.Note that the particle changed directions at t = 2 and t = 6.

30 2

t = 0t = 2

t = 6 Total distance traveled= |s(2) s(0)|+ |s(6) s(2)|+|s(7) s(6)|= 32+ 32+ 7= 71 feet.


Rectilinear Motion

ExampleA ball was thrown from the edge of a cliff such that its directed distance from theground after t seconds is given by s(t) = 16t2 + 32t+ 128 feet. Determine:

1 the acceleration of the ball at t = 32 the maximum height the ball will attain3 the velocity at which the ball was thrown4 how long it would take the ball to hit the ground5 the speed at which the ball would hit the ground


Rectilinear Motion

ExampleA ball was thrown from the edge of a cliff such that its directed distance from theground after t seconds is given by s(t) = 16t2 + 32t+ 128 feet. Determine:1. the acceleration of the ball at t = 3

Solution.s(t) = 16t2 + 32t+ 128= v(t) = 32t+ 32= a(t) = 32

a(3) = 32 feet/second2


Rectilinear Motion

ExampleA ball was thrown from the edge of a cliff such that its directed distance from theground after t seconds is given by s(t) = 16t2 + 32t+ 128 feet. Determine:2. the maximum height the ball will attain

Solution.

Since the graph of s is a parabola opening downward, the maximum value isattained at the vertex.

tcoordinate: 322(16) =

3232

= 1

scoordinate: s(1) = 16(1) + 32(1) + 128 = 144 feet


Rectilinear Motion

ExampleA ball was thrown from the edge of a cliff such that its directed distance from theground after t seconds is given by s(t) = 16t2 + 32t+ 128 feet. Determine:3. the velocity at which the ball was thrown

Solution.

Velocity at which the ball was thrown = v(0)

v(t) = 32t+ 32

v(0) = 32(0) + 32 = 32 feet/second


Rectilinear Motion

ExampleA ball was thrown from the edge of a cliff such that its directed distance from theground after t seconds is given by s(t) = 16t2 + 96t+ 128 feet. Determine:4. how long it would take the ball to hit the ground

5. the speed at which the ball would hit the ground

Solution.s(t) = 0

16(t2 2t 8) = 0 16(t 4)(t+ 2) = 0

Thus, t = 4 or t = 2.

Since t 0, the ball would hit theground after 4 seconds.

It would hit the ground at a speed of|v(4)| = | 32(4) + 32|= 96 feet/second


Exercise

The position function of a particle moving along a coordinate line is

s(t) = 6t2 t3,where s is in feet and t is in seconds.

Determine the velocity and acceleration functions.

When is the particle speeding up?

Answers.v(t) = 12t 3t2 = 3t(t 4)a(t) = 12 6t = 6(t 2)Speeding up at t (0, 2) (4,+)when (i) a(t) > 0 and v(t) > 0 OR (ii) a(t) < 0 and v(t) < 0


Rates of ChangeRectilinear Motion

M53 Lec2.4 Rates of Change and Rectilinear Motion

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