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Applications of Calculus - Contents

1.Rates 0f Change2.Exponential Growth &

Decay3.Motion of a particle4.Motion & Differentiatio

n5.Motion & Integration

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Rates of ChangeThe gradient of a line is a measure of the Rates 0f Change of y in relation to x.

Rate of Changeconstant.

Rate of Changevaries.

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Rates of Change – Example 1/2

R = 4 + 3t2The rate of flow of water is given by

When t =0 then the volume is zero. Find the volume of water after 12 hours.R = 4 + 3t2

i.e dVdt = 4 + 3t2

V = (4 + 3t2).dt∫= 4t + t3 + C

When t = 0, V = 0 0 = 0 + 0 + C C = 0

V = 4t + t3 @ t = 12= 4 x 12 + 123

= 1776 units3

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Rates of Change – Example 1/2

R = 4 + 3t2The rate of flow of water is given by

When t =0 then the volume is zero. Find the volume of water after 12 hours.R = 4 + 3t2

i.e dVdt = 4 + 3t2

V = (4 + 3t2).dt∫= 4t + t3 + C

When t = 0, V = 0 0 = 0 + 0 + C C = 0

V = 4t + t3 @ t = 12= 4 x 12 + 123

= 1776 units3

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Rates of Change – Example 2/2

B = 2t4 - t2 + 2000 a) The initial number

of bacteria. (t =0)

The number of bacteria is given by

B = 2t4 - t2 + 2000 = 2(0)4 – (0)2 + 2000 = 2000 bacteria

b) Bacteria after 5 hours. (t =5)

= 2(5)4 – (5)2 + 2000 = 3225 bacteria

c) Rate of growth after 5 hours. (t =5)

dBdt = 8t3 -2t

= 8(5)3 -2(5) = 990 bacteria/hr

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Exponential Growth and DecayA Special Rate of Change.Eg Bacteria, Radiation, etc

It can be written as dQdt

= kQ

dQdt

= kQcan be solved as Q = Aekt

Growth

Initial Quantity

Growth Constant k(k +ve = growth, k -ve = decay)

Time

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Growth and Decay – Example

Number of Bacteria given by N = Aekt

N = 9000, A = 6000 and t = 8 hoursa) Find k (3 significant figures)

N = A ekt

9000 = 6000 e8k

e8k =90006000

e8k = 1.5

loge1.5 = loge e8k

1.5 =e8k

= 8k loge e= 8k

k =loge1.58

k ≈ 0.0507

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Growth and Decay – Example

Number of Bacteria given by N = Aekt

A = 6000, t = 48 hours, k ≈ 0.0507b) Number of bacteria after 2 days

N = A ekt

= 6000 e0.0507x48

= 68 344 bacteria

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Growth and Decay – Example

Number of Bacteria given by N = Aekt

k ≈ 0.0507, t = 48 hours, N = 68 344c) Rate bacteria increasing after 2 days

dNdt = kN = 0.0507 N

= 0.0507 x 68 3444= 3464 bacteria/hr

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Growth and Decay – Example

Number of Bacteria given by N = Aekt

A = 6000, k ≈ 0.0507d) When will the bacteria reach 1 000 000.

N = A ekt

1 000 000 = 6000 e0.0507t

e0.0507t = 1 000 0006 000

e0.0507t = 166.7

logee0.0507t = loge166.70.0507t logee = loge166.7

0.0507t = loge166.7

t =loge166.70.0507

≈ 100.9 hours

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Growth and Decay – Example

Number of Bacteria given by N = Aekt

A = 6000, k ≈ 0.0507e) The growth rate per hour as a percentage.

dNdt = kN k is the growth constant

k = 0.0507x 100%= 5.07%

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Motion of a particle 1

Displacement (x)

Measures the distance from a point.

To the right is positiveTo the left is negative -

The Origin implies x = 0

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Motion of a particle 2

Velocity (v)

Measures the rate of change of displacement.

To the right is positiveTo the left is negative -

Being Stationary implies v = 0

v = dxdt

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Motion of a particle 3

Acceleration (a)Measures the rate of change of velocity.

+v – a-v + a}

Having Constant Velocity implies a = 0

a = =dvdt

d2xdt2

To the right is positiveTo the left is negative -SlowingDown

+v + a-v - a}Speeding

Up

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Motion of a particle - Example

When is the particle at rest?

t2 & t5

x

tt1 t2 t4 t5 t6 t7t3

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Motion of a particle - Examplex

tt1 t2 t4 t5 t6 t7t3

When is the particle at the origin?

t3 & t6

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Motion of a particle - Examplex

tt1 t2 t4 t5 t6 t7t3

Is the particle faster at t1 or t7?Why?

t1

GradientSteeper

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Motion of a particle - Examplex

tt1 t2 t4 t5 t6 t7t3

Is the particle faster at t1 or t7?Why?

t1

GradientSteeper

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Motion and Differentiation

Displacement

Velocity

Accelerationdtdvx =&& 2

2

dtxd=

x

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Motion and Differentiation - ExampleDisplacement x = -t2

+ t +2 in cm.Find initial velocity (in cm/s).

Initially t = 0

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Motion and Differentiation - ExampleDisplacement x = -t2

+ t + 2Show acceleration is constant.

Acceleration-2 units/s2

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Motion and Differentiation - ExampleDisplacement x = -t2

+ t +2Find when the particle is at the origin.

Origin @ x = 0

&

@ Origin when

t = 2 sec

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Motion and Differentiation - Example

0=dtdx

Displacement x = -t2 + t +2

Find the maximum displacement from origin.Maximum

Displacementwhen v = 012 =+−= t

dtdx

12 −=− t

5.0=t

22 ++−= ttx

25.0)5.0( 2 ++−=x

cmx 25.2=

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Motion and Differentiation - ExampleDisplacement x = -t2

+ t +2Sketch the particles motion

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1

2Initial

Displacementx=2 @ t=0

MaximumDisplacement

x=2.25@ t=0.5

Return toOrigin ‘0’x=0 @ t=2

x

t

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Motion and Integration

Acceleration

Velocity ∫= dtav .

Displacement

a

∫= dtvx .

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Motion and Integration - ExampleVelocity v= 3t2

+ 2t + 1, xo=-2cmFind displacement after 5 secs.

∫= dtvx .

∫ ++= dttt ).123( 2

Ctttx +++= 23

When t=0, x=-2C+++=− 0002 23

2−=C

223 −++= tttxWhen t=5

2555 23 −++=2525125 −++=

153=Displacement is 153cm

to right of origin.

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Motion and Integration - Example

Find displacement after 9 secs.Acceleration a= 6 - , vo=0, xo=+1m2

(t + 1)2

∫= dtav .

( )∫ ⎟⎠⎞⎜

⎝⎛

+−= dtt

.126 2

( )( )∫ −+−= dtt .126 2

( ) Ctt +−+−=

−

1126

1

When t=0, v=0

Ct

t ++

+=)1(

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( ) C+++= )10(2060

C+=20 2−=⇒ C

2)1(26 −+

+=t

tv

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Motion and Integration - Example

Find displacement after 9 secs.Acceleration a= 6 - , vo=0, xo=+1m2

(t + 1)2

2)1(26 −+

+=t

tv

∫= dtvx .

dtt

t .2)1(26∫ ⎟

⎠⎞⎜

⎝⎛ −

++=

cttt e +−++= 2)1(log23 2

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Motion and Integration - Example

Find displacement after 9 secs.Acceleration a= 6 - , vo=0, xo=+1m2

(t + 1)2

ctttx e +−++= 2)1(log23 2

When t=0, x=1

ce +−++= )0(2)10(log2)0(31 2 1=⇒ c

12)1(log23 2 +−++= tttx e When t=91)9(2)19(log2)9(3 2 +−++= ex

)10(log2226 e−=( ))10(log1132 e−= Displacement