Application of Calculus in the Physical World

Maths Extension 1 – Applications of Calculus to the Physical World

Applications of Calculus to the Physical World General Formulae Equations and Deriving Equations Problems

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1. dy

dt

=

2. dQ

dt dN dt

Rates of Change dy dx

dt

Exponential Growth and Decay

= kQ

k Q

= Constant = Aekt

= k ( N − B)

N k B

= B + Aekt = Constant = Constant

n a T F

Period Amplitude Period Frequency

3.

a

×

dx

Motion in 2D

=

d dx

(

1 2

v2 )

4. Simple Harmonic Motion 2 x&& = − n x

v2

= n2 (a 2 − x 2 )

T =

5. x&& x& x

2π

n

=

1

F

Projectile Motion =0 = Vcosθ = Vcosθt

&& y y& y

V

− gx 2 (1 + tan 2 θ ) 2V 2

Time of Flight

=

=

− g − gt + −

gt 2 2

Vsinθ

+

Vsinθt

= -10 = -10t + Vsinθ = -5t2 + Vsinθt

= x& 2 + y& 2

Equation of Motion

=

= =

2V sin θ

g

Maximum Height

+ x tanθ

=

V 2 sin 2 θ 2 g

Maximum Range

=

V 2 sin 2θ g



Equations and Deriving Equations Rates of Change ! What you want to find is on the LHS ! Get into form: dy dy dx

dt

=

dx

×

dt

Exponential Growth and Decay ! Basic Form dQ = kQ dt

k Q

= Constant = Aekt A e k t

!

Complex Form dN = k ( N − B) dt

**Proof** dN = k ( N − B) ___ k, B are constants dt Let u = N – B du d = ( N − B) dt dt dN dB − ____ B is a constant ___ = dt dt dN −0 ___ = dt dN = ku ___ = dt

N k B

= Constant – initial value = = Constant = Time

= B + Aekt = Constant = Constant N = B + Ae kt dN = kAekt dt ___-= k(B + Aekt – B) ___-= k(N – B)

This has solution u = Ae kt, where A is a constant But u = N – B So N – B = Ae kt N = B + Ae kt



Motion in 2D x x x x& v dx

a

&x&

dt dv

Displacement Velocity Acceleration

dt At origin At rest When velocity is constant

x =0 x& = 0 a =0

Special Form: Expressing acceleration in terms of x, not t Velocity d 1 2 dx = (2 v ) a= dx dt d  dx    Acceleration = dx  dt  dv dv dx = = × dt dx dt dv ×v = dx dv d 1 2 × (2 v ) = dx dv d 1 2 = (2 v ) dx



SHM – Simple Harmonic Motion && = − n 2 x Definition: ___ x

v2

= n2 (a 2 − x 2 )

T =

2π

n

=

= = =

Period Amplitude Period Frequency

1

F

Proof x = A cos nt x& = − An sin nt

&& x

n a T F

− An2 cos nt − n 2 A cos nt − n 2 x

&& = x

d dx

(

v2 )

1 2

=

− n 2 x − n2 x 2

+ C 2 At x = a , v = 0 _________ a is the amplitude 0 − n2 x 2 n2 a 2 + C ____ ∴ C = = 2 2 1 2 v − n 2a 2 n 2a 2 2 + = 2 2 2 2 2 2 2 = − n x + n a v = n 2 ( a 2 − x 2 ) dx v = ± n a 2 − x 2 dt 1 2

v

2

dt dx

=

=

t =

1

n a 2 − x 2 1 1

∫

n a 2 − x 2 = 1n cos −1 ( xa ) + C

At x = a, t =

1 2

cos −1 ( xa ) + C _______ ∴ C = 0

nt = cos −1 ( x ) a cos nt =

∴ x

x a

= a cos nt



Projectile Motion V

V

y&

= V sin θ

θ θ

x& =0 = Vcosθ = Vcosθt

&& x x& x

&& y y&

= = =

y

− g − gt + −

gt 2 2

= V cosθ

Vsinθ

+

Vsinθt

At any time t, v the velocity of the particle can be given the equation v MAX HEIGHT __ y&

=0

0 = − gt + V sin θ gt = V sin θ V sin θ = __ Time of Max Height t g

MAX RANGE, TIME OF FLIGHT y = 0

gt 2

+ V sin θ t 0 = − 2 gt 2 = V sin θ t 2 t =

Sub into y equation

2V sin θ _ Time of Flight (Max Range) g

Sub into x equation

 2V sin θ    g  

2

y y y

= = =

 V sin θ  − g  V sin θ    + sin θ  V  g   2  g     −

V 2 sin 2 θ 2 g

V 2 sin 2 θ 2 g

+

V 2 sin 2 θ g

= x& 2 + y& 2

x x x

**

= V cosθ  = =

V 2 sin 2θ g V 2 g

**

_____ only at 45°

Max Angle ___ (using double angles) sin 2θ = 1 2θ = π 2 θ

=

π

4

= 45 °



Cartesian Equation, Trajectory x

= V cosθ t

t =

x

y

=

− gt 2 2

+ V sin θ t

V cosθ Sub into y equation

y

=

y

=

− g  x  2  x    + V sin θ   2  V cosθ   V cosθ  − gx 2 V sin θ x + 2 2

2V cos θ V cosθ 2 2 y = − g sec θ x + tan θ x 2V 2 2 2 y = − gx (1 + tan θ ) + tan θ x 2V 2 This is a quadratic in x and tan θ



Rates of Change Problems Summary Prisms, Spheres

! !

!

Pyramids

Ladder Problem type

dy

=

dy dx

⋅

dt dx dt Sometimes with spheres, use 2 “dummy variables”

!

Use similar triangles V = 13 Ah

!

Try to eliminate unused variables

! !

Solve with Pythagoras Theorem Implicit differentiate with respect to time Sub in values

!

Solve with Cos or Sin rules

!

y

x Boy Chase Girl type

A1 A2 D

B1 B2 !

Shadow Problem type

Solve with similar triangles ____ ratios of length ! Implicit differentiate with respect to time

H1 H2

y

x



Rates of Change is the derivatives of functions with respect to time t. dy dy dx

dt

=

×

dx

dt

Example 1 The Surface Area of a sphere increases at 6cms-1 Find a) The rate of change of radius when r = 5 b) The rate of change of volume when r = 5 A) SA = 4πr 2 dr dr dA = × Find : dt dA dt ________ = 6 ×

dA dr

1

dr

8π r

dA

4π r 3

________ =

Find:

dt

=6

= 8π r =

1 8π r

3

________ =

B) V =

dA

20π

cms-1

3 4 π 3

dV dt

r

=

________ =

dV dr

dr

dr

dt

×

dt

=

3

dV

4π r

dr

= 4π r 2

3 × 4π r 2

4π r ________ = 3r ________ = 15cm3s-1



Exponential Growth and Decay Further Rates of Change

dN dt

= k ( N − P ) _____ k, P are constants

N = P + Ae kt _______-A is constant Example 1 In a certain town, the growth rate in population is given by dN = k ( N − 125) dt a) Show N = 125 + Ae kt is a solution of the differential equation b) If the population is initially 25650, after 5 years it is 31100, find the population after 8 years c) When will the population be 40000? A) N = 125 + Ae kt dN = Akekt dt ___-= k(Aekt) ___-= k(125 + Aekt – 125) ___-= k(N – 125)

C) N = 40000 -40000 = 125 + 25525e 0.0587t ___ 1.6 = e0.0387t ln 1.6 ____- t = 0.0387 _____- ≈ 11.5

B) t = 0, N = 25650 ---25650 = 125 + Ae kt __---_-A = 25525

t = 5, N = 31100 __ 31100 = 125 + 25525e k5 30975 = e k 5 -25525 __- ln1.2 = ln e k 5 ln1.2 _____--k = 5 _-_____- ≈ 0.0387



Example 2 Newton’s Law of Cooling The rate of cooling is proportional to the excess of the temperature of the body over the surrounding medium of the room Temperature of coffee Room Temperature In 10 mins, the temp. is 100° 25° 60° a) What is the temperature of the coffee in 15 minutes? b) What time will it reach 50°? A) dT

= k (T − P ) dt T – P = Ae k0 100 – 25 = Ae k0 A = 75

In this case, P is the room temperature Finding k 60 – 25 = 75Ae k10 35 = e10k __-75 _- ln(157 ) = 10k _____ k =

1 10

ln (157 )

So in 15 minutes time T = 25 + 17e k15 T = 48.9°

B) T = 50 50 – 25 = 75e kt ____- 13 = e kt __ ln (13 ) = kt _____ t =

ln(13 )

k _____--= 14min 28sec



Motion in 2D a

d

= x&& =

dx

(

1 2

v2 )

Example 1 A rocket is projected from Earth, when out of the Earth’s atmosphere, has a retardation of 91000 kms − 2 , where x km is the distance from the centre. 6400km = radius of Earth 2 x a) If the rocket is moving at 4km -1 when it is 600km above the Earth’s surface, find it’s speed after a further 1000km. b) Find also, the total distance traveled before first coming to rest. A) v = 4kms-1, x = 7000km

d dx

(

1 2

v

2

)= −

___- 12 v 2

=

1 2

91000

x 2 91000 x

+ C

(4)2 =

91000

7000 __--8 = 13 + C ___ C = –5

+ C

Finding v _______ x = 8000km 91000 1 2 = −5 v 2 7000 _ v 2 = 12.75 --- v = 3.75 B) v = 0 so v 2 = 0 91000

−5 x 2 5x = 91000 x = 18200 18200 – 6400 = 1182km away form earth 0 =



Simple Harmonic Motion x&& = − n 2 x

Amplitude = a Period 2π = n Frequency 1 n = = T 2π

v 2 = n 2 ( a 2 − x 2 ) x = A sin nt + A cos nt

Example 1 A particle moves along the x-axis according to the law x = 4sin3t a) Show it is SHM. b) Find when the particle is first at 2cm from positive 0. Find the velocity. c) Find the greatest speed of the particle and the interv al at which it moves. A) x = 4 sin 3t x& = 12 cos 3t x&& = −36 sin 3t -- = −9( 4 sin 3t )

C) x& = 12 cos 3t x& is greatest when cos3t = 1 So the greatest velocity is 12cms -1

-- = −32 x ____ is SHM

At x = 0, v = 12

B) 2 = 4 sin 3t 1 = sin 3t 2 3t =

t =

v2

_ 122

π

6

π

18

= n 2 ( a 2 − x 2 )

_ 144

= 32 ( a 2 = 9a2

a2 a

= 16 _ = ±4

− 02 )

= 12 cos 3(18 ) -- = 12 cos 6 -- = 12. 23 -- = 6 3 π

x&

π

The interval at which it moves

-4

0

4



Projectile Motion && = 0 x x& = V cosθ x

&y& y& y

= V cosθ t

Time of Flight

MAX Height

MAX Range

Cartesian Equation

=

=

=

=

= = =

− g − gt + V sin θ − gt + V sin θ t

v

= x& 2 + y& 2

2

V 2 sin 2 θ g 2V sin 2 θ 2 g

V 2 sin 2θ g

− g (1 + tan 2 θ ) 2

2V

x 2

+ tan θ x

Types of projectile Questions: 1) 2) 3) 4)

Level Ground 2 Angles Different Ground Levels No x& , No y&



Level Ground Question A ball is thrown with initial velocity 25ms -1 at angle tan −1 (34 ) . Air resistance is neglected. Take 9 = 10ms-2. Find: a) Maximum height b) Time of flight and range c) Velocity and direction after 0.5secs d) Velocity and direction when it is 10m above the ground e) Find the trajectory of the projectile

25ms-1

25ms-1 (5)

-1

15ms (3)

θ θ

&& x x& x

=0 = 20 = 20t

&y& y& y

20ms-1 (4)

= –10 = –10t + 15 = –5t2 + 15t

A) Max height when y& = 0 0 = –10 t + 15 10 t = 15 t = 1.5 secs

B) Time of flight and range when y = 0 0 = –5t2 + 15t = –5t(t – 3) t = 0 or 3 ______ T = 3

Sub t = 1.5 into y y = –5(1.5)2 + 15(1.5) = 11.25m

Sub t = 3 into x x = 20(3) = 60

C) Velocity and Direction after 0.5 secs x& = 20 y& = –10(0.5) + 15 = 10 Sub into v

v

=

=

x 2 + y 2 202

10 5 ms-1

10ms-1

+ 102

= 10 5

) tan −1 (10 20 θ

θ

=θ

20ms-1

= 26° 34’



D) Velocity and Direction of projectile when it is 10m above the ground 10 = –5t2 + 15t 2 = –t2 + 3t 0 = t2 – 3t + 2 = (t – 1)(t – 2) Projectile is 10m above the ground at t = 1, 2 x& = 20 y& = –10(1) + 15 =5

x& y&

= 20 = –10(2) + 15 = –5 20 θ

5

-5

θ

20

V = 5 17 ___ 14° 2’

V = 5 17 ___ -14° 2’

E) Trajectory of projectile x = 20t x t = 20 Sub into y y = –5t2 + 15t = =

 x 2   x  − 5   + 15   400   20  −

x 2

+

3 x

80 4 x = (60 − x ) 80



2 Angles Question On level ground, projectiles with the same initial velocity may be projected at 2 different angles to reach a particular point x distance away. Except at 45° - the maximum angle. V1

Projectiles can be assumed as parabolic V2

A footballer kicks a ball at 16ms -1. The ball just passes over a wall 4m high when s/he is 16m away. a) Show the angle is 5tan 2θ – 16tanθ + 9 = 0 b) Find the two angles the ball may be kicked.

&& x x& x

=0 = 16cosθ = 16cosθt

&y& y& y

= –10 = –10t + 16sin θ = –5t2 + 16sinθt

4m 16m

A) 16 = 16cosθt 1 = cosθt t = secθ

B) 0 = 5tan2 θ – 16tanθ + 9 tan θ 16 ± 16 2 − 4(5)(9) = 2(5) =

Sub t = secθ and y = 4 into y 2

4 = –5sec θ + 16sinθ.secθ = –5(1 + tan2 θ) + 16tanθ = –5 – 5tan2 θ + 16tanθ 0 = 5tan2 θ – 16tanθ + 9

=

= θ

16 ± 256 − 180 10 16 ± 2 19 10 8 ± 19 5

= 67° 58’ and 36° 4’



Different Ground Levels Question A stone is thrown from the top of a 25m cliff. x& = 20 5 , y& = 20. Find: a) The distance away from the base of the cliff when it hits the sea b) Maximum height above sea level

V

20 25m

θ

20 5

&& x x&

=0

x

= 20 5

&y& y&

= –10 = –10t + 20

= 20 5 t

y

= –5t2 + 20t

A) Where it hits the sea y = –25 –25 = –5t 2 + 20t 0 = t2 – 4t – 5 = (t – 5)(t + 1)

t = 5, -1 Sub t = 5 in x x = 20 5 (5) = 100 5 from cliff base

B) Max height above sea level y& = 0 0 = –10t + 20 t = 2 Sub t = 2 in y y = –5(2)2 + 20(2) = –20 + 40 = 20 20 + 25 = 45m above sea level



No x& , No y& A stone is thrown horizontally from the top of a 25m cliff with i nitial velocity of 40ms -1 Find: a) Equation of motion b) Where the stone hits the sea c) The velocity of impact d) If another stone is dropped from the same place, will it reach the sea at the same time? x&& x& x

-1

40ms

=0 = 40 = 40t

&y& y& y

= –10 = –10t = –5t2

25m

A) x

B) –25 = –5t 2 5 = t2 t = ± 5

= 40t x t = 40

Sub t =

x 40

in y

Sub t =

_ x

2

y

= =

x  − 5    40  − x 2

5 in x

= 40 5

320

C)

D)

x& y&

v

= 40 = =

− 10 40 2

Yes, both will take

5 secs to reach the sea.

5

+ (− 10

5

)

2

= 10 21 θ

= 150° 48’ in the positive direction


Application of Calculus in the Physical World

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