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Periodic Functions And Applications III

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Periodic Functions And Applications III

Significance of the constants A,B,C and D on the graphs of y = A sin(Bx+C) + D, y = A cos(Bx+C) + D

Application of periodic functions

Solution of simple trig equations within a specified domain

Derivatives of functions involving sin x and cos x

Applications of the derivatives of sin x and cos x in life-related situations

Periodic Functions And Applications III

REVISION

FM Page 118 Ex 5.8

New Q

Page 351 Ex 10.1

No. 1-12 (parts a & b only), leave out no.10

- ModelFind all values of x (to the nearest minute) where 0< x <360 for which
- (a) sin x = 0.5
- (b) tan x = -1

sin is positive

angle is in Q1 or Q2

(a) sin x = 0.5

x = 30 or x = 180 - 30

= 30 or 150

30

30

Value of sin x is 0.5

30 off x-axis

tan is negative

angle is in Q2 or Q4

(b) tan x = -1

x = 180 - 45 or x = 360 - 45

= 135 or 315

45

45°

Value of tan x is -1

45 off x-axis

New Q

Ex 10.3

Page 363 2,6

FM Page 119 Ex 5.91 (orally)

General Solution of a Trig Function

So there appears to be more than one solution

cos θ = 0.643

θ = cos-1 (0.643)

θ ≈ 50°

But cos 310° = 0.643 also

So, how many solutions are there?

y=0.643

cos curve

cos θ = 0.634

θ = 50° orθ = 310°

or θ = 50° + 360°orθ = 310° + 360°

or θ = 50° + 2 x 360°orθ = 310° -360°

or θ = 50° + 3 x 360°orθ = 310° - 2 x 360°

or θ = 50° - 360°orθ = 310° - 3 x 360°

or θ = 50° - 2 x 360° etc

θ = 50° + 360° x n θ = 310° + 360° x n

θ = 50° + 360n

θ = 310° + 360°n

For all integer values of n

Model:

Find all values of x (to the nearest minute) for which

(a) sin x = 0.5

sin is positive

angle is in Q1 or Q2

(a) sin x = 0.5

x = 30 or x = 180 - 30

= 30 or 150

general solution is

x = 30 + n x 360 or x = 150 + n x 360

30

30

Value of sin x is 0.5

30 off x-axis

ModelFind all values of x (to the nearest minute) where 0 ≤ x ≤ 360 for which

(a) sin2x = 0.25

(b) tan 3x = -1

sin is pos or neg

angle is in Q1,Q2,Q3 or Q4

(a) sin2x = 0.25

sin x = ± 0.5

x = 30 or x = 150 or x = 210 or x = 330

30

30

30

30

Value of sin x is 0.5

30 off x-axis

tan is negative

angle is in Q2 or Q4

(b) tan 3x = -1

3x = 135° + 360n or 3x = 315 + 360n

x = 45 + 120n or x = 105 + 120n

45, 165, 285, 105, 225, 345

45

45°

Value of tan x is -1

45 off 3x-axis

New Q

Page 371 Ex 10.5

1, 2, 4,10, 14

FM Page 304 Ex 14.161 -23 with GC

- Derivatives of functions involving sin x and cos x

y = sin x

dy = cos x

dx

y = cos x

dy = -sin x

dx

- do some examples on Graphmatica

________________________________________

(a) y = sin 2x

= sin u where u = 2x

dy= cos u du= 2

du dx

dy=dy . du

dx du dx

= 2 cos u

= 2 cos 2x

________________________________________

(b) y = sin32x ( = (sin 2x)3 )

= u3 where u = sin 2x

dy= 3u2du= 2 cos 2x

du dx

dy=dy . du

dx du dx

= 3u2 . 2 cos 2x

= 6 sin22x cos 2x

________________________________________

(c) y = sin2x cos3x

= uv where u = sin2x and v = cos3x

du= 2 sinx cosx dv = -3 sin3x

dx dx

dy= u dv + v du

dx dx dx

= -3 sin3x sin2x + cos3x 2 sin x cos x

= -3 sin3x sin2x + 2 cos3x sin x cos x

________________________________________

(d) y = sin (π-3x)

= sin 3x

dy= 3 cos 3x

dx

NEWQ P50 2.4No. 1(a,b,d,f,h,j), 3 – 8 (all)FM Page 447 Ex 19.51,4,5

- Consider the motion of an object on the end of a spring dropped from a height of 1m above the equilibrium point which takes 2π seconds to return to the starting point.

1m

0m

-1m

s = cos t

v = -sin t

a = -cos t

Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds.(a) How far from the fixed point is the object at the start?(b) How long does it take for the object to return to its starting point?(c) Find the object’s velocity (i) at the start (ii) as it passes the fixed point (iii) after 2 sec(d) Find its acceleration as it passes the fixed point

Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds.(a) How far from the fixed point is the object at the start?

- At the start, t = 0
- When t = 0, s = 4 cos(3x0)
- = 4 cos0
- = 4 x 1
- = 4
- i.e. at the start, object is 4 metres from the fixed point.

Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds. (b)How long does it take for the object to return to its starting point?

Returns to starting point s = 4

4 cos3t = 4

cos3t = 1

3t = 2nπ

t = 2nπ/3

t = 2π/3, 4π/3, 6π/3, …

i.e. first returns to starting point after 2π/3 secs

Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds.(c) Find the object’s velocity (i) at the start(ii) as it passes the fixed point (iii) after 2 sec

s = 4 cos3t

v = -12 sin3t

(i) At the start

When t = 0, v = -12 sin 0

= 0

Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds.(c) Find the object’s velocity (i) at the start (ii) as it passes the fixed point(iii) after 2 sec

- (ii) at what time does it pass the fixed point
s = 0

- 4 cos 3t = 0
- cos 3t = 0
- 3t = π/2, …
- t = π/6, …
- When t = π/6, v = -12 sin 3π/6
- = -12

Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds.(c) Find the object’s velocity (i) at the start(ii) as it passes the fixed point (iii) after 2 sec

- (iii) After 2 sec
- v = -12 sin 3x2
- = -12 sin 6
- = 3.35

Model: The motion of an object oscillates such that its displacement, s, from a fixed point is given by s = 4cos3t where s is in meters and t is in seconds.(d) Find its acceleration as it passes the fixed point

- s = 4 cos3t
- v = -12 sin3t
- a = -36 cos3t
It passes the fixed point when t = π/6,

a = -36 cos 3π/6

= 0

NEWQ P59 Ex 2.6No. 1 – 4, 7