periodic functions and applications iii

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

FM Page 118 Ex 5.8

New Q

51 Ex 10.1No. 1-12 (parts a & b only), leave out

no.10

REVISION

Solving Trig(onometric) Equations

• Model Find all values of x (to the nearest minute) where 0< x <360 for which

• (a) sin x = 0.5• (b) tan x = -1

(a) sin x = 0.5

x = 30 or x = 180 - 30 = 30 or 150

sin is positive

angle is in Q1 or Q2

Value of sin x is 0.5

30 off x-axis

30 30

(b) tan x = -1

x = 180 - 45 or x = 360 - 45 = 135 or 315

tan is negative


Value of tan x is -1

45 off x-axis

45°

45

FM Page 119 Ex 5.9

1 (orally)

New Q

Ex 10.3

Page 363 2,6

General Solution of a Trig Function

cos θ = 0.643θ = cos-1 (0.643)θ ≈ 50°But cos 310° = 0.643 also

So there appears to be more than one

solution

So, how many solutions are there?

cos curve

-1

-0.5

0

0.5

1

-900 -540 -180 180 540 900

y=0.643

cos θ = 0.634

θ = 50° or θ = 310°

or θ = 50° + 360° or θ = 310° + 360°

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

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

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

or θ = 50° - 2 x 360° etc

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

The General Solution forcos θ = 0.643

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

(a) sin x = 0.5

x = 30 or x = 180 - 30 = 30 or 150

general solution isx = 30 + n x 360 or x = 150 + n x

360

sin is positive



30 off x-axis

30 30

Model Find all values of x (to the nearest

minute) where 0 ≤ x ≤ 360 for which

(a) sin2x = 0.25

(b) tan 3x = -1

(a) sin2x = 0.25

sin x = ± 0.5

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

sin is pos or neg

angle is in Q1,Q2,Q3 or Q4


30 off x-axis

30 30

30 30

(b) tan 3x = -1

3x = 135° + 360n or 3x = 315 + 360n

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

45, 165, 285, 105, 225, 345

tan is negative


Value of tan x is -1

45 off 3x-axis

45°

45

FM Page 304 Ex 14.16

1 -23 with GC

New Q

Page 371 Ex 10.5

1, 2, 4,10, 14

Derivatives of functions involving sin x and cos x

• Derivatives of functions involving sin x and cos x

Derivative of sin x and cos x

y = sin x dy = cos x dx

y = cos x dy = -sin x dx

Model : Find the derivative of(a) sin 2x(b) sin32x(c) sin2x cos3x

(d) sin(π-3x) • do some examples on Graphmatica


(d) sin(π-3x) ________________________________________

(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


(d) sin(π-3x) ________________________________________

(b) y = sin32x ( = (sin 2x)3 ) = u3 where u = sin 2x

dy = 3u2 du = 2 cos 2x du dx dy = dy . du dx du dx = 3u2 . 2 cos 2x = 6 sin22x cos 2x


(d) sin(π-3x) ________________________________________

(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) sin(π-3x) ________________________________________

(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 19.5

1,4,5

Model :

Find the gradient of the curve y = sin 2x at the point where x = π/3

?

2sin

dx

dy

xy

Model :

Find the gradient of the curve y = sin 2x at the point where x = π/3

1

cos2,

2cos2

2sin

32

3

dx

dyxWhen

xdx

dy

xy

Trig functions and motion 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


(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


(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 2.6

No. 1 – 4, 7

periodic functions and applications iii

Documents