Maths Quest Maths B Year 12 for Queensland 2e 1

WorkSHEET 6.2 The calculus of periodic functions Name: ___________________________ 1 Find the equation of the tangent to the curve

y = sin x at the point (p, 0).

The tangent function is in the form

𝑦 = 𝑚𝑥 + 𝑐 To find 𝑚, need the gradient of the curve, so find the gradient function,

𝑦' = cos 𝑥 now, at 𝑥 = 𝜋 the slope is

𝑚 = 𝑦' 𝜋 = cos 𝜋 = −1 so we have,

𝑦 = −𝑥 + 𝑐 Because (p, 0) lies on the tangent function, it must satisfy its equation. By sub,

0 = −𝜋 + 𝑐

𝑐 = 𝜋 So the tangent function becomes:

𝑦 = −𝑥 + 𝜋

Maths Quest Maths B Year 12 for Queensland 2e 2

2 Find the equation of the tangent to the curve

y = cos 2x at the point .

Maths Quest Maths B Year 12 for Queensland 2e 3

3 Find the equation of the tangent and the normal to the curve y = 3 sin 2x at the point

where x = 6p .

The Tangent function is in the form

𝑦 = 𝑚𝑥 + 𝑐 First find a point that lies on the tangent line; Since 𝑓 0

1= 3 sin 50

1= 6 6

5

The point 0

1, 6 65

lies on the tangent line. To find 𝑚, need the gradient of the curve, so find the gradient function,

𝑦' = 6 cos 2𝑥 now, at 𝑥 = 0

1 the slope is

𝑚 = 𝑦' 𝜋 = 6 cos2𝜋6 = 3

so we have, 𝑦 = 3𝑥 + 𝑐

Because 0

1, 6 65

lies on the tangent function, it must satisfy its equation. By sub,

3 32 =

3𝜋6 + 𝑐

𝑐 =3 32 −

𝜋2

So the tangent function becomes:

𝑦 = 3𝑥 +3 32 −

𝜋2

** Refer to next page for the Normal function! **

Maths Quest Maths B Year 12 for Queensland 2e 4

4 Find the equation of the tangent and the normal to the curve y = 3 sin 2x at the point

where x = 6p .

Sorry, Ran out of room last page …

We already know at the point 𝜋6 ,3 32

the slope of the function is 3. The slope of the Normal must be −:

6, due to the

rule 𝑚:×𝑚5 = −1. The Normal function is in the form

𝑦 = 𝑚𝑥 + 𝑐 so it becomes,

𝑦 = −13𝑥 + 𝑐

Because 0

1, 6 65

lies on the normal function, it must satisfy its equation. By sub,

3 32 =

−𝜋18 + 𝑐

𝑐 =3 32 +

𝜋18

So the tangent function becomes:

𝑦 = −13𝑥 +

3 32 +

𝜋18

Maths Quest Maths B Year 12 for Queensland 2e 5

5 Find the points on the curve y = 2 sin 3x where the tangent is parallel to the x-axis. *** Bit of a silly question ??? We could use our thorough trig knowledge go straight to a sketch, and come up with the answer without any working, but where’s the fun in that! … So this is a poor exam question, but good for practicing our skills ***

Parallel to the 𝑦 axis means the gradient is zero, so 𝑚 = 0 and the tangent is in the form, 𝑦 = 0𝑥 + 𝑐 Need to determine 𝑐, so must find a point that lies on the tangent. () Need to find where on the given function the slope is zero (parallel to 𝑥-axis), hence find the Gradient function. y = 2 sin 3x ddyx

= 6 cos 3x

Tangent is parallel to the x-axis where ddyx

= 0

6 cos 3x = 0 cos 3x = 0

3x = 2p ,

23p ,

25p ,

27p , ….

x = 6p ,

63p ,

65p ,

67p , ….

By substitution into the given equation, find some coordinate points that lie on the given function, where that function has a gradient of zero.

𝜋6 , 2 ,

3𝜋6 ,−2 ,

5𝜋6 , 2 ,

7𝜋6 ,−2 ,…

Each of these points lie on the tangent line, so find 𝑐 by substitution;

2 = 0×𝜋6 + 𝑐

𝑐 = 2 or

−2 = 0×3𝜋6 + 𝑐

𝑐 = −2 We can clearly see there are two tangents

𝑦 = 2 or

𝑦 = −2

Maths Quest Maths B Year 12 for Queensland 2e 6

6 The depth of water (in metres) at an inlet is given by the equation

D = 8 + 2 cos 4tp

Where t is the number of hours past midnight. (a) Find the depth of water at high tide.

(b) Find when low tide will occur (i.e. when the water depth is a minimum). ***Lets find the first two low tides! ***

(a) *** You should be good at drawing graphs by now.

The maximum value of a periodic function happens where cos 𝑎𝑛𝑦𝑡ℎ𝑖𝑛𝑔 = 1

𝐷 = 8 + 2×1 = 10

Therefore, the maximum depth is 10 metres. (b) Similarly the minimum depth will be 6 (as

verified by your rough sketch of the trig function.

6 = 8 + 2 cos𝜋𝑡4

cos𝜋𝑡4 = −1

(insert exact value triangle) (insert CAST quadrants)

𝜋𝑡4 = 𝜋,3𝜋,5𝜋,𝑒𝑡𝑐

𝑡 = 4,12,20,𝑒𝑡𝑐

So, the first low tide will be at 4:00 am, the second at midday and the third (even though we weren’t required to find it, will be at 8:00pm that night. *** Interesting tides on this planet J

Maths Quest Maths B Year 12 for Queensland 2e 7

7 The velocity of a piston is given by the equation v = 1.5 sin 4t where t is the number of seconds since the engine was started. Find the maximum and minimum velocity of the piston and when they first occur.

Stupid book had a terrible solution involving the use of differentiation … this is why you HAVE to pay attention in class, because you can’t trust the Text Book worked examples, or Solutions manual to give you what I would consider the best way to solve. I’ll leave the text book solution here for you to see how silly it is … v = 1.5 sin 4t ddvt

= 6 cos 4t

Maximum and minimum occur when ddvt

= 0

6 cos 4t = 0 cos 4t = 0

4t = 2p ,

23p , …

t = 8p ,

83p ,

At t = 8p v = 1.5 sin 4(

8p )

= 1.5 sin 2p

= 1.5

At t = 83p v = 1.5 sin 4(

83p )

= 1.5 sin 23p

= -1.5

Maximum velocity = 1.5 m/s after 8p secs

Minimum velocity = -1.5 m/s after 83p secs

*** Can you see that finding where the slope is zero gives you the maximums and minimums all mixed up … how silly! Did you also see they had NO justification for why one answer was a maximum and the other a minimum. How SLACK! Now I’ll do the solutions over the next page.

Maths Quest Maths B Year 12 for Queensland 2e 8

8 The velocity of a piston is given by the equation v = 1.5 sin 4t where t is the number of seconds since the engine was started. Find the maximum and minimum velocity of the piston and when they first occur. Lets even find the first few times these occur!

This Periodic function has a max and min value where the value of sin 4𝑡 = 1𝑜𝑟 − 1 respectively,

Hence the Maximum Velocity is 1.5 and the minimum velocity is -1.5.

Now find when these occur:

Maximums,

sin 4𝑡 = 1 (insert CAST quadrants here)

4𝑡 = 𝜋2 ,

5𝜋2 ,

9𝜋2 ,𝑒𝑡𝑐

therefore maximum velocities occur at;

𝑡 = 𝜋8 ,

5𝜋8 ,

9𝜋8 ,𝑒𝑡𝑐

Minimums,

sin 4𝑡 = −1 (insert CAST quadrants here)

4𝑡 = 3𝜋2 ,

7𝜋2 ,

11𝜋2 ,𝑒𝑡𝑐

therefore minimum velocities occur at;

𝑡 = 3𝜋8 ,

7𝜋8 ,

11𝜋8 ,𝑒𝑡𝑐

Maths Quest Maths B Year 12 for Queensland 2e 9

9 Now a question that requires differentiation: The velocity of a piston is given by the equation v = 1.5 sin 4t where t is the number of seconds since the engine was started. Find the maximum and minimum acceleration of the pistons.

Acceleration is MNMO

, so differentiate;

𝑑𝑣𝑑𝑡 = 6 cos 4𝑡

Maximum and minimum values of this function will occur when cos 4𝑡 = 1𝑜𝑟 − 1 respectively

Hence,

Maximum Acceleration is 6

Minimum Acceleration is -6

10 A body executing simple harmonic motion with a position x(t) is given by the equation x(t) = 4 cos 2t + 1 State (a) the initial position of the body. (b) The period of the motion.

(c) The frequency of the motion (d) The amplitude of the motion.

(e) The value for the centre of the motion.

(a) at t = 0 x(t) = 4 cos 2t + 1 x(0) = 4 cos 4(0) + 1 = 4 cos 0 + 1 = 4 + 1 = 5

(b) 𝑇 = 50

S

= 22p

= p

(c) f = T1

= p1

(d) a = 4 (e) 1

Maths Quest Maths B Year 12 for Queensland 2e 10

11 A body is executing simple harmonic motion has the displacement equation x(t) = 2 sin 3t (a) Find the equation for velocity. (b) Find the equation for acceleration.

(c) Use a graphics calculator to show the three equations on the same axes.

(a) x(t) = 2 sin 3t v(t)= x’(t) = 6 cos 3t

(b) a(t) = v’(t) = x’’(t) = -18 sin 3t

(c)

Maths Quest Maths B Year 12 for Queensland 2e 11

12 The displacement of a particle executing simple harmonic motion can be given by the equation

D(t) = 4 sin 3tp + 8

Find: (a) the initial position (b) the position after 3 seconds

(c) the position after 4 seconds (d) the average velocity in the forth second

(e) an expression for velocity (f) the initial velocity

(g) the velocity at t = 6 (h) when the particle is stationary.

(a) at t = 0

D(0) = 4 sin 3tp + 8

= 4 sin 0 + 8 = 8

(b) at t = 3

D(3 )= 4 sin 3tp + 8

x = 4 sin p + 8 x = 8

(c) at t = 4

D(4) = 4 sin 3tp + 8

x = 4 sin 34p + 8

x = - 2 3+ 8 (d) 2 m/sec (e) 𝑣 𝑡 = T0

6cos 0O

6

(f) at t = 0

𝑣 0 =4𝜋3 cos

𝜋×03

𝑣 0 =4𝜋3

(g) at t = 6

𝑣 4 =4𝜋3 cos

𝜋×63

=4𝜋3 cos 2𝜋

𝑣 0 =4𝜋3

(h) stationary when v = 0

0 =4𝜋3 cos

𝜋𝑡3

cos𝜋𝑡3 = 0

𝜋𝑡3 =

𝜋2 ,3𝜋2 ,

5𝜋2 ,…

t = 1.5, 4.5, 7.5 …

Maths Quest Maths B Year 12 for Queensland 2e 12

13 The population of frogs in a small pond through year can be approximated by SHM. The mean population of frogs is 500, the period of the function is 12 months and the amplitude of the function is 200. Write an equation for P in terms of t, where t is in months.

Equation will be of the form: P = A sin Bt + D 𝐵 = 50

V

𝐵 = 50

:5

B = 6p

A = 200 D = 500

P(t) = 200 sin 6tp + 500