Download HSC Mathematics Workshop 3

HSC Mathematics
Workshop 3
Presented by
Richard D. Kenderdine
BSc, GradDipAppSc(IndMaths), SurvCert, MAppStat, GStat
School of Mathematics and Applied Statistics
University of Wollongong
Moss Vale Centre
October 2009
HSC Mathematics - Workshop 3
Richard D Kenderdine
University of Wollongong
October 2009 1
Trigonometric Functions
A radian is the angle subtended at the centre of a circle by an arc equal in
length to the radius.
Since 1 radian subtends an arc of length r then 2π radians subtend an arc
of length 2πr ie circumference.
Since the angle subtending the circumference is 360◦ :
2π radians = 360 degrees
(1)
π radians = 180 degrees
(2)
Therefore
Note that π 6= 180 because π ≈ 3.14159 always.
◦
◦
Then 1 radian = ( 180
π ) ≈ 57.3
To convert between radians and degrees always start off with the relation
π rad = 180◦ and use ratios.
Example (1) Convert 2.4 radians to degrees.
Example (2) Convert 150◦ to radians.
Arc lengths and areas
Circles were first studied in Year 8. Lengths of arcs and areas of sectors were
calculated by ratioing the circumference and area of a circle. We can still use
these methods if the question uses degrees.
Example (3)
Suppose we have a circle of radius 8cm and a sector with angle 50◦ at the centre.
Find the arc length and area of the sector.
The arc length and area formulae are simplified when we work in radians.
When θ is in radians we replace 360◦ with 2π radians.
1 email:
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1
Arc length
l=
Sector area
A=
θ
(2πr) = rθ
2π
(3)
θ
1
(πr2 ) = r2 θ
2π
2
(4)
Note the relationship between them:
Sector area =
1
r × Arc length
2
(5)
Area of minor segment
This is the difference between the area of the sector and the triangle formed
by the radii and chord.(Area of triangle = 12 ab sin C).
1 2
r [θ − sin θ]
2
A useful formula to calculate the length of the chord:
Area of minor segment =
θ
Length of chord = 2r sin( )
2
ie twice the radius times sin(half the angle).
(6)
(7)
1. Exercise:
Convert to radians, both in terms of π and to 2dp:
(a) 135◦ (b) 240◦ (c) 70◦ (d) 330◦
2. Exercise:
Convert to degrees and minutes:
(a)
2π
3
radians (b)
12π
5
radians c)
5π
6
radians (d) 2.3 radians
3. Exercise:
A chord in a circle of radius 8cm is subtended by an angle of 40◦ at the
centre. Find (a) the length of the chord (b) the area of the minor segment.
4. Exercise:
Find the area of a sector bounded by radii of 15cm and arc of length 20cm.
Remember the triangles used to obtain the exact ratios for angles of
and π3 .
2
π π
6, 4
Remember also the unit circle.
Noting that sin θ = y , cos θ = x and tan θ = xy , we can easily find
sin( π2 ), cos π etc.
The trig. graphs are obtained by plotting the relevant coordinate as the
point moves anti-clockwise around the circle commencing at (1,0). For
example, suppose we plot the y-coordinate. Its starting value is 0 when
θ = 0, increases to 1 at θ = π2 , decreases to 0 at π, reaches its lowest value
of -1 at θ = 3π
2 and then returns to 0 at θ = 2π.
5. Exercise:
Find the trig. ratios for the quadrant angles.
6. Exercise:
Sketch for 0 ≤ θ ≤ 2π: (a) y = cos θ (b) y = tan θ
Graphing trig functions
Now look at more general trig. graphs.
The general sin curve has 4 parameters (a, b, c, d):
y = a sin(bx + c) + d
(8)
The period is calculated from 2π
b and therefore b is the number of cycles
in 2π time periods.
Note the effects of the parameters.
The same parameters apply to the cos curve. The tan curve doesn’t have
an amplitude - a still stretches or compresses in a vertical direction.
For example, y = 2 sin(3x + π4 ) − 1 has an amplitude of 2, a period of 2π
3 ,
π
commences at - 12
and shifted downwards 1 unit compared to the basic
sin curve.
There are two ways of graphing trig functions:
(a) Fit the function to fixed axes
(b) Fit the axes to the function
The second method is easier.
Suppose we need to sketch y = 3 cos(2x) + 1. First draw the basic cos
curve without numbering the axes.
The amplitude is 3, the period is 2π
2 = π and the curve is shifted upward
1 unit. Therefore the curve is centred around y = 1 and lies between -2
and 4. Draw the horizontal axis and insert the relevant numbers.
3
Note: A negative amplitude turns the curve upside down.
7. Exercise:
Sketch the following for 0 ≤ θ ≤ 2π:
(a) 4 sin(3θ) (b) 2 − 2 cos(θ) (c) tan(2θ)
Calculus of trig. functions
Calculus with trig functions requires the argument (θ) to be in radians.
The derivative of sin θ is cos θ.
Then using the function-of-a-function rule we can generalise:
If y = sin[f (x)] then
dy
dx
= f ´(x) cos[f (x)]
In words: differentiate the argument, sin goes to cos and the argument
stays the same.
One way to integrate is to write down the basic form of the integral and
then calculate the constant multiple to give the correct answer.
8. Exercise:
√
Differentiate (a) sin(x3 + 5x) (b) cos( x) (c) tan( x1 )
9. Exercise:
R
R
R
Find the indefinite integrals (a) 10 sin 5xdx (b) −2 cos(3x)dx (c) 6x2 sec2 (x3 )dx
Logarithms
Definition:
The logarithm of a number to a given base is the power to which the base
must be raised to give the number.
logb N = p ⇔ bp = N
(9)
The LHS of the above is the log form while rhe RHS is the index form.
They are equivalent ways of expressing the relationship between three
numbers.
In words, base raised to power gives number
4
A simple example: log2 8 = 3 because 23 = 8
Log laws:
logb xy = logb x + logb y
x
logb = logb x − logb y
y
(10)
logb xn = n logb x
(12)
logb x =
Note that the LHS of (14) is not
(11)
loga x
loga b
(13)
logb x
logb y .
Eqn (13) is the change of base rule
The most useful log law is (15) because it enables us to easily solve exponential equations, for example, 3x = 6. Taking logs to base 10 gives
10 6
x log10 3 = log10 6 so x = log
log 3 = 1.631.
10
10. Exercise:
Solve (a) log2 32 = x (b) log5 x = −2 (c) logx 9 = 0.5
11. Exercise:
Simplify (a) log6 9 + log6 4 (b) log3 5 − log3 45
12. Exercise:
If loga 2 = 0.3562 and loga 3 = 0.5646 find: (a) loga
(c) loga 4a (d) loga a4
2
3
(b) loga 72
13. Exercise:
2
Solve (a) 42x−3 = 13 (b) 5x = 62
The log function is the inverse of the exponential function. Graphically
it is the reflection of the latter in the line y = x.
Note that logb 1 = 0 and that the domain of logb x is x > 0.
The Exponential Function
Consider the compound interest function A = P (1 + i)n .
5
Suppose we invest $1 for 1 year at an interest rate of 100% pa. At the end
of the year we have $2.
Taking it to the limit, in other words interest is compounded continuously,
we have:
1 n
) = 2.71828... = e
(14)
n
Because this limit occurs frequently in maths it is given the special symbol
e.
lim (1 +
n→∞
We can arrive at this value from another starting point. Given the exponential function y = ax with arbitrary base a, can we find a value of a
such that the gradient of the tangent to any point on the curve is equal
to the function value at that point?
That is, y =
dy
dx
for all x.
It turns out that this is true for the case a = 2.71828... which we have
defined to be e.
Thus we have the rules for differentiating and integrating ex :
d x
(e ) = ex
dx
Then using the function-of-a-function rule we have:
(15)
d f (x)
(e
) = f ´(x)ef (x)
(16)
dx
In words, the derivative is the original function multiplied by the derivative of the power.
For all the examples in this course the integral is the opposite, divide the
original function by the derivative of the power. (This is not true in all
cases).
Note
R x2 that primitives do not always exist for exponential functions eg
e dx does not exist.
6
Differentiation of ln x
Now we can derive y = loge x
Now writing ln (natural log) for loge and using the function-of-a-function
rule again we have the more general rule:
d
f ´(x)
loge f (x) =
dx
f (x)
(17)
In words, put the derivative of the argument over the argument.
Note: the rules for differentiating ex and loge x only work when the base
is e. See the examples for handling other cases.
Now we know
R
1
x
= ln x
Therefore if we have a fraction to integrate where the denominator is to
the power of 1, the answer is most likely the ln of the denominator multiplied by a constant.
For example
Another:
R
R
11
6x dx
6x+3
x2 +x dx
√
√
We know ( x)2 = x and x2 = x
This is true because the square and square-root functions are inverses of
each other: do one followed by the other takes you back to the starting
point.
The same is true for ex and ln x:
eln
x
=x
(18)
ln ex = x
(19)
Question:
Z
3
dx = ln(3x)
3x
7
Also
But
Z
3
3x
=
1
x
1
dx = ln(x)
x
so why can’t we use
Z
Z
3
1
dx =
dx = ln(x)?
3x
x
Exercises:
14. What is the domain of y = ln(2 − x)? Sketch the function.
√
15. Differentiate: (a) x2 e2x (b) ln( x) (c) ln(x2 + 1) (d) esin x
R5
R1
16. Evaluate: (a) 2 e5x−2 dx (b) −1
x2
x3 +6 dx
17. Find the equation of the normal to y = x cos(x) at the point x = π.
18. Find the area bounded by y = cos 2θ, the θ-axis and the lines θ =
θ = π3 .
19. Find the volume when the curve y =
from x = 2 to x = 4.
1
√
2 x
π
6
and
is rotated around the x-axis
20. Find the stationary and inflexion points on the curve y = sin x + 2 cos x.
Determine the nature of the SP.
21. Differentiate:
(a) ln(cos x) (b)
tan x
e2x
(c) 2x (d) log10 x3
R
22. Differentiate x ln(x) and hence find ln xdx
8