IB SL MATHS SUMMARY NOTES
LOGS AND EXPONENTIALS
INDEX NOTATION
34
Exponent/ power/ index
Base
INDEX LAWS
FRACTIONAL INDICES
1
𝑛
𝑎 𝑛 = √𝑎
𝑚
𝑎𝑛
=
1 𝑚
𝑛
(𝑎 )
𝑛
= ( √𝑎 )
𝑚
EXPANSION AND FACTORISATION
-
a(b+c) = ab + ac
(a+b)(c+d) = ac+ad+bc+bd
(a+b)(a-b) = a2-b2
(a+b)2 = a2+2ab+b
(a-b)2 = a2-2ab+b2
EXPONENTIAL GRAPHS
𝑓(𝑥) = 𝑎 𝑥 EXPONENTIAL GROWTH FUNCTION
1
𝑎
𝑓(𝑥) = 𝑎− 𝑥 EXPONENTIAL DECAY FUNCTION
1
Graph passes through (−1, ) and (1, 𝑎)
•
•
•
•
Graph passes through (1, ) and (−1, 𝑎)
𝑎
Graph passes through (0,1)
Domain is all real x
Range is all real positive numbers
Y intercept is 1
No x intercepts
BASE e
Base e = irrational number (2.718 …)
LOGARITHMS
𝑏 = 𝑎 𝑥 then loga b = x
e.g. 8 = 23 then log28 = 3
PROPERTIES OF LOGARITHMS
Rule
Logaa = 1
Loga1 = 0
Loga𝑎𝑛 = n
WHEN ARE LOGS UNDEFINED?
Logab if b is negative
Loga0
i.e.
𝑎1 = 𝑎
𝑎0 = 1
𝑎𝑛 = 𝑎𝑛
LOGARITHMIC FUNCTIONS
𝑦 = 𝑎𝑥
𝑦=𝑥
𝑦 = 𝑙𝑜𝑔a𝑥
Properties of 𝑦 = 𝑙𝑜𝑔a𝑥
• Domain: all positive real numbers
• Range: all real numbers
• X intercept 1
LOG BASE 10 AND NATURAL LOGS
Log to the base 10
Log to the base e (natural logs)
Log10x = logx
Logex = lnx
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
a logarithmic function is the inverse of the exponential function
If 𝑓(𝑥) = ⅇ 𝑥 then 𝑓 −1 (𝑥) = 𝑙𝑜𝑔ex
Loga𝑎 𝑥 = alogax = x
Lnex = elnx = x
Log of an exponential/ exponential of a log = x
LOG LAWS
**must learn by <3
Log x + log y = logxy
𝑥
Log x – log y = log
𝑦
Logxn
= nlogx
CHANGE OF BASE
SOLVING EXPONENTIAL EQUATIONS
-
Take the ‘log’ of both sides
SOLVING LOGARITHMIC EQUATIONS
1. Ensure that both side of the equation are logarithms with the same base and the equating
the argument (the expression inside the bracket) i.e. “cancelling” the logs
Or
2. By using an exponent
** For both of these methods you must check that the solution/ solutions are possible.
You cannot have the log of a negative number
Sub your solution back into what was in the brackets of the log and then check the result is
positive
FUNCTIONS
TRANSFORMATIONS
•
•
•
•
Reflection: every point of the image is the same distance from the mirror line
Rotation: image rotated around a point
Translation: moves every point of the image a fixed distance in the same direction
𝑥
(𝑦) x = horizontal y = vertical
Enlargement: increases or decreases the size of an object by a scale factor (produces a
similar figure)
WHAT ARE FUNCTIONS?
Function: a mathematical relation where each x-value has exactly one y-value
•
•
(-2,1), (-1,1) = function (same y values, fine)
(3,8), (3,9) = not a function (same x values, not fine)
➔ Test with the vertical line test (if passes through twice, more than one y value for each x,
therefore not a function)
ASYMPTOTES
Asymptote: a line that a graph approaches but does not intersect
Set notation:
Domain: set of all the x values of the ordered pairs
Range: set of all the y values of the ordered pairs
FUNCTIONS
f(x)
F(g(x))) / (f o g)(x)
𝑓 −1 (𝑥)
Basic function: function of ‘f’ at x
Composite function: f of g of x
Inverse function: reverses the action of f
A reflection of that function in the line y=x
➔ Test using horizontal line test (2 = not)
FINDING INVERSE FUNCTION ALGEBRAICALLY
1.
2.
3.
4.
Replace f(x) with y
Swap the x and y
Make y the subject
Replace y with 𝑓 −1 (𝑥)
TRANSFORMATIONS
RELECTIONS
−𝑓(𝑥)
𝑓(−𝑥)
Reflects in x axis
Reflects in y axis
STRETCHES
𝑓(𝑞𝑥)
Stretches/ compresses horizontally
1
scale factor 𝑞
𝑝𝑓(𝑥)
Stretches/ compresses vertically
Scale factor 𝑝
TRANSLATIONS
𝑓(𝑥 + 1)
𝑓(𝑥) + 1
Horizontal translation
Vertical translation
Outside = x/ horizontal
Inside = y/ vertical
𝑏
𝑏+𝑝
∫𝑎 𝑓(𝑥) = ∫𝑎+𝑝 𝑓(𝑥 − 𝑝)
QUADRATIC FUNCTIONS
SOLVING QUADRATIC EQUATIONS
FACTORISATION
Null factor law: if (𝑥 − 𝑎)(𝑥 − 𝑏) = 0, then if (𝑥 − 𝑎) = 0 or (𝑥 − 𝑏) = 0,
Cross method
COMPLETING THE SQUARE
1.
2.
3.
4.
5.
Halve “ab”
Square ^
Add and subtract (ab/2)^2
Factorise
Square root both sides
QUADRATIC FORMULA (in the formula booklet)
THE DISCRIMINANT
The discriminant: gives information about the roots of the equation
𝛥̇ = 𝑏 2 − 4𝑎𝑐
𝛥>0
2 distinct (different) real roots
𝛥=0
2 equal roots (or double root)
𝛥<0
No real roots
𝛥 is a square
2 rational roots
DIFFERENT FORMS OF QUADRATIC FUNCTIONS
Form
Information
When to use
STANDARD FORM
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑦 intercept → c
𝑦-axis → (0, 𝑐)
−𝑏
Axis of symmetry → 𝑥 = 2𝑎
Don’t know vertex
or intercepts → sub
values into standard
form (simultaneous
equations)
TURNING POINT FORM
𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘
Vertex → (h, k)
Axis of symmetry → x=h
Know vertex
FACTORISED FORM
𝑦 = (𝑥 − 𝑝)(𝑥 − 𝑞)
𝑥 intercepts (roots) → (p, 0) and (q, 0)
𝑝+𝑞
Axis of symmetry → 𝑥 = 2
Know x and y
intercepts
RECIPROCAL FUNCTIONS
𝑘
𝑓(𝑥) = 𝑥 where k is a constant
•
•
•
Hyperbola
Undefined when x = 0
Asymptotes: x -axis and y-axis
TRANSFORMATIONS
𝑘
𝑥
K stretches the hyperbola (low value = more
dip)
−𝑘
𝑥
Reflection in the x axis
𝑘
−𝑥
Reflection in y axis
𝑘
+1
𝑥
𝑘
𝑥+1
Vertical translation
Horizontal translation
RATIONAL FUNCTIONS
𝑓(𝑥 ) =
𝑔(𝑥)
ℎ(𝑥)
when g(x) and h(x) are linear functions
FORMS OF RATIONAL FUNCTIONS
Form
𝑦=
𝑘
𝑥−𝑏
𝑎𝑥 + 𝑏
𝑦=
𝑐𝑥 + 𝑑
Vertical asymptote
When denom = 0
Horizontal asymptote
𝑦=0
𝑦=
𝑎
𝑐
LIMITS
Limit: the fixed value that terms in a sequence approach as the term number increases
(converge upon)
Convergent when -1 < r < 1
**note: when
solving algebraically,
continue to write the
lim until you have
lim 𝑓(𝑥 ) = 𝐿
𝑥→𝑐
𝑥→𝑐
•
•
𝑥 → 𝑐 : as the value of x approaches c (from either direction)
𝑓(𝑥) : the function becomes closer to a fixed value L
subbed in the c value
The limit does not exist when the function doesn’t approach a fixed value
TERMINOLOGY
THE SECANT LINE
𝑔𝑟𝑎𝑑𝑖ⅇ𝑛𝑡 𝑜𝑓 𝑠ⅇ𝑐𝑎𝑛𝑡 𝑙𝑖𝑛ⅇ =
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
e.g. y = x2 + 1 → sub (x2 + 1) into x
**Use gradient of secant line to find
((𝑥 + ℎ)2 + 1) − (𝑥 2 + 1)
𝑚=
ℎ
Average rate of change
(or just change / time)
THE TANGENT LINE
The gradient of a curve = gradient of the tangent line to the curve at that point
*lines have a constant gradient, curves do not
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
𝑔𝑟𝑎𝑑𝑖ⅇ𝑛𝑡 𝑜𝑓 𝑡𝑎𝑛𝑔ⅇ𝑛𝑡 = lim
This gives you the derivative! (which you can just find the normal way instead)
DERIVATIVES
Derivative: the gradient of the function f at any value of x
•
𝑓′(𝑥)
•
𝑑𝑦
•
𝑦′
𝑑𝑥
RULES
•
•
𝑓(𝑥) = 𝑥 𝑛 → 𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1
𝑠𝑢𝑚 𝑜𝑟 𝑑𝑖𝑓𝑓ⅇ𝑟ⅇ𝑛𝑐ⅇ: 𝑑ⅇ𝑟𝑖𝑣ⅇ ⅇ𝑎𝑐ℎ 𝑡ⅇ𝑟𝑚 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙𝑙𝑦
EQUATIONS OF NORMALS AND TANGENTS
Tangent line:
1. Find derivative of f(x)
2. Sub in x value to find MT
3. Sub this into y1 – y = m(x – x1)
Normal line
1. Find MT
2. MN = negative reciprocal of MT
3. Sub into y1 – y = m(x – x1)
DERIVATIVES OF ex and lnx (in formula book)
Original
Derivative
ⅇ𝑥
ⅇ𝑥
1
𝑥
𝑙𝑛𝑥
THE PRODUCT RULE
𝑦 = 𝑢𝑣 → 𝑦 ′ = 𝑢′ 𝑣 + 𝑣′𝑢
THE QUOTIENT RULE
𝑛𝑢𝑚ⅇ𝑟𝑎𝑡𝑜𝑟 ′ × 𝑑ⅇ𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 − 𝑑ⅇ𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 ′ × 𝑛𝑢𝑚ⅇ𝑟𝑎𝑡𝑜𝑟
(𝑑ⅇ𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟)2
THE CHAIN RULE
1. Move power out the front
2. Subtract 1 from the power
3. Multiply by the derivative of inside the bracket
e.g. (2𝑥 + 3)4 → 4(2𝑥 + 3)3 × 2
THE CHAIN RULE WITH LN AND e
𝑦 = ⅇ 𝑓(𝑥)
𝑦 = 𝑙𝑛(𝑓(𝑥 ))
𝑦′ =
𝑓′(𝑥)
𝑓(𝑥)
𝑦 ′ = ⅇ 𝑓(𝑥) × 𝑓′(𝑥)
𝑦 = ln (
𝑓(𝑥)
)
𝑔(𝑥)
𝑦 = ln(𝑛𝑢𝑚ⅇ𝑟𝑎𝑡𝑜𝑟) − ln (𝑑ⅇ𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟)
TRIG DERIVATES
sin′ (𝑥) = cos𝑥
cos′ (𝑥) = −sin (𝑥)
DERIVATIVE OF ANY LOG
𝑑
𝑑𝑥
1
𝑙𝑜𝑔𝑏 𝑥 = (𝑙𝑛𝑏)𝑥
You can show this is true by rewriting it using the change of base law
HIGHER ORDER DERIVATIVES
Words
First derivative
Second
derivative
Third
derivative
Notation
𝑓′(𝑥)
𝑓′′(𝑥)
𝑓′′′(𝑥)
𝑑𝑦
𝑑𝑥
2
𝑑 𝑦
𝑑𝑥 2
𝑑3𝑦
𝑑𝑥 3
Uses
•
•
•
•
•
•
Gradient of tangent line
Instantaneous velocity
𝑠 ′ (𝑡) = 𝑣(𝑡) → velocity function
Graph intervals (increasing or
decreasing)
Relative extrema
𝑠 ′′ 𝑡(𝑡) = 𝑣 ′ (𝑡) = 𝑎(𝑡)
MOTION IN A LINE: DISPLACEMENT
-
Has no direction
Distance from origin
Displacement function s(t)
VELOCITY:
𝒔(𝒕) > 𝟎 : right/ above origin
𝒔(𝒕) < 𝟎: left/ below origin
𝒗(𝒕) > 𝟎 : move right/ upwards
𝒗(𝒕) < 𝟎: move left/downwards
𝒗(𝒕) = 𝟎: at rest
ACCELERATION:
𝑠 ′′ (𝑡) = 𝑎(𝑡)
𝒂(𝒕) > 𝟎 : velocity increasing
𝒂(𝒕) < 𝟎: velocity decreasing
𝒂(𝒕) = 𝟎: velocity is constant
SPEED:
𝑠𝑝ⅇⅇ𝑑 = 𝑎𝑏𝑠𝑜𝑙𝑢𝑡ⅇ 𝑣𝑎𝑙𝑢ⅇ 𝑜𝑓 𝑣ⅇ𝑙𝑜𝑐𝑖𝑡𝑦
𝑠𝑎𝑚ⅇ 𝑠𝑖𝑔𝑛 (𝑣ⅇ𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑛𝑑 𝑎𝑐𝑐ⅇ𝑙ⅇ𝑟𝑎𝑡𝑖𝑜𝑛) = 𝑠𝑝ⅇⅇ𝑑𝑠 𝑢𝑝
𝑑𝑖𝑓𝑓 𝑠𝑖𝑔𝑛 (𝑣ⅇ𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑛𝑑 𝑎𝑐𝑐ⅇ𝑙ⅇ𝑟𝑎𝑡𝑖𝑜𝑛) = 𝑠𝑙𝑜𝑤𝑠 𝑑𝑜𝑤𝑛
INCREASING/ DECREASING FUNCTION
1. find f’(x)
2. Let f’(x) = 0
3. Values for x where f’(x) = 0
Are stationary points
4. Use sign diagram/ table
𝒇′(𝒙) > 𝟎 : f increasing
𝒇′(𝒙) < 𝟎: f decreasing
𝒇′(𝒙) = 𝟎: stationary point at x
RELATIVE EXTREMA
Relative max: + → - (increasing to decreasing)
Relative min - → + (decreasing to increasing)
**note: there may be a stationary point where f doesn’t change from sign ≠ relative
extremum
CONCAVITY
Point of inflexion:
1. F’’(x) = 0
2. F’’(x) changes sign → if it doesn’t change sign, it is called a horizontal point of
inflexion
𝒇′′(𝒙) > 𝟎 : concave up (+)
𝒇′′(𝒙) < 𝟎: concave down (-)
𝒇′′(𝒙) = 𝟎: point of inflexion
OPTIMISATION PROBLEMS
1.
2.
3.
4.
5.
assign variables and draw a sketch
Write an equation to be optimised
Let f’(x)=0 → find the stationary points
Verify whether these are max or min using the second derivative test
Use the appropriate value to find the max or min
*Note: the max of min may occur at the endpoint of the domain
INTEGRATION
How to integrate:
1. Add 1 to the power
2. Divide by the new power
3. + C
Rules: General solution
•
•
•
Bring constants out the front of the integration sign
Integral of constant is constant + C
∫ (𝑓(𝑥) ± 𝑔(𝑥)) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ 𝑔(𝑥) 𝑑𝑥
For ln and e
𝒍𝒏
1
∫ 𝑑𝑥 = 𝑙𝑛𝑥 + 𝐶
𝑥
1
1
∫
𝑑𝑥 = 𝐥𝐧(𝑎𝑥 + 𝑏) + 𝐶
𝑎𝑥 + 𝑏
𝑎
𝒆
∫ ⅇ 𝑑𝑥 = ⅇ 𝑥 + 𝐶
𝑥
∫ ⅇ 𝑎𝑥+𝑏 𝑑𝑥 =
1 𝑎𝑥+𝑏
ⅇ
+𝐶
𝑎
Aka: keep e to power, divide by a, +C
Aka: ln(denominator), divide by a, + C
𝑓 ′ (𝑥)
∫
= 𝑙𝑛[𝑓(𝑥)] + 𝐶
𝑓(𝑥)
Power rule:
∫ (𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 =
1
1
(𝑎𝑥 + 𝑏)𝑛+1 ) + 𝐶
(
𝑎 𝑛+1
Aka: normal integration on brackets, divide by a
SUBSTIUTION: when you are integrating functions w/ × or ÷
1. Define part of function as pronumeral. Let u be:
o The most complex
o In brackets (only what’s inside the brackets, not the power)
o Highest power of x
2. Differentiate 𝑢
𝑑𝑢
3. Rearrange 𝑑𝑥 so that you have a value which you can sub into the equation
4. Optional: Separate the equation
5. Substitute values into the original function
6. Sub u’s value back in
If you use this, with definite integrals, you must adjust the domain by subbing the values in the
equation for 𝑢
DEFINITE INTEGRALS
Formula
explanation
Type into calc to evaluate
𝑏
∫ 𝑓(𝑥)𝑑𝑥
𝑏
𝑎
𝑎
𝑏 to 𝑎 is the opposite sign of 𝑎 to 𝑏
∫ 𝑓(𝑥)𝑑𝑥 = − ∫ 𝑓(𝑥)𝑑𝑥
𝑎
𝑏
VERTICAL TRANSLATIONS
•
•
Think of the rectangle that will be added/ subtracted below x axis
Rewrite, separate f(k) and k
FUNDAMENTAL THEOREM OF CALCULUS
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = [𝐹(𝑥)]𝑏𝑎 = 𝐹(𝑏) − 𝐹(𝑎)
𝑎
AREA UNDER A CURVE
𝑏
∫𝑎 𝑓(𝑥)𝑑𝑥
•
Take the absolute value:
o Check that 𝐹(𝑏) − 𝐹(𝑎) > 0
o If not, taking the absolute value requires you to multiply through with a negative
AREA BETWEEN 2 CURVES
𝑏
∫𝑎 𝑦1 − 𝑦2 𝑑𝑥  where 𝑦1 is the one on the top
*can only be used when the function is continuous within the domain (no asymptotes)
If the curves cross over:
1.
2.
3.
4.
Find the intersection b/w the curves
Find the value for the start and end of the area contained (use as domain)
Calculate each area separately
Sum
*if you get the curves the wrong way around (say that the wrong one is on the top), then you will get
a negative value → fix by taking the absolute value
VOLUME OF A REVOLUTION
𝑏
𝑉 = 𝜋 ∫ 𝑦 2 𝑑𝑥
𝑎
^ 𝜋 × integral of the (function2 )
•
If it is a full rotation, use 2𝜋
•
Check answer w/ the volume of a sphere formula 3 𝜋𝑟 2
4
DEFINITE INTEGRALS WITH DISPLACEMENT & DISTANCE
Displacement (this is not total
distance)
Distance
𝑡2
𝑡1
𝑡2
∫ |𝑣(𝑡)|𝑑𝑡
𝑡1
1.
2.
3.
4.
[𝑠]𝑡𝑡21
∫ 𝑣(𝑡)𝑑𝑡
Need to split up into + and –
directions and then take the
absolute value of the directions
Find the velocity (to know whether it is moving left/ right)
Draw a motion diagram
Use definite integral (of whole domain) to find displacement
Use absolute values of definite integrals to find total distance travelled
CALCULUS WITH TRIG FUNCTIONS
DERIVATIVES OF SIN, COS, AND TAN
If 𝑓(𝑥) = 𝑠𝑖𝑛𝑥
If 𝑓(𝑥) = 𝑐𝑜𝑠𝑥
If 𝑓(𝑥) = 𝑡𝑎𝑛𝑥
𝑓(𝑥) = trig function (𝑎𝑥 + 𝑏)
Product rule
Quotient Rule
Log Functions
𝑓(𝑥) = 𝑢(𝑥) × 𝑣(𝑥)
𝑓(𝑥) =
𝑢(𝑥)
𝑣(𝑥)
𝑦 = 𝑙𝑛𝑓(𝑥)
then 𝑓 ′ (𝑥) = 𝑐𝑜𝑠𝑥
Then 𝑓 ′ (𝑥) = −𝑠𝑖𝑛𝑥
1
Then 𝑓 ′ (𝑥) = 2
𝑐𝑜𝑠 𝑥
𝑓 ′ (𝑥) = (trig function′ (𝑎𝑥 + 𝑏)) × derivative of ( )
𝑓 ′ (𝑥) = 𝑣(𝑥) × 𝑢′ (𝑥) + 𝑢(𝑥) × 𝑣′
𝑓 ′ (𝑥) =
𝑣(𝑥) × 𝑢′ (𝑥) − 𝑢(𝑥) × 𝑣′(𝑥)
[𝑣(𝑥)]2
𝑓′(𝑥)
𝑓(𝑥)
INTEGRATION OF TRIG FUNCTIONS
∫ 𝑠𝑖𝑛𝑥𝑑𝑥 = −𝑐𝑜𝑠𝑥 + 𝑐
∫ 𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝑐
1
∫ 𝑠𝑖𝑛(𝑎𝑥 + 𝑏) 𝑑𝑥 = − cos(𝑎𝑥 + 𝑏) + 𝐶
𝑎
1
∫ cos(𝑎𝑥 + 𝑏) 𝑑𝑥 = sin(𝑎𝑥 + 𝑏) + 𝐶
𝑎
CIRCULAR FUNCTIONS
EXACT VALUES
•
•
•
•
THE UNIT CIRCLE
•
•
•
•
Anticlockwise +𝜃
Clockwise -𝜃
Sin𝜽= y-coordinate
cos𝜽= x-coordinate
•
tan𝜽 =
𝒔𝒊𝒏𝜽
𝒄𝒐𝒔𝜽
•
•
•
(0,1)
Tan is undefined because 1/0
𝑐𝑜𝑠𝜃 = 0
𝑠𝑖𝑛𝜃 = 1
•
•
•
(-1,0)
𝑐𝑜𝑠𝜃 = −1
𝑠𝑖𝑛𝜃 = 0
In which quadrants is each ratio positive?
•
•
•
•
(0, -1)
Tan is undefined because -1/0
𝑐𝑜𝑠𝜃 = 0
𝑠𝑖𝑛𝜃 = −1
TRIG IDENTITIES (in data booklet)
•
•
•
𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1 **note: often rearrange for 𝑠𝑖𝑛2 𝜃 = 1 − 𝑐𝑜𝑠 2 𝜃
𝑐𝑜𝑠2𝜃
= 1 − 2𝑠𝑖𝑛2 𝜃
= 2𝑐𝑜𝑠 2 𝜃 − 1
= 𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛2
𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
*note: you can’t cancel 𝑠𝑖𝑛𝑥 across the = of an equation because you will lose a solution
(1,0)
𝑐𝑜𝑠𝜃 = 1
𝑠𝑖𝑛𝜃 = 0
TRIG CURVES
Amplitude: height of the curve above the horizontal axis
sine curve
• Amplitude: 1
• Period: 2𝜋
Cosine curve
• Amplitude: 1
• Period: 2𝜋
Simply a horizontal
translation of the sine
curve
Tan curve
• Period: 𝜋
• Vertical
asymptotes at:
𝜋 3𝜋
, …
2 2
Finding 𝜃 graphically → plot another equation w/ desired value and find intersection(s)
TRIG FUNCTION TRANSFORMATIONS
𝑦 = 𝒂 sin(𝒃(𝑥 − 𝒄)) + 𝒅
Translations
𝑦 = sin(𝑥) + 𝒅
Vertical
Horizontal
Positive d = up
Negative d = down
𝑦 = sin(𝑥 − 𝒄)
Positive c = to the left
Negative c = to the right
Stretches
Vertical
Horizontal
𝑦 = 𝐚sin(𝑥)
Amplitude = |𝒂|
−𝒂 = reflected over x-axis
𝑦 = sin(𝒃𝑥)
𝒃 repeats of the curve in the original period
2𝜋
Period (sin/cos) = 𝑏
𝜋
Period (tan) = 𝑏
−𝒃 = reflected over y-axis
MODELING WITH SINE AND COSINE FUNCTIONS
Amplitude
Vertical translation
Period
Horizontal translation
Write a function that models the data
Or enter the data points into 2 VAR STAT → Fit
max − min
2
Average of max and min
max + min
2
Difference b/w 2 max points
Look for max point
TRIGONOMETRY
THE COSINE RULE
Finding a side → 𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 𝑐𝑜𝑠𝐴
Finding an angle → 𝑐𝑜𝑠𝐴 =
𝑏 2 +𝑐 2 −𝑎2
2𝑏𝑐
AREA OF A TRIANGLE
𝐴 = 0.5𝑎𝑏 𝑠𝑖𝑛𝐶
RADIANS
Circumference = 2𝜋𝑟 or 2𝜋 radians (c)
2𝜋 = 360°
𝜋 𝑐 = 180°
DEGREE-RADIAN CONVERSIONS
Degrees → Radians = x
Radians → Degrees = x
𝜋
180
180
𝜋
ARC LENGTH
𝐴𝑟𝑐 𝑙ⅇ𝑛𝑔𝑡ℎ = 𝑟𝜃 (𝑡ℎⅇ𝑡𝑎 𝑖𝑛 𝒓𝒂𝒅𝒊𝒂𝒏𝒔)
AREA OF A SECTOR
𝐴𝑟ⅇ𝑎 𝑜𝑓 𝑠ⅇ𝑐𝑡𝑜𝑟 =
OBTUSE ANGLES
•
•
𝑐𝑜𝑠𝜃 = − cos(180° − 𝜃)
𝑠𝑖𝑛𝜃 = sin(180° − 𝜃)
𝜃𝑟 2
(𝑡ℎⅇ𝑡𝑎 𝑖𝑛 𝒓𝒂𝒅𝒊𝒂𝒏𝒔)
2
EXACT VALUES
•
•
•
30 = in radians (pi/6)
45 = in radians (pi/4)
60 = in radians (pi/3)
THE UNIT CIRCLE
-
Measure angles anticlockwise from the origin
Centre: at origin
Radius: 1 unit
Finding point P on a unit circle
Cosθ = x coordinate of P
Sinθ = y coordinate of P
Properties
𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 = 1
𝑡𝑎𝑛𝜃 =
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
𝑚 = 𝑡𝑎𝑛𝜃
THE SINE RULE
𝑎
𝑏
𝑐
=
=
𝑠𝑖𝑛𝐴
𝑠𝑖𝑛𝐵
𝑠𝑖𝑛𝐶
AMBIGUOUS TRIANGLES
•
Two known sides + non-included angle = angle can be obtuse or acute
1. Find acute angle
2. Find obtuse angle
3. Check
a. Is the angle sum 180 degrees?
b. Is the longest side opposite the largest angle?
SEQUENCES AND SERIES
Sequence example: 1,3,4,5,7,9
U1 (first term) = 1
U2 (second term) = 3
TYPES OF FORMULAE
Recursive formula: based on the previous term (previous term → term)
e.g. Un+1 = Un + 6
General formula: based on the term number (term number → term)
e.g. Un = 5n
TYPES OF SEQUENCES and FORMULAE
Arithmetic sequence: each successive term is created by adding a common difference (d)
* formula book
Geometric sequence: each term obtained by multiplying the previous value by a common ratio (r)
* formula book
𝑢2
𝑟=
𝑢1
** learn by heart
USING YOUR CALCULATOR
1.
2.
3.
4.
Function
Plot equation
Numeric view
Number setup → Num start: 1 (n)
SIGMA NOTATION
EQUATIING THE DIFFERENCE: ARITHMETIC
d = U2 – U1
U2 - U1 = U3 - U2
e.g. find k given that 3k+1, k, and -3 are consecutive terms in an arithmetic sequence
k – (3k + 1) = - 3 – k
k – 3k – 1 = - 3 – k
-1 + 3 = - k + 2k
k =2
EQUATING THE DIFFERENCE: GEOMETRIC
𝑢2 𝑢3
=
𝑢1 𝑢2
SUM OF AN ARITHMETIC SERIES
Formula 1
Formula 2
𝑆𝑛 =
𝑛
(𝑢 + 𝑢𝑛 )
2 1
𝑆𝑛 =
𝑛
(2𝑢1 + (𝑛 − 1)𝑑)
2
SUM OF A GEOMETRIC SERIES
When r > 1
When r < 1
𝑢1 (𝑟 𝑛 − 1)
𝑆𝑛 =
𝑟−1
𝑢1 (1 − 𝑟 𝑛 )
𝑆𝑛 =
1−𝑟
CONVERGENT SERIES (FINITE)
When -1 < r < 1
Find a term
Sum
Use formula geometric formula
𝑆𝑛 =
CALCULATING COMPOUND INTEREST
1. Compound interest formula
(amount of money) x (interest)number of periods
𝑢1
𝑟−1
BIVARIATE ANALYSIS
CORRELATION
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Direction
Strength
Linearity
Outliers
Causation
LINE OF BEST FIT/ REGRESSION LINE
1. Calculate the mean point (𝑥̅ , 𝑦̅)
2. Draw a mean point with a balanced number of points on each size
EXTRAPOLATION & INTERPOLATION
LEAST SQUARES REGRESSION LINE
Residual: vertical distance between a data point and the graph of the regression line
The least squares regression line: the line which makes the sum of the squares of the residuals as
small as possible
** use GDC
•
•
•
2 Variable Statistics
C1 C2 Linear (symbolic view)
Plot (if cannot see line of best fit: Menu, fit)
MEASURING CORRELATION
Very weak: 0 – 0.25
Weak: 0.26 – 0.5
Moderate: 0.51 – 0.75
Strong: 0.76 – 1
** use GDC
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•
2 variable statistics
Stats: r
DESCRIPTIVE STATISTICS
STATISTICAL GRAPHS
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Bar charts
Pie charts
Pictograms
Line graphs
Stem and leaf diagrams
TYPES OF DATA
discrete data: only specific values
continuous data: any value
MEASURES OF CENTRAL TENDENCY
Mean: average
•
Population mean 𝜇
∑𝑓𝑖 𝑥𝑖
𝜇=
∑𝑓𝑖
𝑓𝑖 𝑥𝑖 = score/class centre * frequency
•
Sample mean 𝑥̅
Mode: data value that occurs most often
𝑛+1 th
)
2
Median: middle score (
13+1
(
2
) = 7 (median = 7th ordered data value)
14+1
(
2
) = 7.5 (median = average of 7th and 8th ordered data values)
1 ≤ 𝑥 ≤ 4  for 𝑥
1.
2.
Find range/2
Add rang/2 to lower value
BENEFITS OF DIFFERENT MEASURES OF CENTRE
Mean = non-resistant (affected by extreme values)
Median = resistant (unaffected by extreme values)
MEASURES OF DISPERSION
Range = highest – lowest score
Five number summary
-
Minimum
-
Q1 (
-
Median (Qs)
Q3
maximum
𝑛+1
)
4
IQR: shows how spread out the middle 50% of data is (Q3 – Q1)
OUTLIERS
1.5 x IQR below Q1 or above Q3
INTERPRETING A BOX PLOT
Skew = direction of longest line/ where the small boxes are
DRAWING A CUMULATIVE FREQUENCY GRAPH
Height
Frequency
Cumulative Frequency
(add preceding frequencies to
current freq)
150 ≤ ℎ ≤ 155
4
4
155 ≤ ℎ ≤ 160
22
26
160 ≤ ℎ ≤ 165
56
82
*note whether it is 𝑥 ≤ 155 or 𝑥 < 155 and adjust scale accordingly (i.e. make upper value 149.9)
USING THE GDC (FREQUENCY TABLES + STATS)
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•
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•
•
1 Var Statistics
Enter 𝑥𝑖 in D1
Enter 𝑓𝑖 in D2
Go to symb view → select D2 as frequency column
Use stats as normal
STANDARD DEVIATION
•
•
𝜎
Non-resistant
**only expected to do with technology
PROBABILITY
EXPERIMENTAL PROBABILITY
Relative Frequency/ Experimental Probability = frequency of an outcome
Total number of trials
INTERESECTION OF EVENTS
∩ = intersection
∪ = union
FORMULA BOOKLET FORMULAE
SAMPLE SPACE DIAGRAMS
•
•
•
List
Tables and grids
Probability trees
TESTS FOR INDEPENDENT EVENTS
P(AIB) = P(A)
P(BIA) = P(B)
P(A∩B) = P(A) P(B)
PROBABILITY DISTRIBUTIONS
INTERPRETING COMBINATION NOTATION
Interpreting combinations on
calc.
nCr = Reference to Pascal’s triangle
•
•
n = row number
r = how many across (1st number is 0)
1.
2.
3.
4.
𝑛
Can also be written as 𝐶𝑟𝑛 or ( )
𝑟
FINDING COMBINATIONS
tool box
probability
combination
Type (n,r)
Typing factorials (!) on calc.
𝑛!
𝑛
( )=
𝑟
𝑟! (𝑛 − 𝑟)!
1. tool box
2. probability
3. factorial
APPLYING COMBINATIONS
nCr = number of ‘combinations’
•
•
n = no of choices/ possible n number of objects
r = objects are chosen
**in combinations: order is not important (AB = BA)
BINOMIAL THEOREM
Patterns
𝑛
𝑛
( )=1
0
𝑛
( )=𝑛
1
𝑛
(𝑎 + 𝑏)𝑛 = ∑ ( ) 𝑎𝑛−𝑟 𝑏 𝑟
𝑟
𝑟=0
•
•
•
•
a = first term
b = second term
n = power
r = increasing from 0 until it = n
Symmetry
4
4
( )=( )
0
4
e.g. (2a – 5)4
When using binomial theorem to expand expressions with a negative: every second term is
subtracted rather than added (because it is do a neg power and therefore the neg remains)
1
e.g. (ⅇ − 4)4 = +ⅇ 4
1
− 4ⅇ 3 × 𝑒
1
+ 6ⅇ 2 × 𝑒 2
even terms: to the power of 0 and power of 2 = +
odd terms: to the power of 1 and power of 3 = −
1
− 4ⅇ × 𝑒 3
TERMS OF A BINOMIAL EXPANSION
Sometimes we are asked to only find one particular term = don’t need whole expansion
1. find the constant term
•
•
•
write the general term for the binomial expansion
find the r value required for x0
sub this r value into all parts of equation w/o an x
2. find the x3 term of (4x – 1)9
𝑛 𝑛−𝑟 𝑟
)𝑎 𝑏
𝑟
•
sub into formula (
•
what does r need to be such that n-r = power of the term (in this case, 3)
in the case where there is an 𝑥 term out the front of the brackets, simply
include this 𝑥 term into your ‘𝑥 equation’
sub r value into formula
write full term including coefficient and x
•
•
3. Finding n (know the coefficient) In the expansion of (2x+1)n, the coefficient of the
term x3 is 80. Find the value of n.
•
•
•
𝑛
sub into formula ( ) 𝑎𝑛−𝑟 **ignore br (because of symmetry)
𝑟
𝑛
n-r = r e.g. ( ) (2𝑥)3 = 80x3
3
𝑛
solve the equation for ( )
𝑟
𝑛
a. sub ( ) into the binomial coefficient
𝑟
plot on calculator and find intersection
b. or draw Pascal’s triangle
𝑛
what does n have to equal such that ( ) = 10?
3
5
10 = ( )
3
4. Finding the general term
• Write general expression, subbing in n and leaving r
• Simplify for xn-r
• Solve r for when n-r = 0
• Sub r value into equation (ignoring anything that has an x)
5. Finding the coefficient of a specific term
• Figure out term number required to get x and y raised to desired powers
• Find (n,r)
• Coefficient = (n, r) X any other coefficients in the equation e.g. (6x)3
6. Working when there are two brackets
1. Transform (2x+3)(x-2)6 → 2x(x-2)6 + 3(x-2)6
2. Decide what value of x you are trying to find within each of the separate
brackets (given that it may be multiplied by an x outside the brackets)
e.g. find the x3 term
2x(x-2)6
➔ There is an x outside the brackets and therefore, we actually need to
find the x2 term in the brackets
2𝑥(𝑥 2 term) + 3(𝑥 3 term)
3. Ignore the coefficients of the brackets briefly, and work out r (using the
term value you worked out before) for both brackets
4. Sub both r values back into their respective brackets = gives you the x
term that you want
5. Add together
PROBABILITY DISTRIBUTIONS OF DISCRETE VARIABLES
Example: H = no. of heads obtained when 2 coins are tossed. Tabulate the prob. Distribution
of H
X
P(H=x)
0
¼
1
½
2
¼
Sum of all the probabilities will always add up to 1
EXPECTATION
Expected value = mean
2. GDC METHOD
1. Manual method – finding the mean
Multiply x score by P(X=x) score and add all of
these together
•
•
•
•
Statistics 2Var
Enter
X values → C1
Probability→C2
STATS
Look at 𝛴𝑋𝑌
BINOMIAL DISTRIBUTION
•
•
•
Only two possible outcomes
𝑋~𝑁(𝑛, 𝑝)
n = no. of trials
p = probability of success
And probability of success is
•
Toolbox
BINOMIAL (n, k, p)
BINOMIAL_CDF (n, p, k)
BINOMIAL_ICDF (n, p, q)
𝑛
( ) 𝑝𝑟 𝑞 𝑛−𝑟
𝑟
 manually sub values into the equation
EXPECTATION FOR BINOMIAL DISTRIBUTIONS
𝐸 (𝑋) = 𝑛𝑝
Or use graphing calculator…
when
𝑋~𝐵(𝑛, 𝑝)
VARIANCE OF BINOMIAL DISTRIBUTIONS
√𝑣𝑎𝑟𝑖𝑎𝑛𝑐ⅇ = 𝜎
𝑉𝑎𝑟(𝑋) = 𝑛𝑝𝑞 when
𝑋~𝐵 (𝑛, 𝑝)
(k is the same as r = no. of
successes) **
NORMAL DISTRIBUTION
•
•
symmetrical about the mean
mean, mode and median are equal
𝑋~𝑁(𝜇, 𝜎 2 )
** where 𝜎 2 is the variance
Using the graphing
calculator
AREA UNDER A NORMAL DISTRIBUTION CURVE
•
•
•
NORMAL_CDF (𝜇, 𝜎. 𝑍)
area under curve =1
partial areas = probabilities
continuous date therefore (≤ = <)
STANDARD NORMAL DISTRIBUTION
𝜇 = 0 and 𝜎 = 1
𝑍~𝑁(0,1)
**where z = no of standard deviations away from the mean
STANDARDISING NORMAL DISTRIBUTION
Transforming 𝑋~𝑁(𝜇, 𝜎 2 ) → 𝑍~𝑁(0,1)
𝑧=
𝑥−𝜇
𝜎
INVERSE NORMAL DISTRIBUTION
Used when finding the value in the data which has a given cumulative probability
Standardised Distribution = NORMAL_ICDF (cumulative probability)
Non-standardised Distribution = NORMAL_ICDF (𝝁, 𝝈, cumulative probability)
FIND THE MEAN OR THE STANDARD DEVIATION
1. standardise the data (find the z score)
2. use the Inverse Normal Function
VECTORS
Vector: quantity that has both magnitude and direction
Magnitude
Direction
(length)
scalar
vector
Equal vectors: same magnitude and direction
Geometric Negative Vectors: same length, opposite direction
Column Vector Form
𝒙
(𝒚)
𝒛
Base/Unit Vector Form
𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌
⃗⃗⃗⃗⃗
Position vector = 𝑂𝑋
MAGNITUDE OF A VECTOR: like Pythagoras
Magnitude = absolute value of vector (no direction)
⃗⃗⃗⃗⃗ | = (𝑎) = √𝑎2 + 𝑏 2
|𝐴𝐵
𝑏
𝑎
⃗⃗⃗⃗⃗ | = (𝑏) = √𝑎2 + 𝑏 2 + 𝑐 2
|𝐴𝐵
𝑐
FORMULAS
Relative Vectors
Scalar Product of Two Vectors
Algebraic Properties of the scalar product
⃗⃗⃗⃗⃗
𝐴𝐵 = ⃗⃗⃗⃗⃗
𝑂𝐵 − ⃗⃗⃗⃗⃗
𝑂𝐴
𝑣 ⋅ 𝑤 = 𝑣1 𝑤1 + 𝑣2 𝑤2 + 𝑣3 𝑤3
𝑣 ⋅ 𝑣 = |𝑣|2
𝑣 ⋅ (𝑤 + 𝑥) = 𝑣 ⋅ 𝑤 + 𝑣 ⋅ 𝑥
Angle
𝑐𝑜𝑠𝜃 (acute) = 𝑣 ⋅ 𝑤 > 0
𝑐𝑜𝑠𝜃 (obtuse) = 𝑣 ⋅ 𝑤 < 0
PARALLELISM
When one vector is a scalar multiple of the other (not zero vectors)
4
e.g. (10
) = 2(25)  parallel vectors
If vectors are parallel
𝑟1 and 𝑟2 are parallel if 𝑏1 = 𝑘𝑏2
𝑡
Let 𝒑 = (𝑡+2
) and let 𝒒 = (34)
Find t given that p and q are parallel
𝑣 ⋅ 𝑤 = ±|𝑣||𝑤|
1. If 𝑝‖𝑞 then p = kq where 𝑘 ≠ 0
𝑡
3
(
) = 𝑘( )
𝑡+2
4
Write vectors as column vectors
2. Write 𝑣 ⋅ 𝑤 for x and y as separate
equations
𝑡 = 3𝑘 and 𝑡 + 2 = 4𝑘
a. Find 𝑘 and sub into other equation
b. Equate both to 𝑘 and then equate both
to one another
PERPENDICULAR
If vectors are perpendicular
Show vectors are mutually perpendicular
Check if triangle ABC is right angled
Find the form of all vectors which are
𝑥
perpendicular to a given vector (𝑦)
𝑣⋅𝑤 =0
i.e.𝑏1 ⋅ 𝑏2 = 0
Each one is perpendicular to all the others
Show that:
𝑎⋅𝑏 =0
𝑏⋅𝑐 =0
𝑎⋅𝑐 =0
Check one v.w
Only need to find one pair of perpendicular
vectors because a triangle cannot have more
than one right angle
1. Swap x and y of the column
2. Do +/- of bottom based on working out
v.w. for the top row
3. If it is a 3D vector, let one of the terms
= 0 and then swap the others as normal
4. Put 𝑘 out the front (because all scalar
multiples are parallel)
−2
5
e.g. ( ) → 𝑘 ( )
5
2
INTERSECTING LINES/ POINTS OF INTERSECTION
Find the point of intersection of two lines using
their vector equations
1. Write parametric equations for both
vector equations
2. Equate the 𝑥 and 𝑦 components
respectively → rearrange into same
structure
3. Solve simultaneously for 𝑠 or 𝑡
4. Sub 𝑠 or 𝑡 into the 𝑥 and 𝑦 parametric
equations of one of the lines → results
give you (𝑥, 𝑦)
UNIT VECTORS
Unit vector: any vector that has magnitude/ length of 1 unit
1
𝒂
|𝑎|
𝑘
𝒂
|𝑎|
Vector length 1 in direction 𝒂
Vector length k in direction 𝒂
1. Find the unit vector
2. Multiply by k
EQUATIONS
Vector Equation
𝒓 = 𝒂 + 𝒕𝒃
𝒓: general position vector of a point on the line
𝑎: given position vector of a point on the line → specific point
𝑡: parameter
𝑏: direction vector
𝑏
𝑟𝑖𝑠𝑒
𝑦
𝑏
*note: if 𝑏 = ( 1 ) then the gradient of the line 𝑚 = 𝑟𝑢𝑛 = 𝑥 = 𝑏2
𝑏2
1
If you’re only given 2 points: use AB = OB – OA to find b
Parametric equation: 𝑥 and 𝑦 components
𝑥 = 𝑎1 + 𝑡𝑏1
𝑦 = 𝑎2 + 𝑡𝑏2
Cartesian Equation
1. Rearrange parametric so that 𝑡 is the subject
𝑡=
𝑥 − 𝑎1
𝑏1
𝑡=
𝑦 − 𝑎2
𝑏2
2. Equate the expressions for the 𝑥 and 𝑦 components
𝑥 − 𝑎1 𝑦 − 𝑎2
=
𝑏1
𝑏2
Lines in 3D space
•
•
Do not talk about a line’s “gradient”
Do not look at the Cartesian equations
RELATIONSHIPS B/W LINES
2D
Coplanar (same
plane → can
connect a straight
sheet b/w them)
Non-coplanar
•
•
•
3D
Intersecting: 1 POI
Parallel: no POI/solutions
Coincident: same line
Skew: neither parallel nor intersecting
If lines are skew: suppose one line is translated to
intersect w/ the other. Angle b/w lines = angle
b/w intersecting lines
Proving different relationships:
Not coincident
Skew
1. Do not intersect
2. Are not parallel
Choose random point on one line and show it doesn’t exist on
the other
1. When 𝑠 =?, the point on line L1 is (𝑥, 𝑦, 𝑧)
2. L2: for the 𝑦 coordinate to be the coordinate above, 𝑡 =?
3. When 𝑡 =?, the point on L2 is (𝑥, 𝑦, 𝑧)
4. Since (𝑥, 𝑦, 𝑧) ≠ (𝑥, 𝑦, 𝑧), L1 and L2 are not coincident
1: Do not intersect
a. Parametric equations
b. Solve for 𝑡
c. Hence, solve for 𝑠
d. Sub 𝑡 value into a/the remaining parametric equation to
find 𝑠 according to this equation
e. If 𝑠 ≠ 𝑠 → there are not 𝑠 and 𝑡 values where all those
lines will all go through
∴ not coincident
2: not parallel → 𝑏1 ≠ 𝑘𝑏2
CLASSIFYING SHAPES
Given the vector equations of sides of a triangle
Find the points
AB and AC meet at A. Thus, to find A:
1. Equate parametric equations of AB and
AC
2. Use simultaneous to solve for s/t
3. Sub into original vector AB for A
CONSTANT VELOCITY PROBLEMS
𝑟 = 𝑎 + 𝑡𝑏
Initial position
Position at 𝑡
Velocity
Speed
𝑎 (given as coordinates)
Parametric equations (given as coordinates)
𝑏
|𝑏|
Find the position vector at a specific value of 𝑡
Find new velocity if object continues in same
direction but increases/decreases speed
Time when object is due east of (0,0)
1. Write parametric equations
2. Sub 𝑡 values into equations
3. Use the answers as the x and y
components of the vector
New speed = 𝑡(𝑏)
What must 𝑡 equal for the new velocity to
equal the speed?
Position will have a 𝑦 value of 0
Parametric 𝑦 equation = 0
Find t
SHORTEST DISTANCE FROM A LINE TO A POINT
Shortest distance = PERPENDICULAR!!
Find the shortest distance from P to the line (given line’s vector equation and P’s coordinates)
1.
2.
3.
4.
5.
6.
7.
8.
Let N be the point on the line closest to P
Write parametric equations for the line
Use above in coordinates for N (𝑥, 𝑦)
Find vector of line (with shortest distance) - ⃗⃗⃗⃗⃗⃗
𝑃𝑁
⃗⃗⃗⃗⃗⃗
Let line ⋅ 𝑃𝑁 = 0
Solve ^ for 𝑡
⃗⃗⃗⃗⃗⃗ to give you a direction vector
Sub 𝑡 into 𝑃𝑁
⃗⃗⃗⃗⃗⃗ | to find its distance
Take |𝑃𝑁
N
line
⃗⃗⃗⃗⃗⃗
𝑃𝑁
(find shortest length)
P
*Note for Jasmine: this is a more general set of steps for P which accounts for situations where P is
not the origin but you would do the same thing if it was to find the shortest distance from the origin
by just using (0,0) as the point
OBJECTS RELEASED AT DIFFERENT TIMES (torpedos)
•
•
Put 𝑡 in front of both position vectors where 𝑡 = time elapsed since first object released
Thus, in front of the position vector which is released second, you put 𝑡 − 𝑥 with 𝑥 being
how many s/mins/hrs later it is released