NSW Year 12 Maths Advanced
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The NESA Maths Reference Sheet is a great resource… if you know how to use it! Navigate vectors, complex numbers and mechanics with our Ultimate NESA Maths Reference Sheet Guide.
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Click on the following formulas to see what they mean and apply them to a practice question!
Vectors |
\( |{\bf u}|=|x{\bf i}+y{\bf j}|=\sqrt{x^2+y^2} \) \({\bf u} \cdot {\bf v} = |{\bf u}||{\bf v}|\text{cos} \theta = x_1 x_2 + y_1 y_2 , \) \begin{align*} |
Maths Extension 2 only:
Mechanics |
\(\frac{d^2x}{dt^2} = \frac{dv}{dt} = v \frac{dv}{dx} = \frac{d}{dx} \left( \frac{1}{2} v^2 \right)\) |
Use | Formula | Explanation |
Magnitude of the vector | \( |{\bf u}|=|x{\bf i}+y{\bf j}|=\sqrt{x^2+y^2} \) | \begin{align*} |{\bf u}| &= \text{length/magnitude of }{\bf u} \\ (x, \ y) &= \text{ Cartesian coordinates of position vector } {\bf u} \\ {\bf i}&: \text{ unit vector in the direction of the x-axis} \\ {\bf j}&: \text{ unit vector in the direction of the y-axis} \\ \end{align*} |
Dot product | \({\bf u} \cdot {\bf v} = |{\bf u}||{\bf v}|\text{cos} \theta = x_1 x_2 + y_1 y_2 , \)\begin{align*} \text{where } {\bf u} &= x_1 {\bf i} + y_1{\bf j}\\ \text{and } {\bf v} &= x_2{\bf i} + y_2{\bf j}\\ \end{align*} | |
Vector form of a line | \({\bf r}={\bf a} + λ {\bf b}\) | A line can be expressed in many different forms like Cartesian form (\(y =mx+c\)) and vector form. When we express line \({\bf r}\) in vector form, we describe it using a point \({\bf a}= \binom{x}{y}\) that it passes through and its gradient vector \({\bf b}\). By substituting different constant values for \(λ∈R\), the vector equation \({\bf r}={\bf a} + λ {\bf b}\) describes every single point on line \({\bf r}\). |
Example 28:
Express the line y = 4x + 3 in vector form.
Solution 28:
Find a point that y = 4x + 3 passes through: \begin{align*}
Find a gradient vector of the line y = 4x + 3: Since the equation of the line is in the form y = mx + c, the gradient is the coefficient of x. \begin{align*} Now that we know that the line passes through \(\binom{1}{7}\) and has a gradient vector of \(\binom{1}{4}\), we can express the line in vector form: \({\bf r}= \binom{1}{7} + λ \binom{1}{4}\) where \(λ∈R\) |
Maths Extension 2 only:
Use | Formula | Explanation |
Manipulating expressions for acceleration, velocity and displacement | \(\frac{d^2x}{dt^2} = \frac{dv}{dt} = v \frac{dv}{dx} = \frac{d}{dx} \left( \frac{1}{2} v^2 \right)\) | Need help understanding differentiation? See the following Maths Guides! Maths Adv Applications of differentiation Maths Ext 1 |
Expression for displacement \(x\) in terms of time \(t\) | \(x=a \text{cos} (nt+α)+c\) | \begin{align*} x & = \text{displacement} \\ \\ \ \frac{2\pi}{n} &= \text{Period} \\ a &= \text{Amplitude}\\ α &= \text{Phase shift}\\ c &= \text{Centre of motion} \end{align*} |
\(x=a \text{sin} (nt+α)+c\) | ||
Expression for acceleration \( \ddot x\) in terms of displacement \(x\) | \( \ddot x= -n^2(x-c)\) | \(\ddot x = \text{acceleration}\) |
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