2016 – NST First Year Maths

2016 Paper 2 Question 12

11th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Consider $y=0$ , then $x=r$ , so

$\begin{aligned} \frac{z}{a} &= 1 - \left( \frac{r}{b} \right)^2 \\ \frac{r}{b} &= \sqrt{1-\frac{z}{a}} \end{aligned}$

The polar angle is $\alpha$ , thus the cylindrical polar coordinates are

$\left( b\sqrt{1-\frac{z}{a}} , \alpha, z \right)$

(b)

Let

$r_z = b\sqrt{1-\frac{z}{a}}$

Then

$\begin{aligned} V &=\int dV \\ &= \int_{z=0}^{a} \int_{\theta=0}^{\alpha} \int_{r=0}^{r_z} r dr d\theta dz \\ &= \int_{\theta=0}^{\alpha}d\theta \int_{z=0}^{a} \left[ r^2/2 \right]_0^{r_z} dz \\ &= \frac{\alpha}{2} b^2 \int_{z=0}^{a} 1-\frac{z}{a} dz \\ &=\frac{1}{2}\alpha b^2 \left[ z - \frac{z^2}{2a} \right]_0^{a} \\ &= \frac{1}{4} \, \alpha \, a \, b^2 \end{aligned}$

2016 Paper 2 Question 7

11th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

The ‘trick’ is to re-write $2 \sin x \cos x$

$f(x) = \sin x + \sin 2x + \sin 3x$

which is the required Fourier series.

(b)

The derivative of $f(x)$ is

$f^{'}(x) = \cos x + 2\cos 2x + 3\cos 3x$

which again is the required Fourier series.

2016 Paper 2 Question 5

11th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Yes. A real symmetric matrix can be (orthogonally diagonalised).

(b)

No. This is a rotation matrix. The eigenvalues are non-real except for special values of $\theta$ .

2016 Paper 1 Question 11

8th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

When $\lambda = 1$

$|z-i| = |z^*-i|$

Geometrically this means $z$ is as far from $i$ as $z^*$ .

Let $z = x +i\,y$

$\begin{aligned} |x+iy-i| &= |x-iy-i| \\ \Rightarrow x^2 + (y-1)^2 &= x^2 + (y+1)^2 \\ \Rightarrow y^2-2y+1 &= y^2 +2y +1 \\ \Rightarrow y &= 0 \end{aligned}$

$y=0$ is the equation of the straight line in the complex plane. This is consistent with the geometric interpretation.

(b)

When $\lambda = 0$ it must be that $z=i$ . This is another straight, horizontal, line.

2016 Paper 1 Question 12

30th Sep 20201st Nov 2020nstmathsupervisorLeave a comment

(a) With

$u = x^2-y^2, \; v = 2xy$

We can think of $f$ as

$f=f(u,v)$

Then

$\begin{aligned} \frac{\partial f}{\partial x} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \\ &= \frac{\partial f}{\partial u}\, 2x + \frac{\partial f}{\partial v}\, 2y \end{aligned}$

and

$\begin{aligned} \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}(-2y) + \frac{\partial f}{\partial v}(2x) \end{aligned}$

Now substitute in to the given equation

$\begin{aligned} y\frac{\partial f}{\partial x} + x\frac{\partial f}{\partial y} &= y 2x \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} 2y^2 + (-2xy)\frac{\partial f}{\partial u} + 2x^2\frac{\partial f}{\partial v} \\ &= 2 (x^2+y^2) \frac{\partial f}{\partial v} \\ &= 0 \end{aligned}$