April 16, 2021 – Physicspupil

April 16, 2021October 24, 2022

202104162114 Homework 1 (Q3)

Suppose $z=f(x,y)$ is differentiable. Find the expression

$\displaystyle{\bigg( \frac{\partial z}{\partial x} \bigg)^2 + \bigg( \frac{\partial z}{\partial y} \bigg)^2}$

in terms of polar coordinates $r$ and $\theta$ .

Solution.

Setup.

$\begin{aligned} & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}} \\ \end{cases} \\ \Rightarrow & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\cos\theta\frac{\partial z}{\partial x} + \sin\theta\frac{\partial z}{\partial y}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{-r\sin\theta\frac{\partial z}{\partial x} + r\cos\theta\frac{\partial z}{\partial y}} \\ \end{cases} \\ \Rightarrow & \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}}\end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{bmatrix} \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial x}} \\ \displaystyle{\frac{\partial z}{\partial y}} \end{bmatrix} \\ \end{aligned}$

(to be continued)

Lemma.

_{Wikipedia on Cramer’s rule}

(continue)

$\begin{aligned} \frac{\partial z}{\partial x} & = \frac{\begin{vmatrix}\displaystyle{\frac{\partial z}{\partial r}} & \sin\theta \\ \displaystyle{\frac{\partial z}{\partial \theta}} & r\cos\theta \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\ \\ & = \frac{r\cos\theta\displaystyle{\frac{\partial z}{\partial r}}-\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}}}{r\cos^2\theta + r\sin^2\theta} \\\\ & = \cos\theta\displaystyle{\frac{\partial z}{\partial r}} - \frac{1}{r}\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}} \\\\ \frac{\partial z}{\partial y} & = \frac{\begin{vmatrix}\cos\theta & \displaystyle{\frac{\partial z}{\partial r}} \\ -r\sin\theta & \displaystyle{\frac{\partial z}{\partial \theta}} \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\\\ & = \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \end{aligned}$

Thus,

$\begin{aligned} & \quad\bigg( \frac{\partial z}{\partial x}\bigg)^2 + \bigg( \frac{\partial z}{\partial y}\bigg)^2 \\ \\ & = \bigg[ \cos\theta\frac{\partial z}{\partial r} - \frac{1}{r}\sin\theta\frac{\partial z}{\partial \theta} \bigg]^2 + \bigg[ \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \bigg]^2 \\ \\ & = \cos^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 - \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \frac{1}{r^2}\sin^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 \\ & \qquad\qquad + \frac{1}{r^2}\cos^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 + \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \sin^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 \\ \\ & = \bigg( \frac{\partial z}{\partial r} \bigg)^2 + \frac{1}{r^2} \bigg( \frac{\partial z}{\partial \theta} \bigg)^2 \end{aligned}$

April 16, 2021October 24, 2022

202104162147 Homework 1 (Q2)

The angle $a$ of a triangle $ABC$ is increasing at a rate of $3\,\mathrm{^\circ\, s^{-1}}$ , the side of $AB$ is increasing at a rate of $1\,\mathrm{cm\, s^{-1}}$ , and the side of $AC$ is decreasing at a rate of $2\,\mathrm{cm\, s^{-1}}$ . How fast is the side $BC$ changing when $a=30^\circ$ , $AB=10\,\mathrm{cm}$ , and $AC=24\,\mathrm{cm}$ ? Is the length of $BC$ increasing or decreasing?

Solution.

Draw a figure below:

Rephrase the problem.

Given that If , , and , then

By cosine law,

$z^2=x^2+y^2-2xy\cos a$ .

Taking ordinary derivatives w.r.t. time $t$ ,

$\displaystyle{2z\frac{\mathrm{d}z}{\mathrm{d}t} = 2x\frac{\mathrm{d}x}{\mathrm{d}t} + 2y\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} - 2x\cos a\frac{\mathrm{d}y}{\mathrm{d}t} - 2y\cos a\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\begin{aligned} z & =\sqrt{x^2+y^2-2xy\cos a} \\ & = \sqrt{(10)^2+(24)^2-2(10)(24)\cos 30^\circ} \\ & = 16.1341\qquad (4\,\mathrm{d.p.}) \end{aligned}$

Plugging in the value of each,

$\displaystyle{2(\cdot\cdot )\frac{\mathrm{d}z}{\mathrm{d}t} = 2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )(\cdot\cdot )\sin (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big)}$

you will know what $\displaystyle{\frac{\mathrm{d}z}{\mathrm{d}t}}$ is.

But now, I intend to treat it with partial derivatives.

Let $f(x,y,a) = x^2+y^2-2xy\cos a = z^2$ .

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{\partial f}{\partial a}\frac{\mathrm{d}a}{\mathrm{d}t} \\ & = (2x-2y\cos a)\frac{\mathrm{d}x}{\mathrm{d}t} + (2y-2x\cos a)\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} \\ \end{aligned}$

After $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}t}}$ is sought, recognise that

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & =2z\frac{\mathrm{d}z}{\mathrm{d}t} \\ \frac{\mathrm{d}z}{\mathrm{d}t} & = \bigg(\frac{1}{2z}\bigg)\frac{\mathrm{d}f}{\mathrm{d}t} \end{aligned}$

you could have it also.

(to be continued)

April 16, 2021October 24, 2022

202104160754 Homework 1 (Q3)

A ship $A$ , which can sail at a constant speed $60\,\mathrm{km/hr}$ to meet a second ship $B$ which is $100\,\mathrm{km}$ away in the direction of $\mathrm{S60^\circ W}$ and is sailing due east at constant speed $30\,\mathrm{km/hr}$ . Find the sailing direction of $A$ and the time required to meet $B$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

Draw a diagram as follows:

Setup.

By the law of sines,

$\begin{aligned} \frac{V_A}{\sin 30^\circ} & = \frac{V_B}{\sin\theta} \\ \frac{60}{\sin 30^\circ} & = \frac{30}{\sin\theta} \\ \sin\theta & = 0.25 \\ \theta & = 14.5^\circ \end{aligned}$

Direction of $\mathbf{v}_A$ : $\mathrm{S45.5^\circ W}$

$\because 180^\circ -30^\circ -14.5^\circ - 90^\circ = 45.5^\circ$

Calculating $v_{AB}$ :

$\begin{aligned} |\mathbf{v}_{AB}| & = |\mathbf{v}_{A}|\cos\theta + |\mathbf{v}_{B}|\cos 30^\circ \\ & = 60 \cos 14.5^\circ + 30\cos 30^\circ \\ & = 84.1\,\mathrm{km/hr} \end{aligned}$

The time needed to meet ship $B$ is

$\begin{aligned} t & = \frac{100\,\mathrm{km}}{|\mathbf{v}_{AB}|} \\ & = \frac{100\,\mathrm{km}}{84.1\,\mathrm{km/hr}} \\ & = 1.19\,\mathrm{hr} \end{aligned}$

April 16, 2021October 24, 2022

202104160620 Homework 1 (Q4)

A particle is projected from a point $O$ on the horizontal floor. The range of the projectile is $R$ and the maximum height that the particle can reach is $h$ . Show that the equation of trajectory of the particle is

$\displaystyle{\frac{y}{h}=\frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

The trajectory of projectile motion must be a parabola, which can be expressed in the form of a quadratic equation:

$y=ax^2+bx+c$ ;

And since the particle passes through the points $(0,0)$ and $(R,0)$ , the equation of trajectory can be expressed in the form:

$y=A(x-0)(x-R)$ .

When the particle has traveled a horizontal distance $x=\displaystyle{\frac{R}{2}}$ , it reaches the maximum height $y=h$ .

Inserting the point $(\frac{R}{2},h)$ into the trajectory equation, we solve for the unknown $A$ :

$\begin{aligned} h & = A\bigg(\frac{R}{2}-0\bigg)\bigg(\frac{R}{2}-R\bigg) \\ h & = -\frac{AR^2}{4} \\ \Rightarrow \qquad A & = -\frac{4h}{R^2} \end{aligned}$

Thus,

$y = -\displaystyle{\frac{4h}{R^2}}x(x-R)$ ,

or,

$\boxed{\frac{y}{h} = \frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$