Had I done Methods in Physics – the done thing

February 4, 2022October 24, 2022

202204021539 Example 3, Chapter 1.3, Methods in Physics II (2015-2016 Lectures)

If $\mathbf{f}(t)=t\,\hat{\mathbf{i}}+t^3\,\hat{\mathbf{j}}$ , $\mathbf{g}(t)=\cos t\,\hat{\mathbf{i}}+\sin t\,\hat{\mathbf{j}}$ , and $\mathbf{v}=2\,\hat{\mathbf{i}}-3\,\hat{\mathbf{j}}$ . Calculate

(a) $(\mathbf{f}+\mathbf{g})'$ ;

(b) $(\mathbf{v}\cdot\mathbf{f})'$ ;

(c) $(\mathbf{f}\cdot\mathbf{g})'$ .

Settings. (Some properties of vector differentiation)

i. If $\mathbf{f}$ and $\mathbf{g}$ are differentiable vector functions,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}+\mathbf{g})=\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}}$

ii. If $\mathbf{f}$ is a differentiable vector function and $\alpha$ a constant scalar,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\alpha\mathbf{f})=\alpha\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

iii. If $\mathbf{f}(t)$ is a vector function, $\mathbf{v}$ a constant vector, and $\mathbf{v}\cdot\mathbf{f}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{v}\cdot\mathbf{f})=\mathbf{v}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

iv. If $h(t)$ is a scalar function, $\mathbf{f}(t)$ a vector function and $h\mathbf{f}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(h\mathbf{f})=h\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\mathbf{f}\frac{\mathrm{d}h}{\mathrm{d}t}}$

v. If $\mathbf{f}$ and $\mathbf{g}$ are vector functions and $\mathbf{f}\cdot\mathbf{g}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}\cdot\mathbf{g})=\mathbf{f}\cdot\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}+\mathbf{g}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

vi. If $\mathbf{f}$ and $\mathbf{g}$ are vector functions and $\mathbf{f}\times\mathbf{g}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(\mathbf{f}\times\mathbf{g})=\displaystyle{\bigg( \frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}\times\mathbf{g}\bigg) + \bigg( \mathbf{f}\times \frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}\bigg)}$ .

Solution.

(a)

$\begin{aligned} \mathbf{f}+\mathbf{g} & = (t+\cos t)\,\hat{\mathbf{i}}+(t^3+\sin t)\,\hat{\mathbf{j}} \\ (\mathbf{f}+\mathbf{g})' & = (t+\cos t)'\,\hat{\mathbf{i}}+(t^3+\sin t)'\,\hat{\mathbf{j}} \\ & = (1-\sin t)\,\hat{\mathbf{i}}+(3t^2+\cos t)\,\hat{\mathbf{j}} \\ \end{aligned}$

(b)

$\begin{aligned} \mathbf{v}\cdot\mathbf{f} & = (2,-3)\cdot (t,t^3) \\ & = 2t-3t^3 \\ (\mathbf{v}\cdot\mathbf{f})' & = (2t-3t^3)' \\ & = 2-9t^2 \\ \end{aligned}$

(c)

$\begin{aligned} \mathbf{f}\cdot\mathbf{g} & = (t,t^3) \cdot (\cos t,\sin t) \\ & = t\cos t+t^3\sin t \\ (\mathbf{f}\cdot\mathbf{g})' & = \cos t-t\sin t+t^3\cos t + 3t^2\sin t\\ \end{aligned}$

February 4, 2022October 24, 2022

202204021524 Example 1, Chapter 1.3, Methods in Physics II (2015-2016 Lectures)

If $\mathbf{f}(\theta)=\cos\theta \,\hat{\mathbf{i}}+e^{2\theta}\,\hat{\mathbf{j}}$ , find $\mathbf{f}'(\theta)$ .

Setting.

If $\mathbf{f}(t)=f_1(t)\,\hat{\mathbf{i}}+f_2(t)\,\hat{\mathbf{j}}+f_3(t)\,\hat{\mathbf{k}}$ is a vector function and $f_1$ , $f_2$ , and $f_3$ are differentiable, then the derivative of $f(t)$ is defined as

$\displaystyle{\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}=\mathbf{f}'(t)=f_1'(t)\,\hat{\mathbf{i}}+f_2'(t)\,\hat{\mathbf{j}}+f_3'(t)\,\hat{\mathbf{k}}$ .

Solution.

From $\mathbf{f}(\theta)=\cos\theta \,\hat{\mathbf{i}}+e^{2\theta}\,\hat{\mathbf{j}}$ , identify that

$\begin{aligned} f_1(\theta ) & =\cos\theta \\ f_2(\theta ) & = e^{2\theta} \\ f_1'(\theta ) & = -\sin\theta \\ f_2'(\theta ) & = 2e^{2\theta }\\ \end{aligned}$

Thus, $\mathbf{f}'(\theta )= -\sin\theta\,\hat{\mathbf{i}}+2e^{2\theta}\,\hat{\mathbf{j}}$ .

February 4, 2022October 24, 2022

202202041514 Example 1, Chapter 1.2, Methods in Physics II (2015-2016 Lectures)

Find the Cartesian equation of the locus represented by the parametric equation:

$\mathbf{f}(t)=t\,\hat{\mathbf{i}}+e^t\,\hat{\mathbf{j}}$ .

Solution.

$\begin{aligned} x(t) & =t \\ y(t) & = e^t\\ \end{aligned}$

Hence, $y=e^x$ .

April 24, 2021October 24, 2022

202104241713 Homework 1 (Q7)

The height of a mountain is given by $h(x,y)=3000-2x^2-y^2$ , where the $y$ -axis points east, the $x$ -axis points north, and all distances are measured in meters. Suppose a mountain climber is at the point $(30,\, -20,\, 800)$ , will he ascend or descend if he moves in the southwest direction?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

The altitude $h(x,y)$ is given by a function of $x$ and $y$ :

$h(x,y)=3000-2x^2-y^2$ .

Now that the climber moves in the southwest direction

$\begin{aligned} \mathbf{n} & =-1\,\hat{\mathbf{i}}-1\,\hat{\mathbf{j}} \\ \hat{\mathbf{n}} & = \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \end{aligned}$

$\begin{aligned} h_{\hat{\mathbf{n}}}'(x,y) & = \nabla h(x,y)\cdot \hat{\mathbf{n}} \\ & = (-4x\,\hat{\mathbf{i}}-2y\,\hat{\mathbf{j}}) \cdot \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & = \frac{1}{\sqrt{2}} (4x+2y) \end{aligned}$

At point $(30,\, -20,\, 800)$ ,

$\begin{aligned} h_{\hat{\mathbf{n}}}'(30,-20) & = \frac{1}{\sqrt{2}}\big( 4(30)+2(-20)\big) \\ & = \frac{1}{\sqrt{2}}\cdot 80 \qquad (>0) \end{aligned}$

he will ascend southwesterly.

April 22, 2021October 24, 2022

202104221623 Homework 2 (Q1)

Find the gradients of the following functions:

(a) $\phi (x,y,z)=x^2+y^2+z^2$ ;

(b) $\phi (x,y,z)=\exp (x+y)\cos z$ .

Solution.

(a)

$\begin{aligned} & \quad \nabla \phi \\ & = \bigg(\hat{\mathbf{i}}\frac{\partial}{\partial x}+\hat{\mathbf{j}}\frac{\partial}{\partial y}+\hat{\mathbf{k}}\frac{\partial}{\partial z}\bigg)\phi (x,y,z) \\ & = \hat{\mathbf{i}} \frac{\partial}{\partial x}(x^2+y^2+z^2) + \hat{\mathbf{j}} \frac{\partial}{\partial y}(x^2+y^2+z^2) + \hat{\mathbf{k}} \frac{\partial}{\partial z}(x^2+y^2+z^2) \\ & = 2x\,\hat{\mathbf{i}} + 2y\,\hat{\mathbf{j}} + 2z\,\hat{\mathbf{k}} \end{aligned}$

(b) It is left as an exercise to the reader.

April 18, 2021October 24, 2022

202104181519 Homework 1 (Q4)

i. Find the infinitesimal small vector $\mathbf{dr}$ in the cylindrical coordinate induced by an infinitesimal small changes of $\mathrm{d}\rho$ , $\mathrm{d}\theta$ , and $\mathrm{d}z$ in terms of $\rho$ , $\theta$ , $z$ , $\mathrm{d}\rho$ , $\mathrm{d}\theta$ , $\mathrm{d}z$ and the corresponding unit vector.

ii. $f(u_1, u_2, u_3)$ is defined in $\mathbf{r}=(u_1, u_2, u_3)$ coordinate. Its gradient is defined

$\displaystyle{\lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}}_l}$

where $\Delta l_i$ and $\Delta f_i$ are respectively the changes in length and functional value induced purely by the infinitesimal change in $u_i$ . $\hat{\mathbf{u}}_l$ is the unit vector of $\mathbf{u}_i$ . Thus find the gradient of $f$ in cylindrical coordinate.

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

i.

$\mathbf{dr}=\mathrm{d}\rho\,\hat{\boldsymbol{\rho}}+\rho\,\mathrm{d}\theta\,\hat{\boldsymbol{\theta}}+\mathrm{d}z\,\hat{\mathbf{z}}$

Compare to the figure below.

ii.

$\begin{aligned} \nabla f & = \lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}_i} \\ & = \lim_{\Delta\rho\to 0} \frac{\Delta f_\rho}{\Delta \rho}\,\hat{\boldsymbol{\rho}} + \lim_{\Delta\theta\to 0} \frac{\Delta f_\theta}{\rho\Delta\theta}\,\hat{\boldsymbol{\theta}} + \lim_{\Delta z\to 0}\frac{\Delta f_z}{\Delta z}\,\hat{\mathbf{z}} \\ & = \frac{\partial f}{\partial \rho}\,\hat{\boldsymbol{\rho}} + \frac{1}{\rho}\frac{\partial f}{\partial \theta}\,\hat{\boldsymbol{\theta}} + \frac{\partial f}{\partial z}\,\hat{\mathbf{z}} \end{aligned}$

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)

October 13, 2020October 24, 2022

202010132314 Homework 1 (Q1)

The radius $r$ of a right circular cylinder is decreasing at a rate of $12\,\mathrm{cm\, s^{-1}}$ , while its height $h$ is decreasing at a rate of $25\,\mathrm{cm\, min^{-1}}$ . How is the volume changing when $r=180\,\mathrm{cm}$ and $h=500\,\mathrm{cm}$ ? Is the volume increasing or decreasing?

Solution.

The volume of a right circular cylinder is calculated by the formula

$V=\pi r^2h$ .

The volume $V$ (a dependent variable) of a cylinder varies with its radius $r$ and height $h$ (both independent variables). The change of volume, simply put it, is a derivative of volume $V$ with respect to time $t$ :

$\displaystyle{\frac{\mathrm{d}V}{\mathrm{d}t}}$ .

Differentiate $V(r,h,t)$ wrt. time $t$ :

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(r,h,t) & = \frac{\mathrm{d}}{\mathrm{d}t} (\pi r^2 h) \\ & = \pi \frac{\mathrm{d}}{\mathrm{d}t} (r^2h) \\ & = \pi \bigg[ r^2 \frac{\mathrm{d}}{\mathrm{d}t}(h) + h \frac{\mathrm{d}}{\mathrm{d}t} (r^2) \bigg] \\ & = \pi \bigg[ \big( r(t) \big)^2 \frac{\mathrm{d}}{\mathrm{d}t}\big( h(t) \big) + \big( h(t)\big) \bigg( \frac{\mathrm{d}}{\mathrm{d}t}\big( r(t) \big)^2 \bigg) \bigg] \\ & = \pi (r^2h' +2hrr' ) \\ & = \pi r^2h' + 2\pi hrr' \\ \end{aligned}$

Given $r=180\,\mathrm{cm}$ , $h=500\,\mathrm{cm}$ , $r'=-12\,\mathrm{cm\, s^{-1}}=-720\,\mathrm{cm\, min^{-1}}$ , and $h'=-25\,\mathrm{cm\, min^{-1}}$ , you would have it.

In some formalism of partial derivatives,

$\begin{aligned} \frac{\mathrm{d}V(r,h)}{\mathrm{d}t} & = \frac{\partial V}{\partial r}\frac{\mathrm{d}r}{\mathrm{d}t} + \frac{\partial V}{\partial h}\frac{\mathrm{d}h}{\mathrm{d}t} \\ & = (2\pi hr)\frac{\mathrm{d}r}{\mathrm{d}t} + (\pi r^2)\frac{\mathrm{d}h}{\mathrm{d}t}\\ & = 2\pi hrr' +\pi r^2h' \\ \end{aligned}$

you could have it also.

Afterthought.

It just so happens that there are two lines of attack, by taking total/ordinary derivatives and by taking partial derivatives. Is here anyhow the difference? Is there anything the matter?

October 13, 2020October 24, 2022

202010130604 Example 1, Chapter 1.1, Methods in Physics II (2015-2016 Lectures)

Find the trajectory of a cannon ball fired by a cannoneer at $45^\circ$ above his eye level with an initial speed $v$ .

Ans.

Imagine you are in his place, like the picture below:

Then, suppose you are opening fire at such positive $\textrm{(+ve)\enspace }x$ -axis direction as that follows along the meridian. In addition, assume that the earth is to be lying flat beneath the cannon ball during its flight, that every pole ( $z$ -axis) upheld would be so much right-angled (/normal /perpendicular) to the ground as parallel (-transported) anywhere.

We neglect air friction.

1. Along the $x$ -direction, the speed is kept $v\cos 45^\circ$ .

2. Along the $z$ -direction, the speed is initially $v\sin 45^\circ$ . By the laws of gravity, the cannon ball will experience a net force $m\mathbf{g}$ due to gravitational pull by the Earth. If upward direction is taken the positive sign, an equation of motion due to the Galilean transformation (i.e., $v=u+at$ ) will depict that $v_z(t)=v\sin 45^\circ -gt$ .

3. One another equation $s(t)=ut+\displaystyle{\frac{1}{2}}at^2$ depicts how distance $s$ (i.e., the magnitude of displacement $\mathbf{s}$ ) varies with time $t$ during linear motion in constant acceleration $a$ .

One might have already noticed I am setting the original question aside, as it were. Let’s pinpoint the answer now.

The trajectory should be obtained in the form:

$\mathbf{s}(t)=s_x(t)\,\hat{\mathbf{i}} + s_z(t)\,\hat{\mathbf{k}}$ .

where

$\begin{aligned} s_x(t) & = u_x t+\frac{1}{2}a_xt^2 \\ & = (v\cos 45^\circ )\, t + \frac{1}{2}(0)(t^2) \\ & = (v\cos 45^\circ )\, t \\ \end{aligned}$

and

$\begin{aligned} s_z(t) & = u_zt + \frac{1}{2}a_zt^2 \\ & = (v\sin 45^\circ )\, t + \frac{1}{2}(-g)(t^2) \\ \end{aligned}$