Calculus Coursework 1

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Question 1 §

Question 1a §

f(x)=\sin(3x^4-x^2+2)\

$$⟹f(-x)=\sin (3(-x{)}^4-(-x{)}^2+2):\text{testing the behaviour of the function}$$

$$⟹f(-x)=\sin (3x^4-x^2+2)=f(x)$$

$$\therefore f(-x)=f(x):\text{the function is even (it displays reflective symmetry).}$$

Question 1b §

$$g(v)=\frac{v^5-v^3+1}{v^4+v^2+v}$$

$$⟹g(-v)=\frac{(-v{)}^5-(-v{)}^3+1}{(-v{)}^4+(-v{)}^2+(-v)}:\text{testing the behaviour of the function}$$

$$g(-v)=\frac{-v^5+v^3+1}{v^4+u^2-v}\ne \pm g(\mp v)$$

$$\therefore g(-v)\ne \pm g(\mp v):\text{the function is neither odd nor even (asymmetrical).}$$

Question 2 §

Question 2a §

$$\sinh x=\frac{e^x-e^{-x}}{2},\cosh x=\frac{e^x+e^{-x}}{2}$$

$$⟹2\sinh x\cosh x=2\left(\frac{e^x-e^{-x}}{2}\frac{e^x+e^{-x}}{2}\right)$$

$$⟹2\sinh x\cosh x=\frac{(e^x-e^{-x})(e^x+e^{-x})}{2}$$

$$⟹2\sinh x\cosh x=\frac{e^{2x}-e^{-2x}}{2}:\text{by the difference of two squares}$$

$$⟹2\sinh x\cosh x=\frac{e^{2x}-e^{-2x}}{2}=\sinh 2x$$

$$\therefore 2\sinh x\cosh x=\sinh 2x\square$$

Question 2b §

$$\sinh x=\frac{e^x-e^{-x}}{2},\cosh x=\frac{e^x+e^{-x}}{2}$$

$$⟹\cosh x\cosh y+\sinh x\sin y=\left(\frac{e^x+e^{-x}}{2}\frac{e^y+e^{-y}}{2}\right)+\left(\frac{e^x-e^{-x}}{2}\frac{e^y-e^{-y}}{2}\right)$$

$$⟹\cosh x\cosh y+\sinh x\sin y=\frac{(e^x+e^{-x})(e^y+e^{-y})+(e^x-e^{-x})(e^y-e^{-y})}{4}$$

$$⟹\cosh x\cosh y+\sinh x\sin y=\frac{(e^{x+y}+e^{x-y}+e^{y-x}+e^{-x-y})+(e^{x+y}-e^{x-y}-e^{y-x}+e^{-x-y})}{4}$$

$$⟹\cosh x\cosh y+\sinh x\sin y=\frac{2e^{x+y}+2e^{-(x+y)}}{4}$$

$$⟹\cosh x\cosh y+\sinh x\sin y=\frac{e^{(x+y)}+e^{-(x+y)}}{2}=\cosh (x+y)$$

$$\therefore \cosh x\cosh y+\sinh x\sin y=\cosh (x+y)\square$$

Question 3 §

$$z_1=3+i,z_2=7-i$$

Question 3a §

$$z_1{z}_2=(3+i)(7-i)$$

$$⟹z_1{z}_2=22+4i$$

Question 3b §

$$\frac{|z_1|}{|z_2|}=\frac{\sqrt{(3{)}^2+(1{)}^2}}{\sqrt{(7{)}^2+(-1{)}^2}}$$

$$⟹\frac{|z_1|}{|z_2|}=\frac{\sqrt{10}}{\sqrt{50}}=\frac{10}{5\sqrt{2}}$$

Question 3c §

$$\frac{z_1}{z_2}=\frac{3+i}{7-i}\cdot \frac{7+i}{7+i}$$

$$⟹\frac{z_1}{z_2}=\frac{20+10i}{50}=\frac{2+i}{5}$$

$$⟹\frac{z_1}{z_2}=\frac{2}{5}+\frac{1}{5}i$$

Question 3d §

$$\frac{\overline{z_2}}{\overline{z_1}}=\frac{7+i}{3-i}\cdot \frac{3+i}{3+i}$$

$$⟹\frac{\overline{z_2}}{\overline{z_1}}=\frac{20+10i}{10}=2+i$$

Question 4 §

Question 4a §

$$2\sqrt{2}-2\sqrt{2}i⟹(r,\theta )=\left(\sqrt{(2\sqrt{2}{)}^2+(-2\sqrt{2}{)}^2},{\tan }^{-1}\left(\frac{-2\sqrt{2}}{2\sqrt{2}}\right)\right)$$

$$⟹(r,\theta )=\left(4,-\frac{\pi }{4}\right):0\le \theta <2\pi$$

$$\therefore (r,\theta )=\left(4,\frac{7\pi }{4}\right)$$

Calculus Coursework 1 2023-10-26 16.34.17.excalidraw Above (in red) is plotted the complex number $2\sqrt{2}-2\sqrt{2}i$, which in polar form is represented as $4\left(\cos \frac{7\pi }{4}+isin\frac{7\pi }{4}\right)$.

Question 4b §

$$-2\sqrt{3}+2i⟹(r,\theta )=\left(\sqrt{(-2\sqrt{3}{)}^2+(2{)}^2},{\tan }^{-1}\left(\frac{2}{-2\sqrt{3}}\right)\right)$$

$$⟹(r,\theta )=\left(4,-\frac{\pi }{6}\right):0\le \theta <2\pi$$

$$\therefore (r,\theta )=\left(4,\frac{11\pi }{6}\right)\text{, however this is in the wrong direction}$$

$$⟹(r,\theta )=\left(4,\frac{5\pi }{6}\right)$$

Calculus Coursework 1 2023-10-26 16.36.15.excalidraw Above is plotted the complex number $-2\sqrt{3}+2i$, which in polar form is represented as $4\left(\cos \frac{5\pi }{6}+isin\frac{5\pi }{6}\right)$.

Question 5 §

Question 5a §

$$z^3=4(-1+\sqrt{3}i)⟹z=4^{\frac{1}{3}}(-1+\sqrt{3}i{)}^{\frac{1}{3}}$$

First, working out the angle of $z$, which would be the same as the angle of any multiple of $z$, hence:

$$\angle z=\frac{\angle (-1+\sqrt{3}i)+2n\pi }{3}=\frac{{\tan }^{-1}\left(\frac{\sqrt{3}}{-1}\right)+2n\pi }{3}=\frac{-\frac{\pi }{3}+2n\pi }{3}=\frac{(6n-1)\pi }{9}$$

$$⟹\angle z=\frac{(6n-1)\pi }{9}=\theta ,$$

Next, working out the modulus of $z$,

$$|z|=4^{\frac{1}{3}}\cdot |-1+\sqrt{3}i|=4^{\frac{1}{3}}\cdot \sqrt{(-1{)}^2+(\sqrt{3}{)}^2}=4^{\frac{1}{3}}\cdot 2$$

$$⟹|z|=2^{\frac{2}{3}}\cdot 2=2^{\frac{5}{3}}=r.$$

Combining this information gives us the polar form of $z$:

$$\therefore z=(r,\theta )=\left(2^{\frac{5}{3}},\frac{(6n-1)\pi }{9}\right):n\in z$$

$$⟹z=\left(2^{\frac{5}{3}},\frac{5\pi }{9}\right),\left(2^{\frac{5}{3}},\frac{11\pi }{9}\right),\left(2^{\frac{5}{3}},\frac{17\pi }{9}\right)$$

Which we can then convert into exponential form:

$$\therefore z=2^{\frac{5}{3}}\cdot e^{\frac{5\pi }{9}},2^{\frac{5}{3}}\cdot e^{\frac{11\pi }{9}},2^{\frac{5}{3}}\cdot e^{\frac{17\pi }{9}}$$

Question 5b §

$$z^4=\frac{1}{i}\cdot \frac{-i}{-i}=-i⟹z=(-i{)}^{\frac{1}{4}}$$

First, working out the angle of $z$:

$$\angle z=\frac{\angle (-i)+2n\pi }{4}=\frac{\frac{3\pi }{2}+2n\pi }{4}=\frac{(3+4n)\pi }{8}$$

$$⟹\angle z=\frac{(3+4n)\pi }{8}=\theta ,$$

Next, working out the modulus of $z$,

$$|z|=\sqrt{(0{)}^2+(\sqrt{-1}{)}^2}=1=r$$

Combining this information gives us the polar form of $z$:

$$\therefore z=(r,\theta )=\left(1,\frac{(3+4n)\pi }{8}\right):n\in z$$

$$⟹z=\left(1,\frac{3\pi }{8}\right),\left(1,\frac{7\pi }{8}\right),\left(1,\frac{11\pi }{8}\right),\left(1,\frac{15\pi }{8}\right)$$

Which we can then convert into exponential form:

$$\therefore z=e^{\frac{3\pi }{8}},e^{\frac{7\pi }{8}},e^{\frac{11\pi }{8}},e^{\frac{15\pi }{8}}$$

Question 6 §

Question 6a §

$$\underset{x\to 3}{\lim }\frac{x-3}{x^2+x-12}=\frac{0}{0}:\text{undefined, use L’hoˆptial’s Rule - differentiate}$$

$$⟹\underset{x\to 3}{\lim }\frac{1}{2x+1}=\frac{1}{7}$$

Question 6b §

$$\underset{v\to \infty }{\lim }\frac{1-v^2+3v^3-4v^4}{9v^4+2v^3-5}=\underset{v\to \infty }{\lim }\left(\frac{1-v^2+3v^3-4v^4}{9v^4+2v^3-5}\cdot \frac{v^{-4}}{v^{-4}}\right)$$

$$=\underset{v\to \infty }{\lim }\frac{v^{-4}-v^{-2}+3v^{-1}-4}{9+2v^{-1}-5v^{-4}}=-\frac{4}{9}$$

Question 6c §

$$\underset{y\to 0}{\lim }\frac{e^{ay}-1-ay}{y^2}=\frac{0}{0}:\text{undefined, use L’hoˆptial’s Rule - differentiate}$$

$$⟹\underset{y\to 0}{\lim }\frac{a{e}^{ay}-y}{2y}=\frac{a}{0}:\text{undefined, use L’hoˆptial’s Rule - differentiate}$$

$$⟹\underset{y\to 0}{\lim }\frac{a^2{e}^{ay}-1}{2}=\frac{a^2-1}{2}$$

Question 6d §

$$\underset{x\to 1}{\lim }\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)=\underset{x\to 1}{\lim }\left(\frac{(x-1)-(\ln x)}{(\ln x)(x-1)}\right)=\underset{x\to 1}{\lim }\left(\frac{x-1-\ln x}{x\ln x-\ln x}\right)=\frac{0}{0}$$

$$:\text{undefined, use L’hoˆptial’s Rule - differentiate using product rule/chain rule}$$

$$⟹\underset{x\to 1}{\lim }\left(\frac{1-\frac{1}{x}}{\ln x+1-\frac{1}{x}}\right)=\frac{0}{0}:\text{undefined, use L’hoˆptial’s Rule - differentiate}$$

$$⟹\underset{x\to 1}{\lim }\left(\frac{x^{-2}}{x^{-2}+x^{-1}}\right)=\frac{1}{2}$$