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Consider the following guidelines when writing a conclusion for a statement of purpose (SOP):

• Emphasize the key points
• Highlight your strengths
• Be precise and program-focused
• Make it brief and to the point

It might be challenging to write an opening to a statement of purpose (SOP), but it is vital to make a good first impression. This is a good example of an SOP introduction: In the first paragraph, introduce yourself briefly by providing your name, educational background, and any relevant work experience. Instead of introducing yourself, you may employ an enticing hook to spark the reader's interest and persuade them to keep reading. You could also explain why you wish to apply for the programme or degree, as well as why you are interested in the field. Remember to write in a straightforward, honest, and concise manner.

Let ‘x’ be the number of bacteria present in instantaneous time t.

Then, $\frac{dx}{dt}\propto x$

$\Rightarrow \frac{dx}{dt}=kx,where,k=$ constant of proportionality.

$\Rightarrow \frac{dx}{x}=kdt$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dx}{x}={\displaystyle \int kdt}}\\ \Rightarrow logx=kt+c\end{array}$

Given, at $t=0,x=x_0(say)then,$

$log{x}_0=c\left(Initial,x_0=100000\right)$

So, the differential equation is

$\begin{array}{l}logx=kt+log{x}_0\\ \Rightarrow logx-log{x}_0=kt\\ \Rightarrow log\frac{x}{x_0}=kt\end{array}$

As the bacteria number increased by 10% in 2 hours.

The number of bacteria increased in 2hours $=10\mathrm{\%}\times 100000=10000$

Hence, at t=2,

$x=100000+10000=110000$

So, $\begin{array}{l}log\frac{110000}{100000}=2k\\ \Rightarrow k=\frac{1}{2}log\left(\frac{11}{10}\right)\end{array}$

Hence, $log\frac{x}{x_0}=\left[\frac{1}{2}log\frac{11}{10}\right]\times t$

$when,x=200000,$ then we get,

$log\frac{200000}{100000}=\frac{1}{2}log\frac{11}{10}\times t$

$\begin{array}{l}\Rightarrow 2log2=log\left(\frac{11}{10}\right)\times t\\ \Rightarrow t=\frac{2log2}{log\left(\frac{11}{10}\right)}hours\end{array}$

To guarantee that your Statement of Purpose (SOP) accurately represents your individuality, provide real-life examples and perspectives that have influenced your educational and career path. Make your SOP specific to each program, emphasizing elements that support their core principles. Rewrite your SOP to make it more coherent and clearer. Highlight specific accomplishments and objectives; using this method will assist you in developing an engaging and unique Statement of Purpose for US colleges.

Let P and t the principal and time respectively.

Then, increase in principal $\frac{dP}{dt}=P\times 5\mathrm{\%}$

$\Rightarrow \frac{dP}{P}=\frac{5}{100}dt$

Integrating both sides,

$\begin{array}{l}\Rightarrow {\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{1}{20}dt}}\\ \Rightarrow logP=\frac{t}{20}+c\\ \Rightarrow P=e^{\frac{t}{20}+c}\end{array}$

At, t=0, P=1000

So,  $\begin{array}{l}1000=e^{\frac{0}{20}+c}\\ \Rightarrow e^c=1000\end{array}$

And at t=10,

$\begin{array}{l}P=e^{\frac{10}{100}+c}=e^{0.5}.e^c\\ \Rightarrow P=1.648\times 1000=1648\end{array}$

P = ?1648

While sample SOPs are a great source of ideas and direction, it's important to craft a unique and personalized statement that captures your own experiences, goals, and objectives. Understanding the format and tone of a well-written SOP can be facilitated by using an example as a template; however, outright copying should be avoided. To stand out to admissions committees, your Statement of Purpose should highlight your unique qualifications, accomplishments, and future objectives.

Let P, r and t be the principal rate and time respectively.

Then, increase in principal $\frac{dP}{dt}=P\times r\mathrm{\%}$

$\begin{array}{l}\Rightarrow \frac{dP}{dt}=P.\frac{r}{100}\\ \Rightarrow \frac{dP}{P}=\frac{r}{100}dt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{r}{100}dt}}\\ \Rightarrow logP=\frac{rt}{100}+c\\ \Rightarrow P=e^{\frac{rt}{100}+c}\end{array}$

Given at t=0,P=100

So, $100=e^{\frac{r\times 0}{100}+c}$

$\begin{array}{l}\Rightarrow 100=e^0\times e^c\\ \Rightarrow e^c=100\left(?e^0=1\right)\end{array}$

And at $t=10,P=2\times 100=200$

So, $200=e^{\frac{r10}{100}+c}$

$\begin{array}{l}\Rightarrow 200=e^{\frac{r}{10}}.e^e\\ \Rightarrow e^{\frac{r}{10}}=\frac{200}{100}=2\end{array}$

$\begin{array}{l}\Rightarrow \frac{r}{10}=log2\\ \Rightarrow \frac{r}{10}=0.6931\\ \Rightarrow r=6.931\end{array}$

Hence, the rate is 6.931%

111. Given, $y={\left(ta{n}^{-1}x\right)}^2$

So, $y_1=\frac{dy}{dx}=2\left(ta{n}^{-1}x\right)\frac{d}{dx}ta{n}^{-1}x$

$\Rightarrow y_1=2ta{n}^{-1}x\times \frac{1}{1+x^2}$

$\left(x^2+1\right)y_1=2ta{n}^{-1}x$

Differentiating again w r t ‘x’ we get,

$\left(x^2+1\right)\frac{d{y}_1}{dx}+y_1\frac{d}{dx}\left(x^2+1\right)=2\frac{d}{dx}ta{n}^{-1}x$

$\Rightarrow \left(x^2+1\right)y_2+y_1\left(2x\right)=\frac{2}{1+x^2}$

$\Rightarrow {\left(x^2+1\right)}^2{y}_2+2x\left(1+x^2\right)y_1=2$

Hence proved.

Let ‘r’ and U be the radius and volume of the spherical balloon.

Then, $\frac{dU}{dt}=k,$ k = constant

$\begin{array}{l}\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=k\\ \Rightarrow 4\pi r^2\frac{dr}{dt}=k\\ \Rightarrow 4\pi r^{2}dr=kdt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int 4\pi r^{2}dr={\displaystyle \int kdt}}\\ \Rightarrow \frac{4}{3}\pi r^3=kt+c\end{array}$

Given at t = 0, r = 3

So, 4π(3)³ = c

C = 36π

And, at t=3, r=6

So, $\frac{4}{3}\pi {\left(6\right)}^3=3k+36\pi \left(c=36\pi \right)$

$\begin{array}{l}\Rightarrow 288\pi -36\pi =3k\\ \Rightarrow k=\frac{252\pi }{3}=84\pi \end{array}$

Hence, putting value of c and k in,

$\frac{4}{3}\pi r^3=kt+c$ , we get,

$\begin{array}{l}\frac{4}{3}\pi r^3=84\pi .t+36\pi \\ \Rightarrow r^3=\frac{3}{4\pi }\left(84\pi .t+36\pi \right)\\ \Rightarrow r^3=63t+27\\ \Rightarrow r={\left[63t+27\right]}^{\frac{1}{3}}\end{array}$

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110. Kindly go through the solution

The slope of tangent is $\frac{dy}{dx}$ and slope of line joining line (-4,-3) and point say P(x,y)

$\frac{y-\left(-3\right)}{x-\left(-4\right)}=\frac{y+3}{x+4}$

So, $\frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$

$\Rightarrow \frac{dy}{y+3}=\frac{2}{x+4}dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{y+3}={\displaystyle \int \frac{2}{x+4}dx}}\\ \Rightarrow log\left|y+3\right|=2log\left|x+4\right|+log\left|c\right|\\ \Rightarrow log\left|y+3\right|=log{\left(x+4\right)}^2+log\left|c\right|\\ \Rightarrow log\left|y+3\right|=log\left|c{\left(x+4\right)}^2\right|\\ \Rightarrow y+3=c_1{\left(x+4\right)}^2,where,c_1=\pm c\end{array}$

Since, the curve passes through (-2,1) we get,

$\begin{array}{l}y=1,at,x=-2\\ \Rightarrow 1+3=c{\left(-2+4\right)}^2\\ \Rightarrow 4=c\times 4\\ \Rightarrow c=1\end{array}$

$\therefore$ The equation of the curve is $y+3={\left(x+4\right)}^2$

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The slope of the tangent to then curve is $\frac{dy}{dx}$

$\begin{array}{l}\frac{dy}{dx}.y=x\\ \Rightarrow y.dy=xdx\end{array}$

So,

Integrating both sides,

$\begin{array}{l}{\displaystyle \int y.dy={\displaystyle \int xdx}}\\ \Rightarrow {\displaystyle \int \frac{y^2}{2}=\frac{x^2}{2}+c}\\ \Rightarrow y^2=x^2+A,Where,A=2c\end{array}$

As the curve passes through (0, -2) we have,

$\begin{array}{l}{\left(-2\right)}^2=0^2+A\\ \Rightarrow A=4\end{array}$

$\therefore$ The equation of the curve is

$y^2=x^2+4$

109. Given,  $y=500{e^7}^x+600{e^{-7}}^x$

So,  $\frac{dy}{dx}=500\times 7{e^7}^x+600\left(-7\right){e^{-7}}^x$

$\therefore \frac{d^{2}y}{d{x}^2}=500\times 7^2{e^7}^x+600\times 7^2{e^{-7}}^x$

$=49\left[500e^{7x}+600e^{-7x}\right]$

$=49\times y$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=49y$

The Given D.E is

$\begin{array}{l}xy\frac{dy}{dx}=\left(x+2\right)\left(y+2\right)\\ \Rightarrow y\frac{dy}{y+2}=\frac{\left(x+2\right)}{2}dx\\ \Rightarrow \frac{y+2-2}{y+2}dy=\left(\frac{x}{x}+\frac{2}{x}\right)dx\\ \Rightarrow \left(1-\frac{2}{y+2}\right)dy=\left(1+\frac{2}{x}\right)dydx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \left(1-\frac{2}{y+2}\right)dy={\displaystyle \int \left(1+\frac{2}{x}\right)dydx}}\\ \Rightarrow y-2log\left|y+2\right|=x+2log\left|x\right|+c\\ \Rightarrow y-log{\left(y+2\right)}^2=x+log{x}^2+c\\ \Rightarrow y-x=log{\left(y+2\right)}^2+log{x}^2+c\\ \Rightarrow y-x=log\left[{\left(y+2\right)}^2.x^2\right]+c\end{array}$

A the curve passes through (-1,1) then $y=-2,at,x=1$

So, $-1-1=log{\left(-1+2\right)}^2.{\left(1\right)}^2+c$

$\begin{array}{l}\Rightarrow -2=log1+c\\ \Rightarrow c=-2\end{array}$

$\therefore$ The required equation of curve is,

$y-x=log\left[{\left(y+2\right)}^2{x}^2\right]-2$