# Calculus

 Question 1
The closed curve shown in the figure is described by

$r=1+\cos \theta$, where $r=\sqrt{x^{2}+y^{2}}$ $x=r \cos \theta, y=r \sin \theta$

The magnitude of the line integral of the vector field $F=-y \hat{i}+x \hat{j}$ around the closed curve is ___(Round off to 2 decimal places). A 9.42 B 6.36 C 2.45 D 7.54
GATE EE 2023   Engineering Mathematics
Question 1 Explanation:
\begin{aligned} I & =\int_{0}^{2 \pi} \vec{F} \cdot \overrightarrow{d l} \\ & =\int_{0}^{2 \pi}(-y \hat{i}+x)(d x \hat{i}+d y) \\ & =\int_{0}^{2 \pi}(-y d x+x d y) \end{aligned}
Given : $\quad x=r \cos \theta$ and $y=r \sin \theta$
\begin{aligned} \therefore I&=\int_{0}^{2 \pi}[(-r \sin \theta)(-r \sin \theta) d \theta+(r \cos \theta)(r \cos \theta) d \theta] \\ &=\int_{0}^{2 \pi} r^{2} d \theta \\ &=\int_{0}^{2 \pi}(1+\cos \theta)^{2} d \theta \\ &=\int_{0}^{2 \pi}\left(1+\cos ^{2} \theta+2 \cos \theta\right) d \theta \\ &=\int_{0}^{2 \pi}\left(1+\frac{1+\cos 2 \theta}{2}+2 \cos \theta\right) \mathrm{d} \theta \\ &=3 \pi=9.425 \end{aligned}
 Question 2
Consider the following equation in a 2-D realspace.

$\left|x_{1}\right|^{p}+\left|x_{2}\right|^{p}=1$ for $p \gt 0$

Which of the following statement(s) is/are true.
 A When $\mathrm{p}=2$, the area enclosed by the curve is $\pi$. B When $p$ tends to $\infty$, the area enclosed by the curve tends to 4. C When $p$ tends to 0 , the area enclosed by the curve is 1. D When $p=1$, the area enclosed by the curve is 2.
GATE EE 2023   Engineering Mathematics
Question 2 Explanation:
Check option (A),
put $P=2$
$x_{1}^{2}+x_{2}^{2}=1$

Which is equation of circle whose radius is 1 .
$\therefore \quad$ Area $=\pi r^{2}=\pi$

Check option (B),
If $x_{2} \lt 1$ Then, $P \rightarrow \infty,\left|x_{2}\right|^{P} \rightarrow 0$
$\therefore$ For equation $\left|x_{1}\right|^{\mathrm{P}}+\left|\mathrm{x}_{2}\right|^{\mathrm{P}}=1$ satisfaction,
$\mathrm{x}_{1} \rightarrow 1 \text { or } \dot{Y}$

Curve : $\therefore$ Area tends to 4 .

Check option (D),
Put $P=1$,
$\left|x_{1}\right|+\left|\mathrm{x}_{2}\right|=1$

Curve: $\therefore$ Area of curve (square) $=(\sqrt{2})^{2}=2$

 Question 3
One million random numbers are generated from a statistically stationary process with a Gaussian distribution with mean zero and standard deviation $\sigma_{0}$.
The $\sigma_{0}$ is estimated by randomly drawing out 10,000 numbers of samples $\left(x_{n}\right)$. The estimates $\hat{\sigma}_{1}, \hat{\sigma}_{2}$ are computed in the following two ways.
$\hat{\sigma}_{1}^{2}=\frac{1}{10000} \sum_{n=1}^{10000} x_{n}^{2} \quad \hat{\sigma}_{2}^{2}=\frac{1}{9999} \sum_{n=1}^{10000} x_{n}^{2}$

Which of the following statements is true?
 A $E\left(\hat{\sigma}_{2}^{2}\right)=\sigma_{0}^{2}$ B $\mathrm{E}\left(\hat{\sigma}_{2}\right)=\sigma_{0}$ C $E\left(\hat{\sigma}_{1}^{2}\right)=\sigma_{0}^{2}$ D $\mathrm{E}\left(\hat{\sigma}_{1}\right)=\mathrm{E}\left(\hat{\sigma}_{2}\right)$
GATE EE 2023   Engineering Mathematics
Question 3 Explanation:
We know,
$\mathrm{E}\left[(\mathrm{X}-\overline{\mathrm{X}})^{2}\right]=\operatorname{Var}$

Given, $\quad \bar{X}=0$
$\therefore \quad E[X]=\operatorname{Var}$.

We have,
Standard deviation, $\sigma^{2}$ $=\frac{1}{N-1} \sum_{n=1}^{n} x_{n}^{2} \Rightarrow \text { for sample }$

and $\quad \sigma^{2}=\frac{1}{N} \sum_{n=1}^{n} x_{n}^{2} \Rightarrow$ For population

Here, we take samples from $10^{6}$ numbers.
$\therefore \quad \mathrm{E}\left(\sigma_{2}^{2}\right)=\sigma_{0}^{2}$ will be correct.
 Question 4
In the figure, the vectors $u$ and $v$ are related as: $A u=v$ by a transformation matrix $A$. The correct choice of $A$ is A $\left[\begin{array}{cc}\frac{4}{5} & \frac{3}{5} \\ -\frac{3}{5} & \frac{4}{5}\end{array}\right]$ B $\left[\begin{array}{cc}\frac{4}{5} & -\frac{3}{5} \\ \frac{3}{5} & \frac{4}{5}\end{array}\right]$ C $\left[\begin{array}{ll}\frac{4}{5} & \frac{3}{5} \\ \frac{3}{5} & \frac{4}{5}\end{array}\right]$ D $\left[\begin{array}{rr}\frac{4}{5} & -\frac{3}{5} \\ \frac{3}{5} & -\frac{4}{5}\end{array}\right]$
GATE EE 2023   Engineering Mathematics
Question 4 Explanation:
Given : \begin{aligned} & |u|=\sqrt{4^{2}+3^{2}}=5 \\ & |V|=5 \end{aligned}
and $\quad \theta=\tan ^{-1} \frac{3}{4} \Rightarrow \tan \theta=\frac{3}{4}$
$\therefore \quad \sin \theta=\frac{3}{5}$ and $\cos \theta=\frac{4}{5}$
For clockwise relation by $\theta$, transformation matrix,
\begin{aligned} A & =\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{4}{5} & \frac{3}{5} \\ -\frac{3}{5} & \frac{4}{5} \end{array}\right] \end{aligned}

$[\because$ F or clockwise $\theta \rightarrow$ N egative $]$
 Question 5
For a given vector $w=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{\top}$, the vector normal to the plane defined by $\mathbf{w}^{\top} x=1$ is
 A $\left[\begin{array}{lll}-2 & -2 & 2\end{array}\right]^{T}$ B $\left[\begin{array}{lll}3 & 0 & -1\end{array}\right]^{T}$ C $\left[\begin{array}{lll}3 & 2 & 1\end{array}\right]^{T}$ D $\left[\begin{array}{llll}1 & 2 & 3\end{array}\right]^{T}$
GATE EE 2023   Engineering Mathematics
Question 5 Explanation:
Given, $W^{T}=1$

$\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=1$

We have, vector normal to the plane $=\nabla F$
\begin{aligned} & =i \frac{\partial F}{\partial x}+\hat{j} \frac{\partial F}{\partial y}+\hat{k} \frac{\partial F}{\partial z} \\ & =\hat{i}+2 \hat{j}+3 \hat{k} \end{aligned}

$\therefore$ Normal vector $=\left[\begin{array}{lll}1 & 2 & 3\end{array}\right]^{T}$

There are 5 questions to complete.