Question 1 |
A quadratic function of two variables is given as
f\left(x_{1}, x_{2}\right)=x_{1}^{2}+2 x_{2}^{2}+3 x_{1}+3 x_{2}+x_{1} x_{2}+1
The magnitude of the maximum rate of change of the function at the point (1,1) is ____(Round off to the nearest integer).
f\left(x_{1}, x_{2}\right)=x_{1}^{2}+2 x_{2}^{2}+3 x_{1}+3 x_{2}+x_{1} x_{2}+1
The magnitude of the maximum rate of change of the function at the point (1,1) is ____(Round off to the nearest integer).
10 | |
12 | |
8 | |
16 |
Question 1 Explanation:
Given :
\mathrm{F}\left(\mathrm{x}, \mathrm{x}_{2}\right)=\mathrm{x}_{1}^{2}+2 \mathrm{x}_{2}^{2}+\mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{x}_{2}+1
\nabla \cdot F=\left(2 x_{1}+3+x_{2}\right) \hat{a}_{x_{1}}+\left(4 x_{2}+3+x_{1}\right) \hat{a}_{x_{z}}
At (1,1)
\begin{aligned} \nabla \cdot F & =6 \hat{a}_{x_{1}}+7 \hat{a}_{x_{2}} \\ |\nabla \cdot F| & =\sqrt{6^{2}+8^{2}}=10 \end{aligned}
\mathrm{F}\left(\mathrm{x}, \mathrm{x}_{2}\right)=\mathrm{x}_{1}^{2}+2 \mathrm{x}_{2}^{2}+\mathrm{x}_{1}+\mathrm{x}_{1} \mathrm{x}_{2}+1
\nabla \cdot F=\left(2 x_{1}+3+x_{2}\right) \hat{a}_{x_{1}}+\left(4 x_{2}+3+x_{1}\right) \hat{a}_{x_{z}}
At (1,1)
\begin{aligned} \nabla \cdot F & =6 \hat{a}_{x_{1}}+7 \hat{a}_{x_{2}} \\ |\nabla \cdot F| & =\sqrt{6^{2}+8^{2}}=10 \end{aligned}
Question 2 |
Three points in the x-y plane are (-1,0.8) (0,2.2) and (1,2.8). The value of the slope of the best fit straight line in the least square sense is ___ (Round off to 2 decimal places).
0.25 | |
0.5 | |
0.75 | |
1 |
Question 2 Explanation:
Straight line equation, y=a x+b [Let]
where, a= slope
By lest approximation,
\sum \mathrm{y}_{i}=a \sum \mathrm{x}_{\mathrm{i}}+\mathrm{bn}
and \quad \sum x_{i} y_{i}=a \sum x_{1}^{2}+b \sum x_{i}
\begin{array}{|c|c|c|c|} \hline x & y & x^{2} & xy \\ \hline -1 & 0.8 & 1 & -0.8 \\ 0 & 2.2 & 0 & 0 \\ 1 & 2.8 & 1 & 2.8 \\ \hline \sum x=0 & \sum y=5.8 &\sum x^{2}=2 & \sum x y=2 \\ \hline \end{array}
From eqn. (1), we get
\begin{aligned} 5.8 & =a(0)+3 b \Rightarrow 5.8=3 b \\ 2 & =a(2)+b(0) \\ \Rightarrow \quad a & =1 \end{aligned}
where, a= slope
By lest approximation,
\sum \mathrm{y}_{i}=a \sum \mathrm{x}_{\mathrm{i}}+\mathrm{bn}
and \quad \sum x_{i} y_{i}=a \sum x_{1}^{2}+b \sum x_{i}
\begin{array}{|c|c|c|c|} \hline x & y & x^{2} & xy \\ \hline -1 & 0.8 & 1 & -0.8 \\ 0 & 2.2 & 0 & 0 \\ 1 & 2.8 & 1 & 2.8 \\ \hline \sum x=0 & \sum y=5.8 &\sum x^{2}=2 & \sum x y=2 \\ \hline \end{array}
From eqn. (1), we get
\begin{aligned} 5.8 & =a(0)+3 b \Rightarrow 5.8=3 b \\ 2 & =a(2)+b(0) \\ \Rightarrow \quad a & =1 \end{aligned}
Question 3 |
In the following differential equation, the numerically obtained value of y(t), at t=1, is ___ (Round off to 2 decimal places).
\frac{d y}{d t}=\frac{e^{-\alpha t}}{2+\alpha \mathrm{t}}, \alpha=0.01 \text { and } y(0)=0
\frac{d y}{d t}=\frac{e^{-\alpha t}}{2+\alpha \mathrm{t}}, \alpha=0.01 \text { and } y(0)=0
0.25 | |
0.5 | |
0.75 | |
0.85 |
Question 3 Explanation:
Given :
\frac{d y}{d t}=\frac{e^{-\alpha t}}{2+\alpha t}
\alpha=0.01 and y(0)=0
\begin{aligned} \left.\frac{d^{2} y}{d t^{2}}\right|_{\mathrm{t}=0} & =y^{\prime \prime}(0) \\ & =-\frac{\alpha e^{-\alpha t}}{2+\alpha \mathrm{T}}-\left.\frac{\alpha e^{-\alpha t}}{(2+\alpha \mathrm{t})^{2}}\right|_{\mathrm{t}=0} \\ & =-\frac{\alpha}{2}-\frac{\alpha}{4}=-\frac{3 \alpha}{4} \end{aligned}
We know,
Taylor's series
\begin{aligned} y(t) & =y(0)+y^{\prime}(0)+\frac{t^{2} y^{\prime \prime}(0)}{2 !}+\ldots \\ & =0+t\left(\frac{1}{2}\right)+\frac{t^{2}}{2}\left(-\frac{3 \alpha}{4}\right) \\ & =\frac{t}{2}-\frac{3}{8} \alpha t^{2} \end{aligned}
At \mathrm{t}=1
\begin{aligned} y(1) & =\frac{1}{2}-\frac{3}{8}(0.01) \times 1^{2} \\ & =0.5-\frac{3}{8} \times 0.01 \\ & =0.496 \approx 0.5 \end{aligned}
\frac{d y}{d t}=\frac{e^{-\alpha t}}{2+\alpha t}
\alpha=0.01 and y(0)=0
\begin{aligned} \left.\frac{d^{2} y}{d t^{2}}\right|_{\mathrm{t}=0} & =y^{\prime \prime}(0) \\ & =-\frac{\alpha e^{-\alpha t}}{2+\alpha \mathrm{T}}-\left.\frac{\alpha e^{-\alpha t}}{(2+\alpha \mathrm{t})^{2}}\right|_{\mathrm{t}=0} \\ & =-\frac{\alpha}{2}-\frac{\alpha}{4}=-\frac{3 \alpha}{4} \end{aligned}
We know,
Taylor's series
\begin{aligned} y(t) & =y(0)+y^{\prime}(0)+\frac{t^{2} y^{\prime \prime}(0)}{2 !}+\ldots \\ & =0+t\left(\frac{1}{2}\right)+\frac{t^{2}}{2}\left(-\frac{3 \alpha}{4}\right) \\ & =\frac{t}{2}-\frac{3}{8} \alpha t^{2} \end{aligned}
At \mathrm{t}=1
\begin{aligned} y(1) & =\frac{1}{2}-\frac{3}{8}(0.01) \times 1^{2} \\ & =0.5-\frac{3}{8} \times 0.01 \\ & =0.496 \approx 0.5 \end{aligned}
Question 4 |
Consider the initial value problem below. The value of y at x = ln \;2. (rounded off to 3
decimal places) is ________ .
\frac{dy}{dx}=2x-y,\; y(0)=1
\frac{dy}{dx}=2x-y,\; y(0)=1
1.386 | |
0.886 | |
0.452 | |
0.642 |
Question 4 Explanation:
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{d} x}&=2x-y \\ y(0)&=1, \; y \text{ at } x=\ln 2\\ \frac{\mathrm{d} y}{\mathrm{d} x}+y&=2x \\ P&=1, \;\; Q=2x \\ I.F.&=e^{\int Pdx}=e^{\int 1dx}=e^{x}\\ \text{Solution, } Y(I.F)&=\int Q(I.F.)dx\\ye^{x}&=\int 2x\cdot e^{x}dx \\&=2(xe^{x}-e^{x})+C \\ y&=2x-2+ce^{-x} \\ y(0)&=1 \\ 1&=0-2+C \\ C&=3 \\ \therefore \; \; y&=2x-2+3e^{-x} \\ \text{At }x&= \ln 2 \\ y&=2(\ln 2)-2+3e^{-\ln 2} \\ &=1.386-2+\frac{3}{2}\\
&=0.886\end{aligned}
Question 5 |
The partial differential equation
\frac{\partial^2 u}{\partial t^2}-c^2\left ( \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} \right )=0; where c\neq 0
is known as
\frac{\partial^2 u}{\partial t^2}-c^2\left ( \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} \right )=0; where c\neq 0
is known as
heat equation | |
wave equation | |
Poisson's equation | |
Laplace equation |
There are 5 questions to complete.