LinkedIn Insight UD-TRG-Maths2-Mumbai-Winter16-Q6-Ans - Grad Plus

UD-TRG-Maths2-Mumbai-Winter16-Q6-Ans

A poll of students appearing for masters in engineering indicated that 60 % of the students believed that mechanical engineering is a profession unsuitable for women. A research study on women with masters or higher degrees in mechanical engineering found that 99 % of such women were successful in their professions.
Which of the following can be logically inferred from the above paragraph?

(A) Many students have misconceptions regarding various engineering disciplines.
(B) Men with advanced degrees in mechanical engineering believe women are well suited to be mechanical engineers.
(C) Mechanical engineering is a profession well suited for women with masters or higher degrees in mechanical engineering.
(D) The number of women pursuing higher degrees in mechanical engineering is small.

Qu. 6 (A) A condenser of capacitance C is charged through a resistance R by a steady voltage. The charge Q satisfies the DE R\frac{dQ}{dt}+\frac Qc=V.\; If the plate is chargeless find the charge and the current at time ‘t’.

Sol. :- R\frac{dQ}{dt}+\frac Qc=V.\;

\therefore\frac{dQ}{dt}+\frac1{Rc}Q=\frac VR,\;Which\;is\;of\;the\;type\;\frac{dQ}{dt}+AQ=B.\;

Here,

A=\frac1{RC}\;\&\;B=\frac VR\;which\;are\;cons\tan ts.

i.e. fn of only ‘t’.

so given D. E. is linear.

Now,

consider \int A\;dt=\int\frac1{RC}dt\;=\frac1{RC}t

Therefore Integrating factor \left(I.F.\right)=\int_eA\;dt\;=e^{t/RC}

Therefore General Sol. is Q\times I.F.=\int B\times I.F.\;dt+k

\therefore Q\times e^{t/RC}=\int\frac VR\times e^{t/RC}\;\;dt\;+\;K \therefore Q\times e^{t/RC}=\frac VR\int e^{t/RC}\;\;dt\;+\;K

(B) Evaluate \int\int\frac{\left(x^2+y^2\right)^2}{x^2y^2}\;dx\;dy

over the region common to x^2+y^2-ax=0\;\;and\;\;x^2+y^2-by\;=\;0,\;\;a>0,\;\;b>0.

Sol. :- x^2+y^2-ax=0\;\;…..(1)\;is\;a\;circle.

comparing with general form of circle

\begin{array}{l}x^2+y^2+2gx+2fy+c=0\\We\;get,\\2g\;=\;-a,\;2f\;=\;0,\;c\;=\;0\end{array} \therefore g=\frac{-a}2,\;f=0,\;\&\;c=\;0

Therefore centre = \left(-g,\;-f\right)=\left(\frac a2,\;0\right)\;and

\begin{array}{l}Radius\;=\sqrt{g^2+f^2-c}\\=\sqrt{\left(\frac a2\right)^2}+\left(0\right)^2-0\\Radius\;=\;\frac a2\end{array}

ii^{iy},\;\;x^2+y^2-by=0…….(2)\; is a circle with centre = \left(0,\frac b2\right)\;\&\;R\;=\;\frac b2

I=\frac{b^2}2\int_{\theta=0}^\alpha\;sec^2\theta d\theta\;+\;\frac{a^2}2\int_{\theta=\alpha}^{\mathrm\pi/2}\cos ec^2\;\theta\;d\theta I=\frac{b^2}2\left[\tan\;\theta\right]<em>0^\alpha\;+\frac{a^2}2\left[-cot\;\theta\right]</em>\alpha^{\mathrm\pi/2} I=\frac{b^2}2\left[\tan\;\alpha-\tan\;\alpha\right]-\frac{a^2}2\left[cot\;\frac{\mathrm\pi}2-cot\;\alpha\right] I=\frac{b^2}2\left[\frac ab-0\right]\frac{-a^2}2\left[a-\frac ba\right]\;\;\left(from\;(5)\right) \begin{array}{l}=\frac{ab^2}{2b}-\frac{a^2}2\left(\frac{-b}a\right)\\=\frac{ab}2+\frac{ab}2\end{array} \begin{array}{l}I=ab\\\therefore\int\int\frac{\left(x^2+y^2\right)^2}{x^2y^2}\;dxdy\;=ab\end{array}

(C) Find the Volume Common to right circular cylinder x^2+y^2=a^2\;\;\&\;\;x^2+z^2=a^2.

Sol. :- from equ of cylinder \;\;x^2+z^2=a^2, limits of ‘z’ are

z\;=0\;to\;z\;=\;\pm\sqrt{a^2-x^2}

Volume common to bot the cylinder (V)

= 8 x volume in 1st octaut

=8\int\int\int_0^\sqrt{a^2-x^2}dz\;dy\;dx =8\int\int\left[z\right]_0^\sqrt{a^2-x^2}\;dy\;dx =8\int\int\left[\sqrt{a^2-x^2}-0\right]\;dy\;dx

Now, cylinder x^2+y^2\;=\;a^2 cuts the plane z = 0 in a circle with centre (o,o) & radius (r) = a limits of ‘y’ are y = o to y=\sqrt{a^2\;x^2} & limits of ‘x’ are

x = o to x = a

\therefore v\;=8\int_0^a\int_0^\sqrt{a^2-x^2}\sqrt{a^2-x^2}dy\;dx \therefore Qe^{t/rc}=\frac vR\times\frac{e^{t/RC}}{1/RC}+k</p> <p>[latex] \therefore Q=\frac vR\times\frac{RC}1+\;ke^{-t/RC}\;……\;(1)

(Dividing by e^\frac t{RC})

Given, the plate is charge -less i.e. at t = 0, Q = 0

\begin{array}{l}\therefore0\;=\;vc\;+\;ke^0\\K\;=\;-vc……..(2)\end{array}

from (1) and (2)

Q=\;vc\;-\;vce^\frac{-t}{RC}

Differntiating w.r.f. we get,

\frac{dQ}{dt}=0-vc\;\times\;e^\frac t{RC}\times\frac{-1}{Rc} i=\frac VRe^\frac{-t}{RC}\;\left{\therefore t=\frac{dQ}{dt}\right}

Hence, change at time \begin{array}{l}t\;(Q)\;=\;VC\left(1-e^\frac{-t}{RC}\right);\\current\;at\;time\;t\;(i)\;=\frac VRe^\frac{-t}{RC}\end{array}

\therefore V=8\int_0^a\sqrt{a^2-x^2}\left[y\right]_0^\sqrt{a^2-x^2}\;dx =8\int_0^a\sqrt{a^2-x^2\;}\left[\sqrt{a^2-x^2}-0\right]\;dx =8\int_0^a\left(a^2-x^2\right)\;dx =8\left[a^2x-\frac{x^3}3\right]_0^a =8\left[\left{a^3-\frac{a^3}3\right}\right]-\left[\left{0-0\right}\right] =8\left(\frac{3a^3-a^3}3\right)-0 \begin{array}{l}=8\times\frac{2a^3}3\\V=\frac{16a^3}3\end{array}

Therefore volume common to both the cylinders latex=\frac{16a^3}3[/latex]

\tan\;\alpha\;=\frac ab……..

let, I=\int\int\frac{\left(x^2+y^2\right)^2}{x^2\;y^2}\;dx\;dy

=\int\int\frac{\left(r^2\right)^2}{r^2\;\cos^2\;\theta\;\times\;r^2\;\sin^2\;\theta}r\;dr\;d\theta I=\int_{\theta=0}^\alpha\int_{r=0}^{b\sin\;\theta}\frac1{\cos\;^2\;\theta.\;\sin^2\;\theta}\times r\;dr\;d\theta\;+\int_{\theta=\alpha}^{\mathrm\pi/2}\int_{r=0}^{a\cos\;\theta}\frac1{\cos^2\theta\times\sin^2\theta}\times r\;dr\;d\theta =\int_{\theta=0}^\alpha\;\frac1{\cos\;^2\;\theta.\;\sin^2\;\theta}\left[\frac{r^2}2\right]<em>0^{b\sin\;\theta}\;d\theta\;+\int</em>{\theta=\alpha}^{\mathrm\pi/2}\frac1{\cos^2\theta.\;\sin^2\theta}\left[\frac{r^2}2\right]_0^{a\;\cos\;\theta}d\theta

=\int_{\theta=0}^\alpha\;\frac1{\cos\;^2\;\theta.\;\sin^2\;\theta}\times\frac12\left[b^2\;\sin^2\theta-0\right]d\theta +

=\int_{\theta=\alpha}^{\mathrm\pi/2}\;\frac1{\cos\;^2\;\theta.\;\sin^2\;\theta}\times\frac12\left[a^2\;\cos^2\theta-0\right]d\theta K_3=hf\left(to+\frac42,u0+\frac{k_2}2\right)</p> <p>[latex] \begin{array}{l}=0.2\times f\left(0+\frac{0.2}2,\;1-\frac{0.04}2\right)\\=0.2\times f\;\left(0.1,\;0.98\right)\\=0.2\;\times\;-2\;\left(0.1\right)\;\left(0.98\right)^2\end{array} K_3=-0.0384 \begin{array}{l}K_4=hf\;\left(t0+h,\;u0+k_3\right)\\=0.2\;\times\;f\;\left(0+0.2,\;1-0.0384\right)\\=0.2\;\times\;f\;\left(0.2\;,\;0.9616\right)\end{array} \begin{array}{l}=0.2\;\times\;-2\;\left(0.2)\;(\;0.9616\right)^2\\K_4=-0.0740\end{array} \begin{array}{l}K=\frac16\left(k_1+2k_2+2k_3+k_4\right)\\=\frac16\left(0-2\times0.04-2\times0.0384-0.0740\right)\end{array}

K = -0.0385

\begin{array}{l}u_1=u_0+k\\=1-0.0385\\u_1=0.9615\end{array} \therefore\;Exact\;value\;=\int_0^1\frac{x^2}{1+x^3}dx \frac13\int_0^1\frac{3x^2}{1+x^3}dx

Integrating w.r.f. x

\frac13\left[\log\;\left(1+x^3\right)\right]_0^1 \left{\therefore\int\frac{f^1\left(x\right)}{f\left(x\right)}dx=\log\;f\left(x\right)\right} \begin{array}{l}=\frac13\left[\log\;\left(1+1^3\right)-\log\;\left(1+1^3\right)\right]\\=\frac13\left[\log\;2-\log\;1\right]\end{array} \begin{array}{l}=\frac13\;\log\;2\\=0.23104906\end{array}

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Therefore using polar Co-ordinates

\begin{array}{l}put\;X=r\;\cos\theta,\;\;Y=r\;\sin\theta\\dx\;dy\;=rdr\;d\theta\;……(3)\\\end{array} \begin{array}{l}\therefore x^2+y^2=r^2\;\cos^2\theta\;+\;r^2\sin\theta\\=r^2\left(\cos^2\theta\;+\;\sin^2\theta\right)\\=r^2\times1\\\end{array} \begin{array}{l}\therefore x^2+y^2=r^2\;\;………\;(4)\\\end{array}

substituting (3) & (4) in (2)

We get,

\begin{array}{l}r^2-b\;\sin\theta\;=\;0\\\therefore r\left(r-b\sin\;\theta\right)=0\\\end{array} \begin{array}{l}\therefore r\;=\;0\;or\;r-b\sin\;\theta=0\\r=b\sin\;\theta\\\end{array}

At the pt. of intersection

\begin{array}{l}r=a\;\cos\;\theta\;=\;b\sin\;\theta\\\therefore\frac ab=\frac{\sin\;\theta}{\cos\;\theta}\\\therefore\frac ab=\;\tan\;\theta\\\end{array} \begin{array}{l}\therefore\;\theta=\tan\;^{-1}\left(\frac ab\right)\\let,\alpha=\tan^{-1}\left(\frac ab\right)\\\end{array}

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(1) Trapezoidal Rule :-

\begin{array}{l}\int_a^bf\left(x\right)dx=\frac h2\left[\left(y_0+y_5\right)+2\left(y_1+y_2+y_3+y_4\right)\right]\\\end{array} \begin{array}{l}\therefore\int_0^1\frac{x^2}{1+x^3}dx=\frac{0.2}2\left[\left(0+\frac12\right)+2\left(\frac5{126}+\frac{20}{133}+\frac{45}{152}+\frac{80}{189}\right)\right]\\\end{array}

= 0.231878 …..

Therefore Error as compared to exact value

\begin{array}{l}=\left|Exact\;value\;-Calculated\;value\right|\\\end{array} \begin{array}{l}=\left|0.231049-0.231878\right|\;(from\;(1)\;\&\;(2))\\=0.000829\\\end{array}

(2) Simpson's \begin{array}{l}\left(\frac{1\partial d}3\right)\\\end{array} Rule :-

For Simpson's \begin{array}{l}\left(\frac13\right)\\\end{array} rule 'n' should be even Here, n = s

So, We cannot apply Simpson's \begin{array}{l}\left(\frac13\right)\\\end{array} rule

Hence,

Trapezoidal Rule

Simpson's Rule

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Now, put V = 0.1

\begin{array}{l}\therefore Y=1+\left(0.1\right)+\frac52\left(0.1\right)^2+2\left(0.1\right)^3+\frac94\left(0.1\right)^4+\frac{59}{90}\left(0.1\right)^5+…\\\end{array} \begin{array}{l}\therefore Y=1+0.1+0.025+0.002+0.000225+0.000225+..\\\end{array} \begin{array}{l}Y=\;1.127225\\Y=1.1273\\\end{array}

Hence, the value of y for X = 0.1 is 1.1273

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