Summer 2017 - Q.5 - Grad Plus

# Summer 2017 – Q.5

Q.5 a) If $latex z=x^2\tan^{-1}\left(\frac yx\right)+y^2\tan^{-1}\left(\frac xy\right)$ Prove that $latex \frac{\partial^2z}{\partial y}=\frac{x^2-y^2}{x^2+y^2}$ [6M]

$latex z=x^2\tan^{-1}\left(\frac yx\right)+y^2\tan^{-1}\left(\frac xy\right)$

T.P.T

$latex \frac{\partial^2z}{\partial y\partial x}=\frac{x^2-y^2}{x^2+y^2}$

$latex z=x^2\tan^{-1}\left(\frac yx\right)+y^2\tan^{-1}\left(\frac xy\right)$

$latex \begin{array}{l}\frac{\partial z}{\partial x}=2x\tan^{-1}\left(\frac yx\right)+x^2\cdot\frac1{1+{\displaystyle\frac{y^2}{x^2}}}\left(\frac{-y}{x^2}\right)-y^2\cdot\frac1{1+{\displaystyle\frac{x^2}{y^2}}}\left(\frac1y\right)\\\\\;\;\;\;\;\;=2x\tan^{-1}\frac{\displaystyle y}{\displaystyle x}-\frac{\displaystyle x^2y}{x^2+y^2}-\frac{y^3}{x^2+y^2}\\\\\;\;\;\;\;\;=2x\tan^{-1}\frac{\displaystyle y}{\displaystyle x}-\frac{y\left(x^2+y^2\right)}{x^2+y^2}\\\\\;\;\;\;\;\;=2x\tan^{-1}\frac{\displaystyle y}{\displaystyle x}-y\end{array}$

$latex \begin{array}{l}\frac{\partial^2z}{\partial y\partial z}=2x\frac1{1+{\displaystyle\frac{y^2}{x^2}}}\times\frac1x-1\\\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{2x^2}{x^2+y^2}-1\\\\\frac{\partial^2z}{\partial y\partial z}=\frac{x^2-y^2}{x^2+y^2}\end{array}$

Hence proved.

b) Investigate for what values of λ and μ the equations,

$latex \begin{array}{l}2x+3y+5z=9\\\\7x+3y-2z=8\\\\2x+3y+\lambda z=\mu\end{array}$

Have 1) no solution
2) a unique solution
3) an infinite no. of solutions [6M]

$latex \begin{array}{l}2x+3y+5z=9\\\\7x+3y-2z=8\\\\2x+3y+\lambda z=\mu\end{array}$

Let us write the above equation in matrix form.

$latex \begin{bmatrix}2&3&5\\7&3&-2\\2&3&\lambda\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\8\\\mu\end{bmatrix}$

AX=B

c = [A : B]

$latex \begin{array}{l}=\left[\begin{array}{ccc}2&3&5\\7&3&-2\\2&3&\lambda\end{array}\begin{array}{c}:\\:\\:\end{array}\begin{array}{c}9\\8\\\mu\end{array}\right]\\\\=\left[\begin{array}{ccc}2&3&5\\0&\frac{-15}2&\frac{-39}2\\0&0&\lambda-5\end{array}\begin{array}{c}:\\:\\:\end{array}\begin{array}{c}9\\\frac{-47}2\\\mu-9\end{array}\right]\end{array}$

$latex \begin{array}{l}R_2\rightarrow R_2-\frac72R_1\\\\R_3\rightarrow R_3-R_1\end{array}$

1. No solution

Rank (A) ≠ Rank (C)

$latex \begin{array}{l}\lambda-5=0\;or\;\lambda=5\\\\\mu-9\neq0\;or\;\mu\neq9\end{array}$

2. Unique solution

Rank (A) = Rank (C)

=No. of unknowns

$latex \lambda-5\neq0,\;\lambda\neq5$

3. An infinte no. of solutions

Rank (A) = Rank (C) = 2

$latex \begin{array}{l}\lambda-5=0,\;\mu-9=0\\\\\lambda=5,\;\mu=9\end{array}$

c) Using Newton Raphson method, find approximate root of $latex x^3-2x-5=0$ (correct to three places of decimals.) [8M]

$latex \begin{array}{l}f\left(x\right)=x^3-2x-5\\\\f\left(2\right)=8-4-5\\\\\;\;\;\;\;\;=-1\end{array}$

$latex \begin{array}{l}f\left(x\right)=x^3-2x-5\\\\f\left(2.5\right)=\left(2.5\right)^3-4\left(2.5\right)-5\\\\\;\;\;\;\;\;=5.625\end{array}$

we get to know that f(2) and f(2.5) are of opposite signs

So, the root of the equation lies between 2 and 2.5

f(2) is nearer to zero, hence it is more approximate root than f(2.5)

$latex \begin{array}{l}f(x)=3x^2-2\\\\f'(2)=12-2\\\\\;\;\;\;\;\;\;\;=10\end{array}$

Let 2 be an approximate root of (i) by N.R. method.

$latex \begin{array}{l}a_1=a-\frac{f\left(a\right)}{f’\left(a\right)}\\\\\;\;\;\;=2-\frac{f\left(2\right)}{f’\left(2\right)}\\\\\;\;\;\;=2-\frac{\left(-1\right)}{10}\\\\\;\;\;\;=2.1\end{array}$

$latex \begin{array}{l}f\left(2.1\right)=2.1^3-2\times2.1-5\\\\\;\;\;\;\;\;\;\;\;\;=9.261-4.2-5\\\\\;\;\;\;\;\;\;\;\;\;=0.061\end{array}$

$latex \begin{array}{l}f’\left(2.1\right)=3\left(2.1\right)^2-2\\\\\;\;\;\;\;\;\;\;\;\;=11.23\end{array}$

$latex \begin{array}{l}a_2=2.1-\frac{f\left(2.1\right)}{f’\left(2.1\right)}\\\\\;\;\;\;=2.1-\frac{0.061}{11.23}\\\\\;\;\;\;=2.1-0.00543\\\\\;\;\;\;=2.09457\end{array}$

$latex \begin{array}{l}f\left(2.09457\right)=\left(2.09457\right)^3-2\left(2.09457\right)-5\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=8.80558-4.18914-5\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=-0.38356\end{array}$

$latex \begin{array}{l}f’\left(2.09457\right)=3\left(2.09457\right)^2-2\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=13.16167-2\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=11.16167\end{array}$

$latex \begin{array}{l}a_3=2.09457-\frac{f\left(2.09457\right)}{f’\left(2.09457\right)}\\\\\;\;\;\;=2.09457-\frac{-0.38356}{11.16167}\\\\\;\;\;\;=2.09457+0.034364\\\\\;\;\;\;=2.128934\end{array}$

$latex \begin{array}{l}f\left(2.128934\right)=\left(2.128934\right)^3-2\left(2.128934\right)-5\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=9.649095-4.257868-5\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=0.391227\end{array}$

$latex \begin{array}{l}f’\left(2.128934\right)=3\left(2.128934\right)^2-2\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=13.59708-2\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=11.59708\end{array}$

$latex \begin{array}{l}a_4=2.128934=\frac{f\left(2.128934\right)}{f’\left(2.128934\right)}\\\;\;\;\;=2.128934-\frac{0.391227}{11.59708}\\\\\;\;\;\;=2.128934-0.03373\\\\\;\;\;\;=2.09461\end{array}$

This is the required root of the fall equation correct to 3 decimal palces.

Q.6 a) Find tanhx if 5 sinhx – coshx=5 [6M]

Given,

$latex 5\;\sin hx-\cos hx=5$

Divide by coshx

$latex 5\;\tan hx-1=5sechx$

Squaring both sides

$latex \begin{array}{l}25\;\tan h^2x+1-10\tan hx=25sech^2x\\\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=25\left(1-\tan h^2x\right)\\\\50\tan h^2x-24-10\tan hx=0\\\\50\tan h^2x-40\tan hx+30\tan hx-24=0\\\\10\;\tan hx\left(5\;\tan hx-4\right)+6\left(5\tan hx-4\right)=0\\\\\tan hx=\frac{-6}{10}=\frac{-3}5\\\\\tan hx=\frac45\end{array}$

b) If $latex u=\sin^{-1}\left(\frac{x+y}{\sqrt x+\sqrt y}\right)$ prove that [6M]

i. $latex xu_x+yu_y=\frac12\tan u$

ii. $latex x^2u_{xx}+2xy\;u_{xy}+y^2u_{yy}=\frac{-\sin u\;\cos2u}{4\cos^3u}$

We have,

$latex u=\sin^{-1}\frac{x+y}{\sqrt x+\sqrt y}$

$latex \begin{array}{l}z=\sin u\\\\\;\;\;=\frac{x+y}{\sqrt x+\sqrt y}\\\\\;\;\;=\frac{x\left[1+{\displaystyle\frac yx}\right]}{\sqrt x\left[1+\sqrt{\displaystyle\frac yx}\right]}\\\\\;\;\;=\sqrt x\cdot\phi\left(x\right)\\\\z=f\left(u\right)=\sin u\end{array}$

z is a homogeneous function of degree $latex \frac12$

By Euler deduction I

$latex \begin{array}{l}x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=n\frac{f\left(u\right)}{f’\left(u\right)}\\\\x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\frac12\frac{\sin u}{\cos u}\\\\x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\frac12\tan u\\\\xu_x+yu_y=\frac12\tan u\end{array}$

Let, $latex g\left(u\right)=\frac12\tan u$

By Euler deduction II

$latex \begin{array}{l}x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=n\frac{f\left(u\right)}{f’\left(u\right)}\\\\x^2\frac{\partial^2u}{\partial x^2}+\partial xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=g\left(u\right)\left[g^1\left(u\right)-1\right]\\\\=\frac12\tan u\left(\frac12sec^2u-1\right)\\\\=\frac14\frac{\sin u}{\cos u}\left(\frac1{\cos^2u-2}\right)\\\\=\frac14\frac{\sin u}{\cos u}\left(1-2\cos^2u\right)\\\\=\frac{-\sin u\cos2u}{4\cos^3u}\end{array}$

We get,

$latex x^2u_{xx}+2xy\;u_{xy}+y^2u_{yy}=\frac{-\sin u\;\cos2u}{4\cos^3u}$

Hence proved.

c) Solve the following systems of equations by Gauss-seidel method.

$latex \begin{array}{l}20x+y-2z=17\\\\3x+20y-z=-18\\\\2x-3y+20z=25\end{array}$ [8M]

By Gauss-Seidel method

$latex \begin{array}{l}20x+y-2z=17\\\\3x+20y-z=-18\\\\2x-3y+20z=25\end{array}$

Here,

$latex \begin{array}{l}x=\frac1{20}\left(17-y+2z\right)—–\left(i\right)\\\\y=\frac1{20}\left(-18-3x+z\right)—–\left(ii\right)\\\\z=\frac1{20}\left(25-2x+3y\right)—–\left(iii\right)\end{array}$

Let  us approximate $latex x_0=y_0=z_0=0$.

Put $latex y_0=z_0=0$ in equation (i) we get,

$latex \begin{array}{l}x_1=\frac{17}{20}\\\\\;\;\;\;=0.85\end{array}$

put $latex x=x_1\;and\;z=0$ in equation (ii) we get,

$latex \begin{array}{l}y_1=\frac1{20}\left(-18-3\times0.85\right)\\\\\;\;\;\;=-1.0275\end{array}$

Put $latex x=x_1\;and\;y=y_1$ in equation (iii)

$latex \begin{array}{l}z_1=\frac1{20}\left(25-2\times0.85+3\times\left(-1.0275\right)\right)\\\\\;\;\;\;=1.011\end{array}$

By similar method, we get

$latex \begin{array}{l}x_2=1.002\;;\;y_2=-0.9998\;;\;z_2=0.9998\\\\x_3=1.0000\;;\;y_3=-1.0000\;;\;z_3=1.0000\\\\x_4=1.0000\;;\;y_4=-1.0000\;;\;z_4=1.0000\end{array}$

The last 2 sets of roots are almost the same so we get the roots of the given equation by Gauss siedel method as

$latex \begin{array}{l}x=1\\\\y=-1\\\\z=1\end{array}$

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