Differentiation (Part-38) | S N De

In this part, I have discussed about few math problems which I think is essential for those who have opted for WBCS Math (Optional).

1. If $\,\,V=\log(x^3+y^3+z^3-3xyz),\,\,$then prove that
(a) $\,\, \left(\frac{\partial }{\partial x} +\frac{\partial }{\partial y}+\frac{\partial }{\partial z} \right)V=\frac{3}{x+y+z}$
(b) $\,\,\left(\frac{\partial^2 }{\partial x^2} +\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2} \right)V=\frac{-3}{(x+y+z)^2}$

Solution (a): We have, $\,\,V=\log(x^3+y^3+z^3-3xyz)$

$\therefore \frac{\partial V}{\partial x}=\frac{3x^2-3yz}{(x^3+y^3+z^3-3xyz)} \cdots \cdots (1)$

$\text{Similarly,}\,\, \frac{\partial V}{\partial y}=\frac{3y^2-3zx}{(x^3+y^3+z^3-3xyz)} \cdots \cdots (2)$

$\text{and}\,\, \frac{\partial V}{\partial z}=\frac{3z^2-3xy}{(x^3+y^3+z^3-3xyz)} \cdots \cdots (3)$

$\therefore \text{Adding (1),(2),(3), we get}\,\, \left(\frac{\partial }{\partial x} +\frac{\partial }{\partial y}+\frac{\partial }{\partial z} \right)V = \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}=\frac{3}{x+y+z}$

Solution (b): Now, we have, $\frac{\partial V}{\partial x}=\frac{3x^2-3yz}{(x^3+y^3+z^3-3xyz)}$

$\therefore \frac{\partial^2V}{\partial x^2}=\frac{(x^3+y^3+z^3-3xyz)\cdot 6x-(3x^2-3yz)(3x^2-3yz)}{(x^3+y^3+z^3-3xyz)^2}$

$= 3 \cdot \frac{(2x^4+2xy^3+2xz^3-6x^2yz)-3(x^4-2x^2yz+y^2z^2)}{(x^3+y^3+z^3-3xyz)^2}$

$=\frac{-x^4+2xy^3+2xz^3-3y^2z^2}{(x^3+y^3+z^3-3xyz)^2} \cdots \cdots (4)$

$\frac{\partial^2V}{\partial y^2}=\frac{(x^3+y^3+z^3-3xyz)(6y)-(3y^2-3zx)(3y^2-3zx)}{(x^3+y^3+z^3-3xyz)^2}$

$=3 \cdot \frac{(2x^3y+2y^4+2yz^3-6xy^2z)-3(y^4-2xy^2z+z^2x^2)}{(x^3+y^3+z^3-3xyz)^2}$

$=3 \cdot \frac{-y^4+2x^3y+2yz^3-3z^2x^2}{(x^3+y^3+z^3-3xyz)^2} \cdots \cdots (5)$

Similarly, $\,\, \frac{\partial^2 V}{\partial z^2}=3 \cdot \frac{-z^4+2zx^3+2zy^3-3x^2y^2}{(x^3+y^3+z^3-3xyz)^2} \cdots \cdots (6)$

On adding (4), (5), (6) we get, $\left(\frac{\partial^2 }{\partial x^2} +\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2} \right)V$

$=3 \cdot \frac{-x^4+2xy^3+2xz^3-3y^2z^2-y^4+2x^3y+2yz^3-3z^2x^2-z^4+2zx^3+2zy^3-3x^2y^2}{(x^3+y^3+z^3-3xyz)^2}$

$=-3 \cdot \frac{x^4+y^4+z^4-2xy^3-2xz^3+3y^2z^2-2x^3y-2yz^3+3z^2x^2-2zx^3-2zy^3+3x^2y^2}{(x^3+y^3+z^3-3xyz)^2}$

$= -3 \cdot \frac{(x^2+y^2+z^2-xy-yz-zx)^2}{(x+y+z)^2(x^2+y^2+z^2-xy-yz-zx)^2} = \frac{-3}{(x+y+z)^2}$

2. If $\,\,r^2=x^2+y^2+z^2,\,\, V=r^3,\,\,$, then prove that
(i) $\left(\frac{\partial^2 }{\partial x^2} +\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2} \right)V=12r$
(ii) $\frac{1}{zx} \frac{\partial^2V}{\partial z \partial x}+\frac{1}{yz} \frac{\partial^2V}{\partial y \partial z}+\frac{1}{xy} \frac{\partial^2V}{\partial x \partial y}=\frac{9}{r}$

Solution: Since $\,\,r^2=x^2+y^2+z^2$

$\therefore 2r \cdot \frac{\partial r}{\partial x}=2x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}; \,\, \frac{\partial r}{\partial y}=\frac{y}{r}; \,\, \text{and}\,\, \frac{\partial r}{\partial z}=\frac{z}{r}$

$\therefore V=r^3 \Rightarrow \frac{\partial V}{\partial x}=3r^2 \cdot \frac{\partial r}{\partial x}=3r^2 \cdot \frac{x}{r}=3rx;$

$\therefore \frac{\partial^2V}{\partial x^2}=\frac{\partial}{\partial x}(3rx)=3r+ 3x \cdot \frac{\partial r}{\partial x} =3r+3x \cdot \frac{x}{r} =3 \cdot \frac{r^2+x^2}{r} =\frac{3}{r}(r^2+x^2)\cdots \cdots (1)$

$\therefore \frac{\partial^2V}{\partial y^2}= \frac{3}{r}(r^2+y^2)\cdots \cdots (2), \,\, \text{and} \,\, \frac{\partial^2V}{\partial z^2}=\frac{3}{r}(r^2+z^2)\cdots \cdots (3)$

Adding (1), (2), (3): $\left(\frac{\partial^2 }{\partial x^2} +\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2} \right)V = \frac{3}{r}(3r^2 + x^2+y^2+z^2) = \frac{3}{r}(3r^2 + r^2) = 12r$

Now, $\,\, \frac{\partial^2V}{\partial y \partial z}=\frac{\partial }{\partial y}\left(\frac{\partial V}{\partial z}\right)=\frac{\partial }{\partial y}(3rz)=3 \cdot \frac{\partial r}{\partial y} \cdot z =3 \cdot \frac{y}{r} \cdot z$

$\therefore \frac{1}{yz} \frac{\partial^2V}{\partial y \partial z}=\frac{3}{r} \cdots \cdots (4)$

$\text{Similarly,}\,\, \frac{1}{zx} \frac{\partial^2V}{\partial z \partial x}=\frac{3}{r}\cdots \cdots (5) \,\, \text{and}\,\, \frac{1}{xy} \frac{\partial^2V}{\partial x \partial y}=\frac{3}{r}\cdots \cdots (6)$

So, adding (4), (5), and (6) we get: $\frac{1}{yz}\frac{\partial^2V}{\partial y \partial z}+\frac{1}{zx}\frac{\partial^2V}{\partial z \partial x}+\frac{1}{xy}\frac{\partial^2V}{\partial x \partial y} =\frac{3}{r}+\frac{3}{r}+\frac{3}{r} =\frac{9}{r}$

Read more : TOP BOOKS FOR WBCS MAINS EXAMINATION

3. If $\,\,u=f(r),\,\,\text{where,}\,\, r=\sqrt{x^2+y^2+z^2}, \,\, \text{prove that}\,\,\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=f''(r)+\frac{2}{r}f'(r).$
In particular, prove that if $\,\,f(r)=r^m$, then the left side $=m(m+1)r^{m-2}$

Solution: $\,\,u=f(r) \Rightarrow \frac{\partial u}{\partial x}=f'(r) \cdot \frac{\partial r}{\partial x}=f'(r) \cdot \frac{x}{r}$

$\therefore \frac{\partial^2u}{\partial x^2}=f'(r) \cdot \frac{r \cdot 1 - x \cdot \frac{\partial r}{\partial x}}{r^2}+\frac{x}{r} \cdot f''(r) \cdot \frac{\partial r}{\partial x} = \frac{f'(r)}{r^2}\left(r-x \cdot \frac{x}{r}\right)+\frac{x}{r} \cdot f''(r) \cdot \frac{x}{r}$

$=\frac{f'(r)}{r^3}(r^2-x^2)+\frac{f''(r)}{r^2}x^2$

$\text{Similarly,}\,\, \frac{\partial^2u}{\partial y^2}=\frac{f'(r)}{r^3}(r^2-y^2)+\frac{f''(r)}{r^2}y^2 \,\, \text{and}\,\, \frac{\partial^2u}{\partial z^2}=\frac{f'(r)}{r^3}(r^2-z^2)+\frac{f''(r)}{r^2}z^2$

$\text{Now, adding we get,}\,\,\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2} = \frac{f'(r)}{r^3}[3r^2-(x^2+y^2+z^2)]+\frac{f''(r)}{r^2}(x^2+y^2+z^2)$

$= \frac{f'(r)}{r^3}[3r^2-r^2]+\frac{f''(r)}{r^2} \cdot r^2 = \frac{f'(r)}{r^3} \cdot 2r^2+f''(r) = f''(r)+\frac{2}{r}f'(r)$

Second part: $\,\, f(r)=r^m \Rightarrow f'(r)=mr^{m-1} \,\, \text{and}\,\, f''(r)=m(m-1)r^{m-2}$

$\text{Now, L.H.S.}=f''(r)+\frac{2}{r}f'(r) = m(m-1)r^{m-2}+\frac{2}{r} \cdot mr^{m-1} = m(m-1)r^{m-2}+2mr^{m-2}$

$=r^{m-2}[m(m-1)+2m] = r^{m-2}(m^2-m+2m) = r^{m-2}(m^2+m) = m(m+1)r^{m-2}$

4. If $\,\,z=f(x,y)\,\text{and}\,\, x=r \cos{\theta},\,\,y=r \sin{\theta}, \,\, \text{Prove that,}\,\, \left( \frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2=\left(\frac{\partial z}{\partial r}\right)^2+\frac{1}{r^2} \left(\frac{\partial z}{\partial \theta}\right)^2$

Sol. $\,\, \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial z}{\partial x}\cos{\theta} +\frac{\partial z}{\partial y}\sin{\theta} \cdots \cdots (1)$

$\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}=\frac{\partial z}{\partial x}(-r \sin{\theta})+\frac{\partial z}{\partial y}(r \cos{\theta})$

$\therefore \frac{1}{r}\frac{\partial z}{\partial \theta}=-\frac{\partial z}{\partial x}\sin{\theta}+\frac{\partial z}{\partial y}\cos{\theta} \cdots \cdots (2)$

Squaring (1) and (2), and adding them: $\left(\frac{\partial z}{\partial r}\right)^2+\frac{1}{r^2} \left(\frac{\partial z}{\partial \theta}\right)^2 = \left(\frac{\partial z}{\partial x} \cos{\theta} +\frac{\partial z}{\partial y} \sin{\theta} \right)^2+\left(-\frac{\partial z}{\partial x} \sin{\theta} +\frac{\partial z}{\partial y} \cos{\theta} \right)^2$

$= \left( \frac{\partial z}{\partial x}\right)^2(\cos^2{\theta}+\sin^2{\theta})+\left( \frac{\partial z}{\partial y}\right)^2(\sin^2{\theta}+\cos^2{\theta}) = \left( \frac{\partial z}{\partial x}\right)^2+\left( \frac{\partial z}{\partial y}\right)^2$

5. Solve (By Method of Separation of Variables) $\,\,\frac{\partial u}{\partial x}= 2 \frac{\partial u}{\partial t}+u,\,\,\text{where,}\,\, u(x,0)=6e^{-3x}.$

Solution: Let the solution of the given equation be $u(x,t)= X(x) T(t).$

$\therefore X' T=2XT'+XT \Rightarrow (X'-X)T=2XT' \Rightarrow \frac{X'-X}{X}=2 \frac{T'}{T}=k(\neq 0), \text{say}$

$\frac{X'-X}{X}=k \Rightarrow \frac{X'}{X}=k+1 \Rightarrow \log{X}=(k+1)x+\log C_1 \Rightarrow X=C_1e^{(k+1)x}$

$\frac{T'}{T}=\frac{k}{2} \Rightarrow \log{T}=\frac{k}{2}t+ \log{C_2} \Rightarrow T=C_2e^{kt/2}$

$\therefore u(x,t)=XT = (C_1C_2)e^{(k+1)x+kt/2}$

$\text{When}\,\, t=0, u(x,0)=C_1C_2e^{(k+1)x}=6e^{-3x} \Rightarrow C_1C_2=6,\,\, k+1=-3 \Rightarrow k=-4$

$\therefore u(x,t)=6e^{-3x-4t/2}=6e^{-3x-2t}=6e^{-(3x+2t)}$

6. If $\,\, x^n-1=(x^2-1) \prod_{k=1}^{(n-2)/2} \left(x^2-2x\cos{\frac{2k\pi}{n}}+1 \right),$ if $n$ be an even positive integer. Also, Deduce that $\,\,\sin{\frac{\pi}{32}}\sin{\frac{2\pi}{32}}\sin{\frac{3\pi}{32}}\cdots \sin{\frac{15\pi}{32}}=\frac{1}{2^{13}}$

Sol. $\,\, x^n-1=0 \Rightarrow x^n=1=\cos{2k\pi}+i\sin{2k \pi}$

$\Rightarrow x= \cos{\frac{2k \pi}{n}}+i \sin{\frac{2k \pi}{n}}, \,\, \text{where}\,\, k=0,1,2,\dots,(n-1)$

When $n$ is even, roots are $\pm 1$ and $\cos{\frac{2r \pi}{n}} \pm i \sin{\frac{2r\pi}{n}},$ yielding the product form $(x^2-1) \prod_{k=1}^{(n-2)/2} \left(x^2-2x\cos{\frac{2k\pi}{n}}+1 \right).$

2nd part: With $\,\,n=32,\,\,$ $\frac{x^{32}-1}{x^2-1}=\prod_{k=1}^{15}(x^2+1-2x \cos{\frac{k\pi}{16}})$

Taking limit as $x \to 1$: $\lim_{x \to 1} \frac{x^{32}-1}{x^2-1}= \lim_{x \to 1} \prod_{k=1}^{15}(2-2\cos{\frac{k\pi}{16}}) \Rightarrow 16 = \prod_{k=1}^{15}\left(4\sin^2{\frac{k\pi}{32}}\right) \Rightarrow \frac{1}{2^{13}}=\prod_{k=1}^{15}\sin{\frac{k\pi}{32}}$

7. Using $\,\,x^n-1=(x-1)\prod_{k=1}^{(n-1)/2} \{x^2-2x\cos{\frac{2k\pi}{n}}+1\}\,\,$ if $n$ be odd positive integer, deduce that $\,\,\sin{\frac{\pi}{25}}\sin{\frac{2\pi}{25}}\sin{\frac{3\pi}{25}}\dots\sin{\frac{12\pi}{25}}=\frac{5}{2^{12}}$

Solution. Put $\,\,n=25\,\,$. Then $\,\frac{x^{25}-1}{x-1}=\prod_{k=1}^{12} \{x^2-2x \cos{\frac{2k\pi}{25}}+1\}$

Taking limits as $x \to 1$ and applying L'Hospital's Rule: $\frac{25}{1} = \prod_{k=1}^{12} \left(2-2\cos{\frac{2k\pi}{25}}\right) = \prod_{k=1}^{12} 4\sin^2{\frac{k\pi}{25}}$

$\Rightarrow 25 = 4^{12} \left(\prod_{k=1}^{12}\sin{\frac{k\pi}{25}}\right)^2 \Rightarrow \frac{5}{2^{12}}=\prod_{k=1}^{12} \sin{\frac{k\pi}{25}}$

8. Show that the solutions of the equation $\,\,(1+x)^{2n+1}-(1-x)^{2n+1}=0\,\,$ are $\,\,x=0, \pm i \tan{\frac{r \pi}{2n+1}},r=1,2,3,\dots,n$

Sol. $\,\, (1+x)^{2n+1}=(1-x)^{2n+1} \Rightarrow \left( \frac{1+x}{1-x} \right)^{2n+1}=\cos{2r\pi}+i\sin{2r \pi}$

$\Rightarrow \frac{1+x}{1-x}=\cos{\frac{2r\pi}{2n+1}}+i \sin{\frac{2r\pi}{2n+1}},\,\, r=1,2,\dots,n$

Applying Componendo-dividendo: $\frac{2}{2x} = \frac{1 + \frac{1+x}{1-x}}{1 - \frac{1+x}{1-x}} \Rightarrow x = i\tan{\frac{r\pi}{2n+1}}$

LAGRANGE'S METHOD OF UNDETERMINED MULTIPLIERS :

Procedure: Let it be required to find the stationary values of a function of 3 variables, say $\,\,u=f(x,y,z)\,$ subject to the conditions $\,\,\phi(x,y,z)=0\,$ and $\,\,\psi(x,y,z)=0.$

Construct the function: $\,F=f+\lambda_1 \phi +\lambda_2 \psi \,$ where $\lambda_1, \lambda_2$ are non-zero multipliers.

Solve the equations $\frac{\partial F}{\partial x}=0, \frac{\partial F}{\partial y}=0, \frac{\partial F}{\partial z}=0$ alongside the given conditions to determine stationary points.

9. Show that the stationary value of $\,\,a^3x^2+b^3y^2+c^3z^2\,\,\text{where}\, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\,$is given by $\,\,x=\frac{a+b+c}{a}, y=\frac{a+b+c}{b}, z=\frac{a+b+c}{c}$

Solution. Let $\, u = a^3x^2+b^3y^2+c^3z^2$ and $\phi(x,y,z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1 =0$

Construct $\,F = f+\lambda \phi = (a^3x^2+b^3y^2+c^3z^2)+\lambda \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)$

$\frac{\partial F}{\partial x}=0 \Rightarrow 2a^3x-\frac{\lambda}{x^2}=0 \Rightarrow 2a^3x^3=\lambda \cdots (3)$

Similarly, $2b^3y^3=\lambda$ and $2c^3z^3=\lambda$, giving $ax=by=cz=t \implies x=\frac{t}{a}, y=\frac{t}{b}, z=\frac{t}{c}$

Substituting into the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, we get $t=a+b+c$, hence $x=\frac{a+b+c}{a}, y=\frac{a+b+c}{b}, z=\frac{a+b+c}{c}$

10. Find the stationary values of $\,x^2+y^2+z^2\,\,$ subject to the conditions $\,\,ax^2+by^2+cz^2+2fyz+2gzx+2hxy=1\,\,\text{and}\,\, lx+my+nz=0$

Sol. Let $\,\,u=x^2+y^2+z^2$, $\phi = ax^2+by^2+cz^2+2fyz+2gzx+2hxy-1=0$, and $\psi= lx+my+nz=0$

Construct $F = u + \lambda_1\phi + \lambda_2\psi$

Partial derivatives with respect to $x, y, z$ equated to zero yield equations leading to $\lambda_2 = -u$

Eliminating $x, y, z, \lambda_2$ using Cramer's rule gives the determinant condition:
$\begin{vmatrix} 1-au & -hu & -gu & l \\ -hu & 1-bu & -fu & m \\ -gu & -fu & 1-cu & n \\ l & m & n & 0 \end{vmatrix}=0$

In the next article, we will discuss few more mathematical problems and their solutions, which may be beneficial for students who are preparing for WBCS math optional papers.