Question

$\rm Find \: the \: value \: of \: \dfrac{\partial^2r}{\partial x^2} + \dfrac{\partial^2r}{\partial y^2} \\ \rm when \: r = {ax}^{2} + 2hxy + {by}^{2}$​

Answer 1

Solution:

Given That:

$\rm\longrightarrow r =ax^{2}+2hxy+by^{2}$

Partial differentiating both sides with respect to x, we get:

$\rm\longrightarrow \dfrac{\partial r}{\partial x} =\dfrac{\partial}{\partial x}(ax^{2}+2hxy+by^{2})$

$\rm\longrightarrow \dfrac{\partial r}{\partial x} =2ax + 2hy + 0$

$\rm\longrightarrow \dfrac{\partial r}{\partial x} =2ax + 2hy$

Again partial differentiating both sides with respect to x, we get:

$\rm\longrightarrow \dfrac{\partial^{2} r}{\partial x^{2}} =2a+0$

$\rm\longrightarrow \dfrac{\partial^{2} r}{\partial x^{2}} =2a-(i)$

Now, partial differentiating both sides with respect to y, we get:

$\rm\longrightarrow \dfrac{\partial r}{\partial y} =\dfrac{\partial}{\partial y}(ax^{2}+2hxy+by^{2})$

$\rm\longrightarrow \dfrac{\partial r}{\partial y} =0+2hx+2by$

$\rm\longrightarrow \dfrac{\partial r}{\partial y} =2hx+2by$

Partial differentiating both sides with respect to y, we get:

$\rm\longrightarrow \dfrac{\partial^{2} r}{\partial y^{2}} =0+2b$

$\rm\longrightarrow \dfrac{\partial^{2} r}{\partial y^{2}} =2b-(ii)$

Adding (i) and (ii), we get:

$\rm\longrightarrow \dfrac{\partial^{2} r}{\partial x^{2}} +\dfrac{\partial^{2} r}{\partial y^{2}} =2a+2b$

$\rm\longrightarrow \dfrac{\partial^{2} r}{\partial x^{2}} +\dfrac{\partial^{2} r}{\partial y^{2}} =2(a+b)$

Which is our required answer.

Answer:

$\rm\hookrightarrow \dfrac{\partial^{2} r}{\partial x^{2}} +\dfrac{\partial^{2} r}{\partial y^{2}} =2(a+b)$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: r = {ax}^{2} + 2hxy + {by}^{2}$

On differentiating partially, w. r. t. x, we get

$\rm \: \frac{\partial }{\partial x}r = \frac{\partial }{\partial x}({ax}^{2} + 2hxy + {by}^{2})$

$\rm \: \frac{\partial r}{\partial x} = 2ax + 2hy$

On differentiating partially w. r. t. x, we get

$\rm \: \frac{\partial^{2} r}{\partial {x}^{2} } = 2a - - - (1)$

Again,

$\rm \: r = {ax}^{2} + 2hxy + {by}^{2}$

On differentiating partially w. r. t. y, we get

$\rm \: \frac{\partial }{\partial y}r = \frac{\partial }{\partial y}({ax}^{2} + 2hxy + {by}^{2})$

$\rm \: \frac{\partial r}{\partial y} = 2hx + 2by$

On differentiating partially w. r. t. y, we get

$\rm \: \frac{\partial^{2} r}{\partial{y}^{2}} = 2b - - - (2)$

So, on adding equation (1) and (2), we get

$\rm \: \dfrac{\partial^2r}{\partial x^2} + \dfrac{\partial^2r}{\partial y^2} = 2a + 2b$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{\partial^2r}{\partial x^2} + \dfrac{\partial^2r}{\partial y^2} = 2(a + b )\: \: }}$

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$ADDITIONAL \: INFORMATION$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$