Question

Solve Q 48

x = a sint - b cost

y = a cost + b sint

Prove that d^2y/dx^2 = -(x^2+y^2)/y^3

Answer 1

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

Given that

$\rm \: x = a \: sint - b \: cost - - - (1)$

and

$\rm \: y = a \: cost + b \: sint - - - (2)$

Now, Consider

$\rm \: {x}^{2} + {y}^{2}$

$\rm \: = \: (asint - bcost)^{2} + (acost + bsint)^{2}$

$\rm \: = {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t - 2absint \: cost \: + {a}^{2} {cos}^{2}t + {b}^{2} {sin}^{2}t + 2absint \: cost$

$\rm \: = {a}^{2} {sin}^{2}t + {b}^{2} {cos}^{2}t + {a}^{2} {cos}^{2}t + {b}^{2} {sin}^{2}t$

$\rm \: = {a}^{2} ({sin}^{2}t + {cos}^{2}t) + {b}^{2} ({cos}^{2}t + {sin}^{2}t)$

$\rm \: = \: {a}^{2} + {b}^{2}$

So, we get

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

On differentiating both sides w. r. t. x, we get

$\rm \: \: \dfrac{d}{dx}({x}^{2} + {y}^{2}) = \: \dfrac{d}{dx}( {a}^{2} + {b}^{2} )$

$\rm \: 2x + 2y\dfrac{dy}{dx} = 0$

$\rm \: x + y\dfrac{dy}{dx} = 0$

$\rm \: y\dfrac{dy}{dx} = - x$

$\rm \: \dfrac{dy}{dx} = - \dfrac{x}{y}$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = -\dfrac{d}{dx} \dfrac{x}{y}$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y\dfrac{d}{dx}x - x\dfrac{d}{dx}y}{ {y}^{2} } \bigg]$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y - x\dfrac{dy}{dx}}{ {y}^{2} } \bigg]$

can be further rewritten as using equation (1),

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y + x \times \dfrac{x}{y}}{ {y}^{2} } \bigg]$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{y + \dfrac{ {x}^{2} }{y}}{ {y}^{2} } \bigg]$

$\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \bigg[\dfrac{ \dfrac{ {y}^{2} + {x}^{2} }{y}}{ {y}^{2} } \bigg]$

$\rm\implies \:\rm \: \dfrac{ {d}^{2} y}{d {x}^{2} } = - \: \dfrac{ {x}^{2} + {y}^{2} }{ {y}^{3} }$

Hence, Proved

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FORMULAE USED

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

$\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: }}$

$\boxed{\tt{ \dfrac{d}{dx}x \: = \: 1 \: }}$

$\boxed{\tt{ \dfrac{d}{dx}k \: = \: 0\: }}$

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ADDITIONAL INFORMATION

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

Answer 2

Refer the Given Attachments.