Question

$\sf \: if \: x = { \cos }^{3} \theta.y = { \sin }^{1} \theta \: then \: \displaystyle \sqrt{1 + \bigg(\sf\frac{dy}{dx} \bigg) } {}^{2}$

Answer 1

Appropriate Question :-

$\rm \: If \: x = {cos}^{3}\theta \: and \: y = {sin}^{3}\theta , \: then \: \sqrt{1 +\bigg( {\dfrac{dy}{dx}}\bigg)^{2} }$

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

Given that,

$\rm \: x = {cos}^{3}\theta$

So,

$\rm \: \dfrac{d}{d\theta }x = \dfrac{d}{d\theta }{cos}^{3}\theta$

We know,

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

So, using this, we get

$\rm \: \dfrac{dx}{d\theta } = 3 {cos}^{2}\theta \dfrac{d}{d\theta }cos\theta$

We know,

$\boxed{\sf{ \:\dfrac{d}{dx}cosx = - sinx \: }}$

So, using this result, we get

$\rm \: \dfrac{dx}{d\theta } = {3cos}^{2}\theta ( - sin\theta )$

$\rm\implies \: \dfrac{dx}{d\theta } = \: - \: {3cos}^{2}\theta \: sin\theta$

Again, given that,

$\rm \: y = {sin}^{3}\theta$

So,

$\rm \: \dfrac{d}{d\theta }y = \dfrac{d}{d\theta } {sin}^{3}\theta$

$\rm \: \dfrac{dy}{d\theta } = {3sin}^{2}\theta \: \dfrac{d}{d\theta }sin\theta$

$\rm \: \dfrac{dy}{d\theta } = {3sin}^{2}\theta \: \cos\theta$

So,

$\rm\implies \:\dfrac{dy}{d\theta } = 3 {sin}^{2}\theta \: cos\theta$

So,

$\rm \: \dfrac{dy}{dx}$

$\rm \: = \: \dfrac{dy}{d\theta } \div \dfrac{dx}{d\theta }$

$\rm \: = \: \dfrac{ {3sin}^{2}\theta cos\theta }{ - {3cos}^{2} \theta sin\theta }$

$\rm \: = \: - \: tan\theta$

So,

$\rm\implies \:\dfrac{dy}{dx} = \: - \: tan\theta$

Now, Consider

$\rm \: \sqrt{1 + {\bigg(\dfrac{dy}{dx}\bigg) }^{2} }$

$\rm \: = \: \sqrt{1 + {( - tan\theta )}^{2} }$

$\rm \: = \: \sqrt{1 + {tan}^{2} \theta }$

$\rm \: = \: \sqrt{ {sec}^{2} \theta }$

$\rm \: = \: sec\theta$

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: \sqrt{1 + {\bigg(\dfrac{dy}{dx}\bigg) }^{2} } \: = \: sec\theta \: \: }}$

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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

Step-by-step explanation:$\color{red} \tt\dfrac{d y}{d x}=\cfrac{\cfrac{d y}{d \theta}}{\cfrac{d x}{d \theta}}\\ \\ \color{navy}\tt=\cfrac{\cfrac{d}{d \theta}\left( \sin ^{3} \theta\right)}{\cfrac{d}{d \theta}\left(\cos ^{3} \theta\right)}\\ \\ \color{green}\tt=\cfrac{3 \sin ^{2} \theta \cos \theta}{-3 \cos ^{2} \theta \sin \theta}\\ \\ \color{orange}\tt=-\tan \theta$$\color{blue}\tt\sqrt{\left[1+\left(\dfrac{d y}{d x}\right)\right]^2}=\sqrt{1+(-\tan \theta)^{2}}=\sqrt{1+\tan ^{2} \theta}=|\sec \theta| .$