Question

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Answer 1

Question :

$\displaystyle \sf x=\cos^3 \theta \ \ , \ y=\sin^3 \theta \ then \ \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2}$

Answer:

$\displaystyle \sf \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2}=\sec \theta$

Step-by-step explanation:

Given :

$\displaystyle \sf x=\cos^3 \theta$

Diff. w.r.t. θ :

$\displaystyle \sf \longrightarrow \frac{dx}{d \theta} =- \sin \theta. \cos^2 \theta$

Also we have :

$\displaystyle \sf y=\sin^3 \theta$

Diff. w.r.t. θ :

$\displaystyle \sf \longrightarrow \frac{dy}{d \theta} = \sin^2 \theta. \cos \theta$

Now as we do in parametric function :

$\displaystyle \sf \frac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}$

Putting above values we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{\sin^2 \theta.\cos \theta}{-\sin \theta. \cos^2 \theta}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = -\frac{\sin \theta}{ \cos \theta}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = - \tan \theta$

Squaring on both side we get :

$\displaystyle \sf \longrightarrow \left(\frac{dy}{dx}\right)^2= \left(- \tan \theta\right)^2$

$\displaystyle \sf \longrightarrow \left(\frac{dy}{dx}\right)^2= \left(\tan^2 \theta\right)$

Adding + 1 both side :

$\displaystyle \sf \longrightarrow 1 + \left(\frac{dy}{dx}\right)^2= 1 + \left(\tan^2 \theta\right)$

Now squaring root of that :

$\displaystyle \sf \longrightarrow \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2}= \sqrt{1 + \left(\tan^2 \theta\right)}$

We know :

$\displaystyle \sf \sec^2 \theta -\tan^2 \theta=1$

$\displaystyle \sf \longrightarrow \sec^2 \theta = 1 + \tan^2 \theta$

$\displaystyle \sf \longrightarrow \sec \theta = \sqrt{1 + \tan^2 \theta} \ ...(i)$

Using ( i ) we get :

$\displaystyle \sf \longrightarrow \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2}=\sec \theta$

Therefore , we get required answer.

Answer 2

$\large \underline{ \red{ \boxed{ \bf \orange{Required \: Answer}}}}$