Question

if X=acos^3t,y=asin^3t, then find dy /dx​

Answer 1

Answer:

• $\large\bold\red{\frac{dy}{dx} = - \tan(t) }$

Step-by-step explanation:

Given,

$x = a { \cos }^{3} t \: \: and \: \: y = a { \sin}^{3} t$

First of all,

We have to find $\frac{dx}{dt}$

For this,

Let us assume,

$\cos(t) = m$

On differentiating both sides wrt t,

We get,

$= > - \sin(t) dt = dm \\ \\ = > \frac{dm}{dt} = - \sin(t) \: \: \: ........(i)$

Therefore,

We have,

$x = a {m}^{3}$

Again,

Differentiating both sides wrt m,

We get,

$= > dx = 3a {m}^{2} dm \: \: \: (°.° \frac{d}{dx} {x}^{n} = n {x}^{n - 1}) \\ \\ = > \frac{dx}{dm} = 3a {m}^{2} \: \: \: \: ...........(ii)$

Now,

From Equation $(i)\:and\:(ii),$

We get,

$= > \frac{dx}{dt} = \frac{dx}{dm} \times \frac{dm}{dt} \\ \\ = > \frac{dx}{dt} = 3a {m}^{2} \times ( - \sin \: t)$

Putting the value of $m=Cos t$

We get,

$= > \frac{dx}{dt} = - 3a \: { \cos }^{2} t \: \sin(t) \: \: \: ...........(iii)$

Now,

For Solving $\frac{dy}{dt}$

Let's assume,

$\sin(t) = n$

Differentiating both sides wrt t,

We get,

$= > \cos(t) dt = dn \\ \\ = > \frac{dn}{dt} = \cos(t) \: \: \: ........(iv)$

Therefore,

We have,

$y = a {n}^{3}$

Again,

Differentiating both sides wrt n,

We get,

$= > \frac{dy}{dn} = 3a {n}^{2} \: \: \: ........(v)$

From Equation $(iv)\:and\:(v)$,

We get,

$\frac{dy}{dt} = \frac{dy}{dn} \times \frac{dn}{dt} \\ \\ = > \frac{dy}{dt} = 3a {n}^{2} \times \cos(t)$

Putting the value of $n=sint$,

We get,

$= > \frac{dy}{dt} = 3a \: { \sin }^{2}t \: \cos \: t \: \: \: ..........(vi)$

Now,

From Equation $(iii)\:and\:(vi)$,

We get,

$= > \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \\ \\ = > \frac{dy}{dx} = \frac{3a { \sin }^{2} t \cos(t) }{ - 3a { \cos }^{2} t \sin(t) } \\ \\ = > \frac{dy}{dx} = - \frac{ \sin(t) }{ \cos(t) } \\ \\ = > \large\bold{\frac{dy}{dx} = - \tan(t) }$