Question

Find the value of dy/dx , if x= 2 cos square t, y= 2 sin square t

Answer 1

ANSWER :–

$\\ \longrightarrow \: \: \large { \red{ \boxed{ \bold{ \dfrac{dy}{dx} = - 1 }}}}$

EXPLANATION :–

GIVEN :–

• ${ \bold{ x = 2 \: {cos}^{2}t }}$ and ${ \bold{ y = 2 \: {sin}^{2}t }}$

TO FIND :–

• The value of ${ \bold{ \: \dfrac{dy}{dx} = ? }}$

SOLUTION :–

• We know that –

$\\ \implies { \pink{ \boxed{ \bold{ \dfrac{dy}{dx} = \dfrac{ \dfrac{dy}{dt} }{ \dfrac{dx}{dt} } }}}}$

• So , let's find ${ \bold{ \: \dfrac{dy}{dt} \: \: - }}$

$\\ { \bold{ \implies y = 2 \: {sin}^{2}t }}$

• Now Differentiate with respect to 't' –

$\\ { \bold{ \implies \frac{dy}{dt} = 2 \times \: 2{sin}t \times cost }}$

$\\ { \bold{ \implies \frac{dy}{dx} = 4 \: {sin}t . cost \: \: \: \: - - - -eq.(1) }}$

$\\ { \bold{ \implies x = 2 \: {cos}^{2}t }}$

• Now Differentiate with respect to 't' –

$\\ { \bold{ \implies \frac{dx}{dt} =2 \times 2 \: {cos}t \times ( - sint ) }}$

$\\ { \bold{ \implies \frac{dx}{dt} =4\: {cos}t \times ( - sint ) \: \: \: - - - - eq.(2) }}$

• Now divided eq.(1) and eq.(2) –

$\\ \because \: \: \: { \bold{ \dfrac{dy}{dx} = \dfrac{ \dfrac{dy}{dt} }{ \dfrac{dx}{dt} } }}$

• Put the values from eq.(1) and eq.(2) –

$\\ \implies \: \: \: { \bold{ \dfrac{dy}{dx} = \dfrac{ 4 sin(t) cos(t) }{4 cos(t) [ - sin(t) ] } }}$

$\\ \implies \: \: \large { \red{ \boxed{ \bold{ \dfrac{dy}{dx} = - 1 }}}}$

$\\ \: \: \large { \bold{ \underline{Used \: \: formula} : - }}$

$\\ \: \: { \bold{(1) \: \: \frac{d( {x}^{n} )}{dx} = n {x}^{n - 1} }}$

$\\ \: \: { \bold{(2) \: \: \frac{d[sin( \theta)]}{dx} = cos( \theta) }}$

$\\ \: \: { \bold{(3) \: \: \frac{d[cos( \theta)]}{dx} = - \: sin( \theta) }}$