# If $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$, show that at $t=\dfrac{\pi }{4}$,

$\dfrac{dy}{dx}=\dfrac{b}{a}$.

Answer

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Hint: Take $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$ and differentiate both of them w.r.t $t$ .After that, divide each other and substitute $t=\dfrac{\pi }{4}$. You will get the answer.

Complete step by step solution :

We are given $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$.

So now differentiating $x$ w.r.t $t$ and differentiating $y$ w.r.t $t$ we get,

For $x$,

\[\begin{align}

& \dfrac{dx}{dt}=\dfrac{d}{dt}\left( a\sin 2t(1+\cos 2t) \right) \\

& \dfrac{dx}{dt}=a\sin 2t\dfrac{d}{dt}(1+\cos 2t)+a(1+\cos 2t)\dfrac{d}{dt}\sin 2t \\

& \dfrac{dx}{dt}=a\sin 2t(-2\sin 2t)+2a(1+\cos 2t)(\cos 2t) \\

& \dfrac{dx}{dt}=-2a{{\sin }^{2}}2t+2a(1+\cos 2t)(\cos 2t) \\

\end{align}\]

\[\dfrac{dx}{dt}=-2a{{\sin }^{2}}2t+2a(1+\cos 2t)(\cos 2t)\] …………… (1)

For $y$,

\[\begin{align}

& \dfrac{dy}{dt}=\dfrac{d}{dt}b\cos 2t(1-\cos 2t) \\

& \dfrac{dy}{dt}=b\cos 2t\dfrac{d}{dt}(1-\cos 2t)+b(1-\cos 2t)\dfrac{d}{dt}\cos 2t \\

& \dfrac{dy}{dt}=b\cos 2t(2\sin 2t)+b(1-\cos 2t)(-2\sin 2t) \\

& \dfrac{dy}{dt}=2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t) \\

\end{align}\]

\[\dfrac{dy}{dt}=2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t)\]……………. (2)

Now dividing (2) by (1) we get,

\[\begin{align}

& \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t)}{-2a{{\sin

}^{2}}2t+2a(1+\cos 2t)(\cos 2t)} \\

& \dfrac{dy}{dx}=\dfrac{2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t)}{-2a{{\sin }^{2}}2t+2a(1+\cos 2t)(\cos 2t)}

\\

\end{align}\]

Now substituting $t=\dfrac{\pi }{4}$in (1) and (2), we get,

\[\begin{align}

& \dfrac{dy}{dx}=\dfrac{2b\cos 2\left( \dfrac{\pi }{4} \right)\sin 2\left( \dfrac{\pi }{4} \right)-2b(1-\cos

2\left( \dfrac{\pi }{4} \right))(\sin 2\left( \dfrac{\pi }{4} \right))}{-2a{{\sin }^{2}}2\left( \dfrac{\pi }{4}

\right)+2a(1+\cos 2\left( \dfrac{\pi }{4} \right))(\cos 2\left( \dfrac{\pi }{4} \right))} \\

& \dfrac{dy}{dx}=\dfrac{2b\cos \left( \dfrac{\pi }{2} \right)\sin \left( \dfrac{\pi }{2} \right)-2b(1-\cos

\left( \dfrac{\pi }{2} \right))(\sin \left( \dfrac{\pi }{2} \right))}{-2a{{\sin }^{2}}\left( \dfrac{\pi }{2}

\right)+2a(1+\cos \left( \dfrac{\pi }{2} \right))(\cos \left( \dfrac{\pi }{2} \right))} \\

\end{align}\]………….

Now taking $2a$and $2b$ common we get,

\[\begin{align}

& \dfrac{dy}{dx}=\dfrac{2b}{2a}\left[ \dfrac{0-(1-0)(1)}{-1+0} \right] \\

& \dfrac{dy}{dx}=\dfrac{b}{a}\left[ \dfrac{-1}{-1} \right] \\

\end{align}\]

\[\dfrac{dy}{dx}=\dfrac{b}{a}\]

So we get, \[\dfrac{dy}{dx}=\dfrac{b}{a}\].

Hence proved.

Note: Read the question carefully. Don’t confuse yourself. Your concept regarding differentiation should be clear. Also, take care that while simplifying no terms are missed. Do not make any silly mistakes. While solving, take care that no signs are missed.

Complete step by step solution :

We are given $x=a\sin 2t(1+\cos 2t)$ and $y=b\cos 2t(1-\cos 2t)$.

So now differentiating $x$ w.r.t $t$ and differentiating $y$ w.r.t $t$ we get,

For $x$,

\[\begin{align}

& \dfrac{dx}{dt}=\dfrac{d}{dt}\left( a\sin 2t(1+\cos 2t) \right) \\

& \dfrac{dx}{dt}=a\sin 2t\dfrac{d}{dt}(1+\cos 2t)+a(1+\cos 2t)\dfrac{d}{dt}\sin 2t \\

& \dfrac{dx}{dt}=a\sin 2t(-2\sin 2t)+2a(1+\cos 2t)(\cos 2t) \\

& \dfrac{dx}{dt}=-2a{{\sin }^{2}}2t+2a(1+\cos 2t)(\cos 2t) \\

\end{align}\]

\[\dfrac{dx}{dt}=-2a{{\sin }^{2}}2t+2a(1+\cos 2t)(\cos 2t)\] …………… (1)

For $y$,

\[\begin{align}

& \dfrac{dy}{dt}=\dfrac{d}{dt}b\cos 2t(1-\cos 2t) \\

& \dfrac{dy}{dt}=b\cos 2t\dfrac{d}{dt}(1-\cos 2t)+b(1-\cos 2t)\dfrac{d}{dt}\cos 2t \\

& \dfrac{dy}{dt}=b\cos 2t(2\sin 2t)+b(1-\cos 2t)(-2\sin 2t) \\

& \dfrac{dy}{dt}=2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t) \\

\end{align}\]

\[\dfrac{dy}{dt}=2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t)\]……………. (2)

Now dividing (2) by (1) we get,

\[\begin{align}

& \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t)}{-2a{{\sin

}^{2}}2t+2a(1+\cos 2t)(\cos 2t)} \\

& \dfrac{dy}{dx}=\dfrac{2b\cos 2t\sin 2t-2b(1-\cos 2t)(\sin 2t)}{-2a{{\sin }^{2}}2t+2a(1+\cos 2t)(\cos 2t)}

\\

\end{align}\]

Now substituting $t=\dfrac{\pi }{4}$in (1) and (2), we get,

\[\begin{align}

& \dfrac{dy}{dx}=\dfrac{2b\cos 2\left( \dfrac{\pi }{4} \right)\sin 2\left( \dfrac{\pi }{4} \right)-2b(1-\cos

2\left( \dfrac{\pi }{4} \right))(\sin 2\left( \dfrac{\pi }{4} \right))}{-2a{{\sin }^{2}}2\left( \dfrac{\pi }{4}

\right)+2a(1+\cos 2\left( \dfrac{\pi }{4} \right))(\cos 2\left( \dfrac{\pi }{4} \right))} \\

& \dfrac{dy}{dx}=\dfrac{2b\cos \left( \dfrac{\pi }{2} \right)\sin \left( \dfrac{\pi }{2} \right)-2b(1-\cos

\left( \dfrac{\pi }{2} \right))(\sin \left( \dfrac{\pi }{2} \right))}{-2a{{\sin }^{2}}\left( \dfrac{\pi }{2}

\right)+2a(1+\cos \left( \dfrac{\pi }{2} \right))(\cos \left( \dfrac{\pi }{2} \right))} \\

\end{align}\]………….

Now taking $2a$and $2b$ common we get,

\[\begin{align}

& \dfrac{dy}{dx}=\dfrac{2b}{2a}\left[ \dfrac{0-(1-0)(1)}{-1+0} \right] \\

& \dfrac{dy}{dx}=\dfrac{b}{a}\left[ \dfrac{-1}{-1} \right] \\

\end{align}\]

\[\dfrac{dy}{dx}=\dfrac{b}{a}\]

So we get, \[\dfrac{dy}{dx}=\dfrac{b}{a}\].

Hence proved.

Note: Read the question carefully. Don’t confuse yourself. Your concept regarding differentiation should be clear. Also, take care that while simplifying no terms are missed. Do not make any silly mistakes. While solving, take care that no signs are missed.

Last updated date: 27th May 2023

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