Question

If we are given two expression as $x=\dfrac{{{\sin }^{3}}t}{\sqrt{\cos 2t}},y=\dfrac{{{\cos }^{3}}t}{\sqrt{\cos 2t}}$ then find $\dfrac{dy}{dx}$.

Answer 1

Hint: In this question, we are given values of x and y in terms of parameter t and we need to find the value of $\dfrac{dy}{dx}$. For this, we will differentiate x and y separately with respect to t and then use $\dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$ to find the value of $\dfrac{dy}{dx}$. We will use the following rules of differentiation.
(i) Quotient rule on any two functions u and v given by ${{\left( \dfrac{u}{v} \right)}^{l}}=\dfrac{u'v-v'u}{{{v}^{2}}}$
(ii) $\dfrac{d}{dt}\left( {{\cos }^{n}}t \right)=n{{\cos }^{n-1}}t\dfrac{d}{dt}\left( \cos t \right)=-n{{\cos }^{n-1}}t\sin t$.
(iii) $\dfrac{d}{dt}\left( {{\sin }^{n}}t \right)=n{{\sin }^{n-1}}t\dfrac{d}{dt}\left( \sin t \right)=n{{\sin }^{n-1}}t\cos t$.
(iv) $\dfrac{\cos t}{\sin t}=\cot t$.
(v) $1+\cos 2t=2{{\cos }^{2}}t$.
(vi) $\cos 3t=4{{\cos }^{3}}t-3\cos t$.
(vii) $\sin 3t=3\sin t-4{{\sin }^{3}}t$.
(viii) $\sin 2t=2\cos t\cos t$.

Complete step-by-step solution
Here we are given the value of x in terms of t as $x=\dfrac{{{\sin }^{3}}t}{\sqrt{\cos 2t}}$ and value of y in terms of t as $y=\dfrac{{{\cos }^{3}}t}{\sqrt{\cos 2t}}$.
We need to find the value of $\dfrac{dy}{dx}$.
We can write $\dfrac{dy}{dx}$ as,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dy}{dx}\times \dfrac{dt}{dt} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx} \\
\end{align}$
It can also be written as,
$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$.
So we just need to find the value of $\dfrac{dy}{dt}\text{ and }\dfrac{dx}{dt}$.
Now, x function is given as $x=\dfrac{{{\sin }^{3}}t}{\sqrt{\cos 2t}}$.
Differentiating x with respect to t, we get: $\dfrac{dx}{dt}=\dfrac{d}{dt}\left( \dfrac{{{\sin }^{3}}t}{\sqrt{\cos 2t}} \right)$
As we know according to quotient rule ${{\left( \dfrac{u}{v} \right)}^{l}}=\dfrac{u'v-v'u}{{{v}^{2}}}$ so applying we get:
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{\sqrt{\cos 2t}\dfrac{d}{dt}\left( {{\sin }^{3}}t \right)-{{\sin }^{3}}t\dfrac{d}{dt}\left( \sqrt{\cos 2t} \right)}{{{\left( \sqrt{\cos 2t} \right)}^{2}}}\]
We know that $\dfrac{d}{dx}\left( {{\sin }^{n}}x \right)=n{{\sin }^{n-1}}x\cdot \dfrac{d}{dx}\left( \sin x \right)=n{{\sin }^{n-1}}x\cos x$ so applying we get:
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{\sqrt{\cos 2t}\left( 3{{\sin }^{2}}t\cos t \right)-{{\sin }^{3}}t\left( \dfrac{1}{2}{{\left( \cos 2t \right)}^{\dfrac{-1}{2}}}\cdot \left( -\sin 2t \right)\left( 2 \right) \right)}{\cos 2t}\]
Canceling 2 with 2 in the second term of the numerator, we get:
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{\sqrt{\cos 2t}\left( 3{{\sin }^{2}}t\cos t \right)+{{\sin }^{3}}t\dfrac{1}{\sqrt{\cos 2t}}\sin 2t}{\cos 2t}\]
Taking LCM of $\sqrt{\cos 2t}$ we get:
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{\dfrac{\sqrt{\cos 2t}\sqrt{\cos 2t}\left( 3{{\sin }^{2}}t\cos t \right)+{{\sin }^{3}}t\sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}\]
Combining both denominator and using $\sqrt{x}\sqrt{x}=x$ we get:
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{3\cos 2t{{\sin }^{2}}t\cos t+{{\sin }^{3}}t\sin 2t}{\left( \cos 2t \right)\sqrt{\cos 2t}}\cdots \cdots \cdots \cdots \left( 1 \right)\]
Now, y is given as $y=\dfrac{{{\cos }^{3}}t}{\sqrt{\cos 2t}}$.
Differentiating y with respect to t, we get:
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( \dfrac{{{\cos }^{3}}t}{\sqrt{\cos 2t}} \right)$
Applying quotient rule again we get:
\[\Rightarrow \dfrac{dy}{dt}=\dfrac{\sqrt{\cos 2t}\dfrac{d}{dt}\left( {{\cos }^{3}}t \right)-{{\cos }^{3}}t\dfrac{d}{dt}\left( \sqrt{\cos 2t} \right)}{{{\left( \sqrt{\cos 2t} \right)}^{2}}}\]
We know that $\dfrac{d}{dx}\left( {{\cos }^{n}}x \right)=n{{\cos }^{n-1}}x\cdot \dfrac{d}{dx}\left( \cos x \right)=x{{\cos }^{n-1}}\left( -\sin x \right)$ so applying it we get:
\[\Rightarrow \dfrac{dy}{dt}=\dfrac{\sqrt{\cos 2t}\left( 3{{\cos }^{2}}t\left( -\sin t \right) \right)-{{\cos }^{3}}t\left( \dfrac{1}{2}{{\left( \cos 2t \right)}^{\dfrac{-1}{2}}}\left( -\sin 2t \right)\left( 2 \right) \right)}{\cos 2t}\]
Cancelling 2 in the second term of the numerator we get:
\[\Rightarrow \dfrac{dy}{dt}=\dfrac{-\sqrt{\cos 2t}\left( 3{{\cos }^{2}}t\sin t \right)+\dfrac{{{\cos }^{3}}t}{\sqrt{\cos 2t}}\sin 2t}{\cos 2t}\]
Taking LCM of $\sqrt{\cos 2t}$ we get:
\[\Rightarrow \dfrac{dy}{dt}=\dfrac{\dfrac{-\sqrt{\cos 2t}\sqrt{\cos 2t}\left( 3{{\cos }^{2}}t\sin t \right)+{{\cos }^{3}}t\sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}\]
Combining both denominator and simplifying the numerator we get:
\[\Rightarrow \dfrac{dy}{dt}=\dfrac{-3\cos 2t{{\cos }^{2}}t\sin t+{{\cos }^{3}}t\sin 2t}{\left( \cos 2t \right)\sqrt{\cos 2t}}\cdots \cdots \cdots \cdots \left( 2 \right)\]
Dividing 1 by 2 we get:
\[\Rightarrow \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{-3\cos 2t{{\cos }^{2}}t\sin t+{{\cos }^{3}}t\sin 2t}{\left( \cos 2t \right)\sqrt{\cos 2t}}}{\dfrac{3\cos 2t{{\sin }^{2}}t\cos t+{{\sin }^{3}}t\sin 2t}{\left( \cos 2t \right)\sqrt{\cos 2t}}}\]
As we know, $\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{dy}{dx}$ so applying here, also cancelling the denominator we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-3\cos 2t{{\cos }^{2}}t\sin t+{{\cos }^{3}}t\sin 2t}{3\cos 2t{{\sin }^{2}}t\cos t+{{\sin }^{3}}t\sin 2t}\]
We know that $\sin 2\theta =2\cos \theta \sin \theta $ so we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-3\cos 2t{{\cos }^{2}}t\sin t+2{{\cos }^{3}}t\sin t\cos t}{3\cos 2t{{\sin }^{2}}t\cos t+2{{\sin }^{3}}t\sin t\cos t}\]
Simplifying we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-3\cos 2t{{\cos }^{2}}t\sin t+2{{\cos }^{4}}t\sin t}{3\cos 2t{{\sin }^{2}}t\cos t+2{{\sin }^{4}}t\cos t}\]
Taking ${{\cos }^{2}}t\sin t$ common in the numerator and ${{\sin }^{2}}t\cos t$ common in the denominator we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{{{\cos }^{2}}t\sin t\left( -3\cos 2t+2{{\cos }^{2}}t \right)}{{{\sin }^{2}}t\cos t\left( 3\cos 2t+2{{\sin }^{2}}t \right)}\]
Cancelling $\cos t\sin t$ from the numerator and the denominator we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\cos t\left( -3\cos 2t+2{{\cos }^{2}}t \right)}{\sin t\left( 3\cos 2t+2{{\sin }^{2}}t \right)}\]
Now we know that, $1+\cos 2t=2{{\cos }^{2}}t$ so using $\cos 2t=2{{\cos }^{2}}t-1$ we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\cos t\left( -3\left( 2{{\cos }^{2}}t-1 \right)+2{{\cos }^{2}}t \right)}{\sin t\left( 3\left( 2{{\cos }^{2}}t-1 \right)+2{{\sin }^{2}}t \right)}\]
Simplifying we get:
\[\Rightarrow\dfrac{dy}{dx}=\dfrac{\cos t\left( -6{{\cos }^{2}}t+3+2{{\cos }^{2}}t \right)}{\sin t\left( 6{{\cos }^{2}}t-3+2{{\sin }^{2}}t \right)} \]
Further simplifying we get:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\cos t\left( -4{{\cos }^{2}}t+3 \right)}{\sin t\left( -4{{\sin }^{2}}t+3 \right)}$ $[ \because {{\cos}^{2}} t = 1-{{\sin}^{2}} t ]$
Multiplying $\cos t\sin t$ inside the bracket we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-4{{\cos }^{3}}t+3\cos t}{-4{{\sin }^{3}}t+3\sin t}\]
Taking negative sign common from the numerator we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-\left( 4{{\cos }^{3}}t-3\cos t \right)}{3\sin t-4{{\sin }^{3}}t}\]
Now applying the formula of $\cos 3t\text{ and }\sin 3t$ given by $\cos 3t=4{{\cos }^{3}}t-3\cos t\text{ and }\sin 3t=3\sin t-4{{\sin }^{3}}t$ we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-\cos 3t}{\sin 3t}\]
We know that $\dfrac{\cos \theta }{\sin \theta }=\cot \theta $ so we get:
\[\Rightarrow \dfrac{dy}{dx}=-\cot 3t\]
Hence $-\cot 3t$ is the required value of $\dfrac{dy}{dx}$.

Note: Students should take care of signs while applying the trigonometric formula. In the quotient rule, follow the exact order otherwise mistakes could be made in signs. Try to simplify the answer as much as you can by applying trigonometric identities.