Question

If \[x=a{{\cos }^{3}}t\] and \[y=a{{\sin }^{3}}t\] then find the value of \[\dfrac{dy}{dx}\] after eliminating \[t\]

Answer 1

Hint: We solve this problem first by finding the value of \[\cos t,\sin t\] from the given two equations.
Then we use the standard identity of the trigonometric ratios to eliminate \[t\]
We have the standard identity of sine ratio and cosine ratio as
\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]
Then we differentiate the equation with respect to \[x\] in order to get \[\dfrac{dy}{dx}\]
We use the simple formula of differentiation that is
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

Complete step by step answer:
We are given that \[x=a{{\cos }^{3}}t\] and \[y=a{{\sin }^{3}}t\]
Now, let us take the first equation and try to find out the value of \[\cos t\] as
\[\begin{align}
  & \Rightarrow x=a{{\cos }^{3}}t \\
 & \Rightarrow {{\left( \cos t \right)}^{3}}=\dfrac{x}{a} \\
 & \Rightarrow \cos t={{\left( \dfrac{x}{a} \right)}^{\dfrac{1}{3}}} \\
\end{align}\]
Now, let us take the second equation and try to find out the value of \[\sin t\] as follows
\[\begin{align}
  & \Rightarrow y=a{{\sin }^{3}}t \\
 & \Rightarrow {{\left( \sin t \right)}^{3}}=\dfrac{y}{a} \\
 & \Rightarrow \sin t={{\left( \dfrac{y}{a} \right)}^{\dfrac{1}{3}}} \\
\end{align}\]
Now, let us use the trigonometric identity of sine ratio and cosine ratio
We know that the standard identity of sine ratio and cosine ratio as
\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]
By using the above identity we get
\[\Rightarrow {{\cos }^{2}}t+{{\sin }^{2}}t=1\]
By substituting the required values in above equation we get
\[\Rightarrow {{\left[ {{\left( \dfrac{x}{a} \right)}^{\dfrac{1}{3}}} \right]}^{2}}+{{\left[ {{\left( \dfrac{y}{a} \right)}^{\dfrac{1}{3}}} \right]}^{2}}=1\]
We know that the standard formula of exponents that is
\[{{\left( {{a}^{b}} \right)}^{c}}={{a}^{b\times c}}\]
By using the this formula in above equation we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{x}{a} \right)}^{\dfrac{2}{3}}}+{{\left( \dfrac{y}{a} \right)}^{\dfrac{2}{3}}}=1 \\
 & \Rightarrow {{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}} \\
\end{align}\]
Now, let us differentiate the above equation on both sides with respect to \[x\] then we get
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{\dfrac{2}{3}}} \right)+\dfrac{d}{dx}\left( {{y}^{\dfrac{2}{3}}} \right)=\dfrac{d}{dx}\left( {{a}^{\dfrac{2}{3}}} \right)\]
We know that the simple formula of differentiation that is
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]
By using the above formula we get
\[\begin{align}
  & \Rightarrow \dfrac{2}{3}{{x}^{\dfrac{2}{3}-1}}+\dfrac{2}{3}{{y}^{\dfrac{2}{3}-1}}\dfrac{dy}{dx}=0 \\
 & \Rightarrow {{y}^{-\dfrac{1}{3}}}\dfrac{dy}{dx}=-{{x}^{-\dfrac{1}{3}}} \\
 & \Rightarrow \dfrac{1}{{{y}^{\dfrac{1}{3}}}}\dfrac{dy}{dx}=-\dfrac{1}{{{x}^{\dfrac{1}{3}}}} \\
\end{align}\]

Now, by cross multiplying the terms in above equation we get
\[\Rightarrow \dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}\]
Therefore we can conclude that the value of \[\dfrac{dy}{dx}\] after eliminating \[t\] is
\[\therefore \dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}\]

Note: We can solve this problem by using the other method.
We are given that \[x=a{{\cos }^{3}}t\] and \[y=a{{\sin }^{3}}t\]
We have the standard formula of differentiation that is
\[\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{dt} \right)}{\left( \dfrac{dx}{dt} \right)}\]
By using the above formula we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( \dfrac{d}{dt}\left( a{{\sin }^{3}}t \right) \right)}{\left( \dfrac{d}{dt}\left( a{{\cos }^{3}}t \right) \right)}\]
We know that the formulas of differentiation that is
\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)\]
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
\[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\]
By using the above two formula in the \[\dfrac{dy}{dx}\] we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{3a{{\sin }^{2}}t\cos t}{-3a{{\cos }^{2}}t\sin t} \\
 & \Rightarrow \dfrac{dy}{dx}=-\tan x......equation(i) \\
\end{align}\]
We are given that \[x=a{{\cos }^{3}}t\] and \[y=a{{\sin }^{3}}t\]
By dividing the value of \[y\] with \[x\] then we get
\[\begin{align}
  & \Rightarrow \dfrac{y}{x}=\dfrac{a{{\sin }^{3}}t}{a{{\cos }^{3}}t} \\
 & \Rightarrow {{\left( \tan t \right)}^{3}}=\dfrac{y}{x} \\
 & \Rightarrow \tan t={{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}} \\
\end{align}\]
By substituting this value in equation (i) we get
\[\Rightarrow \dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}\]
Therefore we can conclude that the value of \[\dfrac{dy}{dx}\] after eliminating \[t\] is
\[\therefore \dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}\]