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

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

& \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{aligned}$$ Now, let us take the second equation and try to find out the value of $$\sin t$$ as follows $$\begin{aligned} & \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{aligned}$$ Now, let us use the trigonometric identity of sine ratio and cosine ratio We know that the standard identity of sine ratio and cosine ratio as $${{\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{aligned} & \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{aligned}$$ Now, let us differentiate the above equation on both sides with respect to $$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{aligned} & \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{aligned}$$ Now, by cross multiplying the terms in above equation we get $$\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{aligned} & \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{aligned}$$ We are given that $$x=a{{\cos }^{3}}t$$ and $$y=a{{\sin }^{3}}t$$ By dividing the value of $$y$$ with $$x$$ then we get $$\begin{aligned} & \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{aligned}$$ By substituting this value in equation (i) we get $$\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}}}$$