Question

If \[x=a{{\cos }^{3}}t,y=b{{\sin }^{3}}t\], then at the point \[\left( \dfrac{a}{2\sqrt{2}},\dfrac{a}{2\sqrt{2}} \right),\dfrac{dy}{dx}=\]
(a) $\dfrac{b}{a}$
 (b) $-\dfrac{b}{a}$
(c) $\dfrac{a}{b}$
 (d) $-\dfrac{a}{b}$

Answer 1

Hint:- Consider ‘x’ and ‘y’ separately, then differentiate them with respect to ‘t’ separately. Then apply the parametric form of derivation, i.e., \[\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\]. Then substitute the value of the given point and find out the value of ‘t’ and solve further to obtain the desired result.

Complete step-by-step solution -
As per the given information, \[x=a{{\cos }^{3}}t,y=b{{\sin }^{3}}t.\]
For this first we will find \[\dfrac{dx}{dt},\dfrac{dy}{dt}.\]
So, deriving $'x'$ with respect to $'t'$, we get
\[\dfrac{dx}{dt}=\dfrac{d}{dt}(a{{\cos }^{3}}t)\]
Taking out the constant term, we get
\[\begin{align}
  & \dfrac{dx}{dt}=a\dfrac{d}{dt}({{\cos }^{3}}t) \\
 & \Rightarrow \dfrac{dx}{dt}=a(3{{\cos }^{2}}t)\dfrac{d}{dt}(\cos t) \\
\end{align}\]
We know derivative of $\cos x$ is $-\sin x$ , so we get
\[\begin{align}
  & \dfrac{dx}{dt}=a\left( 3{{\cos }^{2}}t \right)(-\sin t) \\
 & \Rightarrow \dfrac{dx}{dt}=-3a{{\cos }^{2}}t\sin t..........(i) \\
\end{align}\]
Now deriving $'y'$ with respect to $'t'$, we get
\[\dfrac{dy}{dt}=\dfrac{d}{dt}(b{{\sin }^{3}}t)\]
Taking out the constant term, we get
\[\begin{align}
  & \dfrac{dy}{dt}=b\dfrac{d}{dt}(si{{n}^{3}}t) \\
 & \Rightarrow \dfrac{dy}{dt}=b(3{{\sin }^{2}}t)\dfrac{d}{dt}(sint) \\
\end{align}\]
We know derivative of $\sin x$ is $\cos x$ , so we get
\[\begin{align}
  & \dfrac{dy}{dt}=b\left( 3{{\sin }^{2}}t \right)(\cos t) \\
 & \dfrac{dy}{dt}=3b{{\sin }^{2}}t\cos t..........(ii) \\
\end{align}\]
Now dividing equations (ii) by (i), we have
\[\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3b{{\sin }^{2}}t\cos t}{-3a{{\cos }^{2}}t\sin t}\]
Cancelling like terms, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-b\sin t}{a\cos t}\]
But we know $\tan x=\dfrac{\sin x}{\cos x}$ , so above equation becomes
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-b}{a}\tan t........(iii)\]
Now, let us find the value of $'t'$ at \[(x,y)=\left( \dfrac{a}{2\sqrt{2}},\dfrac{a}{2\sqrt{2}} \right)\].
As per the given information,
\[x=a{{\cos }^{3}}t\]
By substituting \[x=\dfrac{a}{2\sqrt{2}},\]we get,
\[\dfrac{a}{2\sqrt{2}}=a{{\cos }^{3}}t\]
\[\Rightarrow {{\cos }^{3}}(t)=\dfrac{1}{2\sqrt{2}}=\dfrac{1}{\sqrt{{{2}^{3}}}}\]
Taking cube root on both sides, we get
\[\Rightarrow \cos t=\dfrac{1}{\sqrt{2}}\]
We know $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ , so
\[\Rightarrow t=\dfrac{\pi }{4}\]
Now, substitute this value in \[\dfrac{dy}{dx}\], we get
\[\dfrac{dy}{dx}=-\dfrac{b}{a}\tan \left( \dfrac{\pi }{4} \right)\]
We know $\tan \left( \dfrac{\pi }{4} \right)=1$ , so
\[\dfrac{dy}{dx}=\dfrac{-b}{a}\]
Hence the correct answer is option (b).

Note: In this problem we may get stuck after finding \[\dfrac{dy}{dx}\] as this function is not in (x, y). So, we need to recheck how to get the value of $'t'$ from the given information and go further, i.e., \[x=a{{\cos }^{3}}t\]
By substituting \[x=\dfrac{a}{2\sqrt{2}},\]we get,
\[\dfrac{a}{2\sqrt{2}}=a{{\cos }^{3}}t\]
And find the value of $'t'$ .