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

& \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{aligned}$$ We know derivative of $\cos x$ is $-\sin x$ , so we get $$\begin{aligned} & \dfrac{dx}{dt}=a\left( 3{{\cos }^{2}}t \right)(-\sin t) \\\ & \Rightarrow \dfrac{dx}{dt}=-3a{{\cos }^{2}}t\sin t..........(i) \\\ \end{aligned}$$ Now deriving $'y'$ with respect to $'t'$, we get $$\dfrac{dy}{dt}=\dfrac{d}{dt}(b{{\sin }^{3}}t)$$ Taking out the constant term, we get $$\begin{aligned} & \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{aligned}$$ We know derivative of $\sin x$ is $\cos x$ , so we get $$\begin{aligned} & \dfrac{dy}{dt}=b\left( 3{{\sin }^{2}}t \right)(\cos t) \\\ & \dfrac{dy}{dt}=3b{{\sin }^{2}}t\cos t..........(ii) \\\ \end{aligned}$$ Now dividing equations (ii) by (i), we have $$\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 $'t'$ .