If \[x=a\cos 2t,y=b{{\sin }^{2}}t\], then at \[(-a,b),\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] is equal to (a) $1$ (b) \[\dfrac{1}{2}\] (c) $0$ (d) \[\dfrac{1}{\sqrt{2}}\]
Answer
Verified
Hint: Use parametric equations and chain rule. As per the given information, \[x=a\cos 2t,y=b{{\sin }^{2}}t\] For this problem, first we shall find \[\dfrac{dy}{dt}\And \dfrac{dx}{dt}.\] So, consider, \[y=b{{\sin }^{2}}t\] Now take differentiation on both sides with respect to ‘t’, we get \[\dfrac{dy}{dt}=\dfrac{d}{dt}(b{{\sin }^{2}}t)\] Taking out the constant term, we get \[\begin{align} & \dfrac{dy}{dt}=b.\dfrac{d}{dt}({{\sin }^{2}}t) \\ & \dfrac{dy}{dt}=b(2\sin t)\dfrac{d}{dt}(\sin t) \\ \end{align}\] We know derivative of $sinx$ is $\cos x$ , so we get \[\dfrac{dy}{dt}=b(2\sin t)(cost)\] But we know, $\sin 2t=2\sin t\cos t$ , so above equation becomes, \[\dfrac{dy}{dt}=b\sin 2t........(i)\] Now consider,\[x=a\cos 2t\] Now take differentiation on both sides with respect to ‘t’, we get \[\dfrac{dx}{dt}=\dfrac{d}{dt}(a\cos 2t)\] Taking out the constant term and derivative of $\cos x$ is $-\sin x$ , so we get \[\begin{align} & \dfrac{dx}{dt}=a(-\sin 2t)\dfrac{d}{dt}(2t) \\ & \dfrac{dx}{dt}=-2a.\sin 2t..........(ii) \\ \end{align}\] Now dividing equations (ii) by (i), we have \[\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{b\sin 2t}{-2a.\sin 2t}\] Cancelling like terms, we get \[\dfrac{dy}{dx}=\dfrac{-b}{2a}\] Now as this is free from the ‘t’ term, we can directly differentiate to find the second derivative. Now take differentiation on both sides with respect to ‘t’, we get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{-b}{2a} \right)\] We know derivative of constant term is zero, so we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0\] As the second derivative is free of variable terms, so its value is zero at any point. Hence the correct answer is option (c).
Note: Instead of deriving with respect to ‘t’ and then dividing, we can find the value of ‘t’ from the given values of ‘x’. We know, $\cos 2t=1-{{\sin }^{2}}t$ Substituting this in value of ‘x’, we get \[\begin{align} & x=a\cos 2t \\ & \Rightarrow x=a(1-{{\sin }^{2}}t) \\ & \Rightarrow \dfrac{x}{a}+1={{\sin }^{2}}t \\ \end{align}\] Now substituting this value in the value of ‘y’, we get \[\begin{align} & y=b{{\sin }^{2}}t \\ & \Rightarrow y=b\left( \dfrac{x}{a}+1 \right) \\ \end{align}\] Now we can see that this can be differentiated directly with respect to ‘x’. But this is a complicated process.
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