If x=a(cost+tsint) and y=a(sint-tcost),find \(\frac{d^2y}{dx^2}\)
If x=a(cost+tsint) and y=a(sint-tcost),find \(\frac{d^2y}{dx^2}\)
Solution and Explanation
It is given that x=a(cost+tsint) and y=a(sint-tcost)
∴\(\frac{dx}{dt}\)=a.\(\frac{d}{dt}\)(cost+sint)
=a[-sint+sint.\(\frac{d}{dt}\)(t)+t.\(\frac{d}{dt}\)(sint)]
=a[-sint+sint+tcost]=atcost
\(\frac{dy}{dt}\)=a.\(\frac{d}{dt}\)(sint-tcost)
=a[cost-{cost.\(\frac{d}{dt}\)(t)+t.\(\frac{d}{dt}\)(cost)}]
=a[cost-{cost-tsint}]=atsint
∴\(\frac{dy}{dx}\)=\(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)=\(\frac{atsint}{atcost}\)=tant
Then,\(\frac{d^2y}{dx^2}\)=\(\frac{d}{dx}\)(\(\frac{dy}{dx}\))=\(\frac{d}{dt}\)(tant)=sec2t.\(\frac{dt}{dx}\)
=sec2t.\(\frac{1}{atcos\,t}\) [\(\frac{dx}{dt}\)=atcost\(\Rightarrow\)\(\frac{dt}{dx}\)=\(\frac{1}{atcos\,t}\)]
=\(\frac{sec^3t}{at}\), 0<t<\(\frac{\pi}{2}\)
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.