If (x-a)2+(y-b)2=c2, for some c>0 prove that
[1+(\(\frac{dy}{dx}\))2]\(^{\frac{3}{2}}\)/\(\frac{d^2y}{dx^2}\) is a constant independent of a and b
If (x-a)2+(y-b)2=c2, for some c>0 prove that
[1+(\(\frac{dy}{dx}\))2]\(^{\frac{3}{2}}\)/\(\frac{d^2y}{dx^2}\) is a constant independent of a and b
Solution and Explanation
It is given that,(x-a)2+(y-b)2=c2
Differentiating both sides with respect to x, we obtain
\(\frac{d}{dx}\)[(x-a)2]+\(\frac{d}{dx}\)[(y-b)2]=\(\frac{d}{dx}\)(c2)
⇒ 2(x-a).\(\frac{d}{dx}\)(x-a)+2(y-b).\(\frac{d}{dx}\)(y-b)=0
⇒2(x-a).1+2(y-b).\(\frac{dy}{dx}\)=0
⇒ \(\frac{dy}{dx}\)=\(\frac{-(x-a)}{y-b}\) ...(1)
∴\(\frac{d^2y}{dx^2}\)=\(\frac{d}{dx}\)[\(\frac{-(x-a)}{y-b}\)]
=-[(y-b).\(\frac{d}{dx}\)(x-a)-(x-a).\(\frac{d}{dx}\)\(\frac{(y-b)}{(y-b)^2}\)]
=-c, where c is constant and is independent of a and b
Hence, proved.
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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.