Prove that (x-a)^2+(y-b)^2=c^2. [1+(dy/dx)^2]^3/2 ---------------- [d^2y/dx^2] Is a constant a and b?

2 Answers
Feb 28, 2018

#((1+((dy)/dx)^2)^(3/2))/((d^2y)/dx^2)=-c#

Explanation:

#(x-a)^2+(y-b)^2=c^2#
#((1+((dy)/dx)^2)^(3/2))/((d^2y)/dx^2)# is a constant a and b)

Differentiating wrt x

#2(x-a)+2(y-b)dy/dx=0#

#(x-a)+(y-b)dy/dx=0#

#(y-b)dy/dx=-(x-a)#

#dy/dx=-(x-a)/(y-b)#

#(x-a)+(y-b)dy/dx=0#

Differentiating again wrt x

#1+(y-b)(d^2y)/(dx^2)+dy/dxdy/dx=0#

#1+(y-b)(d^2y)/(dx^2)+(dy/dx)^2=0#

#1+(y-b)(d^2y)/(dx^2)+((x-a)/(y-b))^2=0#

#(y-b)^2+(y-b)^3(d^2y)/(dx^2)+(x-a)^2=0#

#(y-b)^3(d^2y)/(dx^2)+(x-a)^2+(y-b)^2=0#

#(x-a)^2+(y-b)^2=c^2#

#(y-b)^3(d^2y)/(dx^2)+c^2=0#

#(d^2y)/(dx^2)=-c^2/(y-b)^3#

#(1+((dy)/dx)^2)^(3/2)=(1+((x-a)/(y-b))^2)^(3/2)#

#=(((y-b)^2+(x-a)^2)/(y-b)^2)^(3/2)#

#=(c^2/(y-b)^2)^(3/2)#

#(1+((dy)/dx)^2)^(3/2)=c^3/(y-b)^3#

#(d^2y)/(dx^2)=-c^2/(y-b)^3#

#((1+((dy)/dx)^2)^(3/2))/((d^2y)/dx^2)=(c^3/(y-b)^3)/(-c^2/(y-b)^3)#

#((1+((dy)/dx)^2)^(3/2))/((d^2y)/dx^2)=-c#

Mar 2, 2018

Realising that, #(x-a)^2+(y-b)^2=c^2# represents a Circle with

Centre #(a,b)# and Radius #c#, let us use its

parametric eqns. :

#x=a+c*cos t, y=b+c*sint; t in [0,2pi)" (preferably)"#.

Then, we have, #dx/dt=-c*sint, dy/dt=c*cost#,

so that, #dy/dx=-cott...(1)#.

#:. (d^2y)/dx^2=d/dx{dy/dx}=d/dx{-cott}#,

#=d/dt{-cott}*dt/dx.........[because," the Chain Rule]"#,

#=-(-csc^2t)*{1/(dx/dt)}...[because, dt/dx=1/(dx/dt)]#,

#=csc^2t/(-c*sint)#.

#rArr (d^2y)/dx^2=-1/c*csc^3t......................(2)#.

Hence, by #(1) and (2),#

#{1+(dy/dx)^2}^(3/2)/{(d^2y)/dx^2}#,

#={1+(-cott)^2}^(3/2)/{-1/c*csc^3t}#,

#=-c*(csc^2t)^(3/2)/csc^3t#,

#=-c," a constant, as desired!"#

#"BONUS : "#

For the curve #y=f(x), kappa={(d^2y)/dx^2}/{1+(dy/dx)^2}^(3/2)#,

is called the curvature of the curve, and, #1/kappa#, its radius.

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