+4.31. Let a(s) be a unit speed curve with kt = 0. Prove there is a curve
1 answer:
Answer:
It is proved
Step-by-step explanation:
A curve immersed in the three-dimensional sphere is said to be a Bertrand curve if there exists another curve and a one-to-one correspondence between and such that both curves have common principal normal geodesics at corresponding points.
See attachment for the step by step solution of the given problem.
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Domain is the numbers yo can use
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0
deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)
so
dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))
domain is all real numbers greater than 0
derivitive is
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0
deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)
so
dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))
domain is all real numbers greater than 0
derivitive is