for y=f(x), the unit tangent = v/|v| where v= (1,f\'(x),0) and |v|=sqrt(1+f\'(x)
ID: 1891627 • Letter: F
Question
for y=f(x),the unit tangent = v/|v| where v= (1,f'(x),0) and |v|=sqrt(1+f'(x)^2) - prove that T = f'(x)
Explanation / Answer
First, what do you mean by a "complex function"? My first thought was "a function that returns complex numbers", but the set of complex numbers includes the real numbers. Do you mean "a function that returns, for some (x, y), non-real values"? Yes, thank you. I don't know if f(x,y) returns a real or complex value. So I think that means I cannot evaluate its complex conjugate. I think I'm looking for some algebraic properties unique to complex numbers (that include the reals) that I can use to prove my function is complex(including reals). Maybe something like f(x,y)?f(y,x), but f2(x,y)=f2(y,x), or something like that? Or maybe I can somehow show that there is no sense in saying f(x,y)>0? y' = t(1+y^2) Euler's method for y' = f(y,t): y(t+dt) = y(t) + dt f(y(t),t), dt = 0.2 y(0.2) = y(0) +0.2*[ 0*(1+y(0)^2) ] = 1 y(0.4) = y(0.2) + 0.2*[ 0.2*(1+y(0.2)^2) ] = 1 + 0.2*0.2*2 = 1.08 y(0.6) = y(0.4) + 0.2*[ 0.4*(1+y(0.4)^2) ] = 1.08 + 0.2*0.4*(1+1.08^2) = 1.2533 Analytic solution: dy / (1+y^2) = t dt ==> atan(y) = t^2/2 + C ==> y(t) = tan(t^2/2 + C) , y(0) = 1 ==> 1 = tan(C) ==> C = pi/4 ==> y(0.6) = tan(0.6^2/2 + pi/4) = 1.445
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