Question #fdebb

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Answer 1 · 2016-09-11T12:32:53

I got $r=sqrt(a^2+b^2)$ .

$d/dx(e^(ax)cos(bx)) = ae^(ax)cos(bx)-be^(ax)sin(bx)$

$= e^(ax)[acos(bx)-bsin(bx)]$

And

$re^(ax)[cos(bx+tan^-1(b/a))] = re^(ax)[cos(bx) cos(tan^-1(b/a))-sin(bx) sin(tan^-1(b/a))]$

Now, use trigonometry to get

$cos(tan^-1(b/a)) = a/sqrt(a^2+b^2)$ and $sin(tan^-1(b/a)) = b/sqrt(a^2+b^2)$

So we have

$re^(ax)[cos(bx+tan^-1(b/a))] = re^(ax)[cos(bx) (a/sqrt(a^2+b^2)))-sin(bx) (b/sqrt(a^2+b^2))]$

$= (re^(ax))/sqrt(a^2+b^2)[bcos(bx)-asin(bx)]$

Setting the derivative above equal to this expression we get

$e^(ax)[acos(bx)-bsin(bx)] = (re^(ax))/sqrt(a^2+b^2)[bcos(bx)-asin(bx)]$ .

Which leads immediately to $r=sqrt(a^2+b^2)$ .