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First we take the derivative of the vector using the product rule

$\frac{\partial}{\partial t}(e^{at}\cos{bt},e^{at}\sin{bt},e^{at}) = (ae^{at}\cos{bt} - be^{at}\sin{bt},ae^{at}\sin{bt}+be^{at}\cos{bt},ae^{at})$

Now we will find the magnitude of the result

We begin by squaring the i terms.

$(ae^{at}\cos{bt} - be^{at}\sin{bt})(ae^{at}\cos{bt} - be^{at}\sin{bt}) = a^2e^{2at}\cos^2{bt} - 2abe^{2at}\sin{bt}\cos{bt}+b^2e^{2at}\sin^2{bt}\,$

And now we square the j terms.

$(ae^{at}\sin{bt}+be^{at}\cos{bt})(ae^{at}\sin{bt}+be^{at}\cos{bt}) = a^2e^{2at}\sin^2{bt}+2abe^{2at}\sin{bt}\cos{bt}+b^2e^{2at}\cos^2{bt}\,$

Now squaring the k terms give

$a^2e^{2at}\,$

Now we add all three results and use the identity $\sin^2 x + \cos^2 x = 1 \,$

$\sqrt{a^2e^{2at} + b^2e^{2at} + a^2e^{2at}} = \sqrt{e^{2at}(2a^2+b^2)} = \sqrt{e^{2at}}\sqrt{2a^2+b^2}$

now we integrate the result

$\sqrt{2a^2+b^2}\int_0^t \sqrt{e^{2av}} \partial v = \frac{\sqrt{2a^2+b^2}}{a} \sqrt{e^{2at}} = s(t)$

now we isolate for t.

$s = \frac{\sqrt{2a^2+b^2}}{a}\sqrt{e^{2at}}$

$s^2 = \frac{2a^2+b^2}{a^2}e^{2at}$

$2 \ln{s} = \ln{\frac{2a^2+b^2}{a^2}}2at$

$t = \frac{1}{a}\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}$

And finally we plug this result into the original equation.

$x(t) = e^{\Bigg(\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)}\cos{b\Bigg(\frac{1}{a}\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)}i, e^{\Bigg(\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)}\sin{b\Bigg(\frac{1}{a}\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)}j,e^{\Bigg(\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)}k$

But that's ugly and hard to read so here's the ln of the result

$\ln{x(t)} = \Bigg(\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg) \ln{\Bigg(\cos{b\Bigg(\frac{1}{a}\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)}}\Bigg) i, \Bigg(\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)\ln{\Bigg(\sin{b\Bigg(\frac{1}{a}\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)}}\Bigg) j, \Bigg(\frac{\ln{s}}{\ln{\frac{2a^2+b^2}{a^2}}}\Bigg)k$