Vector question

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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


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

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