# Question fdebb

Sep 11, 2016

I got $r = \sqrt{{a}^{2} + {b}^{2}}$.

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({e}^{a x} \cos \left(b x\right)\right) = a {e}^{a x} \cos \left(b x\right) - b {e}^{a x} \sin \left(b x\right)$

$= {e}^{a x} \left[a \cos \left(b x\right) - b \sin \left(b x\right)\right]$

And

$r {e}^{a x} \left[\cos \left(b x + {\tan}^{-} 1 \left(\frac{b}{a}\right)\right)\right] = r {e}^{a x} \left[\cos \left(b x\right) \cos \left({\tan}^{-} 1 \left(\frac{b}{a}\right)\right) - \sin \left(b x\right) \sin \left({\tan}^{-} 1 \left(\frac{b}{a}\right)\right)\right]$

Now, use trigonometry to get

$\cos \left({\tan}^{-} 1 \left(\frac{b}{a}\right)\right) = \frac{a}{\sqrt{{a}^{2} + {b}^{2}}}$ and $\sin \left({\tan}^{-} 1 \left(\frac{b}{a}\right)\right) = \frac{b}{\sqrt{{a}^{2} + {b}^{2}}}$

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))]#

$= \frac{r {e}^{a x}}{\sqrt{{a}^{2} + {b}^{2}}} \left[b \cos \left(b x\right) - a \sin \left(b x\right)\right]$

Setting the derivative above equal to this expression we get

${e}^{a x} \left[a \cos \left(b x\right) - b \sin \left(b x\right)\right] = \frac{r {e}^{a x}}{\sqrt{{a}^{2} + {b}^{2}}} \left[b \cos \left(b x\right) - a \sin \left(b x\right)\right]$.

Which leads immediately to $r = \sqrt{{a}^{2} + {b}^{2}}$.