Question

Question #fdebb

Answer 1

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

Explanation:

`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)`.