Question #fdebb

1 Answer
Sep 11, 2016

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)#.