My method: I did |}{dt}=0)
and ended up with \left(sin(2bt)\right)=0)
I figured that 
and 
.
Subbing it in, I got |_{max}=\pi\sqrt{2})
Am I right as well?
your fallacy is in that when you solve for
, a and b are constants, the only variable is t. Hence the solution is
Substituting that into
|)
| = \sqrt{a^2b^2\cos^2 (n\pi/2) + b^2 \sin^2(n\pi/2) + b^2})
When n is even (n=2k), we have
 = \pm 1 \implies \cos^2(k\pi) = 1)
,
 = 0)
, thus
| = \sqrt{a^2b^2 + b^2} = \sqrt{(a^2+1)b^2} = b\sqrt{a^2+1})
When n is odd (n=2k+1), we have
 = 0,\; \sin(n\pi/2) = \pm 1)
, thus
| = \sqrt{0 + b^2 + b^2} = \sqrt{2}b)
both of these are maximum speeds, so we need to determine which one is greater. This involves restraints on the constants:
if
, hence if a>1 (since it is non-zero positive), the first expression is valid. Otherwise the second expression will be greater.
So your method does work, but you solved for the wrong thing
And also, the method shown in the exam solutions (simplification of trig terms in the sqrt) is an easier method than this. Especially since questions in the actual exam won't be nearly as complicated.
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