STEP 2002 Q3
By Alex Holyoake on Friday, August 23,
2002 - 08:58 pm:
Find the stationary points on the curve
y = (a2 cos2 q+ b2 sin2 q)1/2 +(a2 sin2 q+ b2 cos2 q)1/2
where a and b are constants. State with brief reasons which points are
maxima and which are minima.
By Andre Rzym on Sunday, August 25, 2002 -
11:50 pm:
My approach is as follows:
first, square both sides, collect terms and use sin2 q+cos2 q = 1
y2 = a2 +b2 +2[sin2 (q)cos2 (q)(a4 +b4)+a2 b2 (sin4q+ cos4 q)]1/2
Now use cos4q = cos2 q(1-sin2 q), ditto for sin, and
again use sin2 q+cos2 q = 1 to get
y2 = a2 +b2 +2[a2 b2 +sin2 (q)cos2 (q)(a4 +b4 -2a2 b2 )]1/2
But sin(2q)=2sinqcosq so substitute and take roots:
y=[a2 +b2 +[4a2 b2 +sin2 (2q)(a4 +b4 -2a2 b2 )]1/2]1/2
Now we have two approaches to finding the stationary
points:
(i) the squareroot function is monotonically increasing, so assuming a
and b are not both zero, y is stationary when the argument of the outer
squareroot is stationary. But a and b are constants so we are looking for
inner root to be stationary. The same argument tells us that the argument of
the inner root must be stationary, and because a, b are constants, we must
have sin2 q stationary, i.e. at 0, pi/4 etc. You should also be able
to argue for the 0, pi/2 etc being minima, the others maxima.
(ii) Use calculus. But this time it is easy, if we assume a, b not both
zero, because we can cancel lots of stuff as we work through the differential
(make sure you understand why):
0=dy/dq = d/dq[a2 +b2 +[4a2 b2 +sin2 (2q)(a4 + b4 -2a2 b2 )]1/2 ]1/2
0=d/dq[a2 +b2 +[4a2 b2 +sin2 (2q)(a4 +b4 -2a2 b2 )]1/2 ]
0=d/dq[4a2 b2 +sin2 (2 q)(a4 +b4 -2a2 b2 )]1/2
0=d/dq[4a2 b2 +sin2 (2 q)(a4 +b4 -2a2 b2 )]
0=d/dq[sin2 (2q)]
Hope this makes sense,
Andre