Let there be the function #f(alpha)= (L( v_a - v_0 sin alpha))/(v_0 cos alpha)# Where #L# ,#v_a# and #v_o# are constants. Determine #alpha# such that #f(alpha)# is minimal ?

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Answer 1 · 2018-07-05T22:17:58

#f(alpha)= (L( v_a - v_0 sin alpha))/(v_0 cos alpha)#

# = L( v_a/v_o sec alpha - tanalpha)#

#L# is just a scaling factor so it can be removed, thus writing:

To optimise #F(alpha)#, differentiate wrt #alpha#:

#implies underbrace(sec^2alpha)_(gt 0, forall alpha)( v_a/v_o sinalpha - 1 ) qquad = square qquad = 0 #

Critical points can, therefore, only occur under these rules:

# sinalpha = v_o/v_a qquad qquad implies {(v_o lt= v_a),(cos alpha = sqrt(v_a^2 - v_o^2)/v_a),(tanalpha = v_o/sqrt(v_a^2 - v_o^2) ):}#

To determine the max or min nature of the Critical Point, grab the second derivative of #square#, using the product rule:

#F_(alpha alpha) = (2 sec^2alpha tanalpha )( underbrace(v_a/v_o sinalpha - 1)_(= 0) ) + sec^2alpha(v_a/v_o cos alpha)#

#= sec^2 alpha (v_a/v_o cos alpha) = v_a/v_o sec alpha#

#:. F_(alpha alpha) = color(blue)(v_a/v_o) * 1/color(green)(sqrt(1 - v_o^2/v_a^2))#

With assumption #v_o lt= v_a#, then the green term is positive; whole expression is positive provided #v_a, v_o gt 0#.

Accordingly, there is a min when this occurs:

This min will be periodic.

Answer 2 · 2018-07-10T02:42:38

I have solved this way. See the answer below:
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