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/sci/ - Science & Math

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Without resorting to calculus to find min/max, how could one determine that the correct plot is (b) and not (c)?

Additionally, how could one determine that (c)
is incorrect (again, without resorting to calculus to find min/max)?
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It shouldn't rise any higher than its first crest.
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>>9218953
What are the local extrema of sin(x)?
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>>9218953
sin has a max of 1 so the max of 1^2 is 1 so it wouldn't rise like in c
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>>9218953
sin attains a max value of 1 so the max value its square could have is also 1.
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>>9218953
You know it is not a) because a takes negative values.
You know it is not c) because it grows beyond it's first maximum.

Proving it can't be d) is the most tricky as you finally need to get into what the function looks like. We need to find the first maximum of the function. It is clearly $x = \frac{\pi^2}{4}$. Then the second max would happen at $x = (\frac{\pi}{2} + \pi)^2$ and the next at $x = (\frac{\pi}{2} + 2 \pi)^2$. So you can clearly see that the maximums of the curve are not equidistant, which contradicts d.
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It can't be c because a sin would have the same max everywhere, it can't be d because sqrt(x)'s derivative is decreasing.
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>>9218953

Sin^2 X = (1 - cos 2x)/2 by the power reduction formula.
SQRT X can't take negative values, but there is nothing that would make the amplitude dampen or change max.

Without knowing anything else, it must be B because the period changes but the amplitude doesn't.
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>>9218953
>how could one determine that the correct plot is (b) and not (c)?
The image of the sin^2 function is [1,0] since the sqrt() function is strictly monotonous (and its image is [0,\infty)) it is obvious that the amplitude needs to be always one.
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The squaring makes the range non-negative which eliminates A.

sin has a maximum of 1 and so too does sin squared. Therefore there is a uniform max height eliminating C.

The zeroes occur at root x = pi, 2pi, 3pi etc. This gives the zeroes at x = pi^2, 4pi^2, 9pi^2 etc. Therefore the zeroes are not uniformly distributed eliminating D.

Therefore the answer is B.
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Amlitud is constant, it's positive, the period is not.
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>>9218953
Conversion to Unit Circle | Rotational Symmetry
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>>9218953
>how could one determine that the correct plot is (b) and not (c)

sin^2(x) is bounded by 0 and 1 killing everything but (b) and (d)
pi*n = sqrt(T) means the period is T=(pi*n)^2 so it increases quadratically rather than linearly killing everything but (b) and (c)

The intersection is (b)
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Because we're squaring the sin, it must be strictly positive. We can ignore A.

The square root function is strictly increasing, and sine squared is cyclic. We can immediately ignore C because sine squared will always have peaks at 1.
---------------------------

So, that leaves us B and D.

We can think of the square root function as providing the "crank" by which we're turning the sine function. If it were just X, that would mean we get the regular sine squared function, which we know will look something like D.

Graphically, you can look at the square root function and notice that it starts off rapidly increasing close to 0, then tapers off and grows ever-slower as X goes. So, we can think of this as rapidly turning the crank for sine, then getting slower as we go on. That means we can expect the frequency to start out high-ish, and get lower as time goes on. Which is exactly the behavior we see in B.
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>>9218953
>Without resorting to calculus to find min/max, how could one determine that the correct plot is (b) and not (c)?
you cannot as the axis ticks are not given
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>>9220683
$\arg (\sin (z+i y))=\frac{1}{2} \pi \text{sgn}\left(\text{sgn}(\sinh (y) \cos (z))+\frac{1}{2}\right) (1-\text{sgn}(\cosh (y) \sin (z)))+\tan ^{-1}(\tanh (y) \cot (z))$
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>>9220692
If we can't use calculus, we sure as hell can't use complex analysis. And the arg function is straight out of the complex analysis toolbox.
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>>9220706 I was translating >>9220683

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