# precalculus applications

How do I find the equation of a sinusoidal curve from just knowing the lowest point and highest point of it?

For example:
A weight attached to the end of a long spring is bouncing up and down. If we do not worry about friction, its distance from the floor varies sinusoidally with time. You start a stopwatch. When the stopwatch reaches 0.3 seconds, the weight reaches a highpoint for the first time at 60 cm. above the floor. The next time the weight reaches a lowpoint is 1.8 seconds later and that lowpoint is at 40 cm. above the floor.

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1. The distance between the high and low points is 60-40 or 20 cm, so the amplitude of the curve will be 10

The time between the high and the low points is 1.8-.3 or 1.5 seconds, so the period of the curve is 3 seconds.

If we had y = 10cos(2pi/3)x we would fulfill those two conditions, but our first highpoint would be at 0 sec.
So we need a horizontal shift of .3 sec to the right.
that would be 9/(20pi)

so lets see what we have so far

y = 10cos(2pi/3)[x - 9/(20pi)]

(check: phase shift = (2pi/3)*9/(20pi) = .3)

Finally we have to move the curve vertically so that its centre line runs along y = 50, the midway between 40 and 60

so we would have

y = 10cos(2pi/3)[x - 9/(20pi)] + 50

You could do the same thing with a sine curve, but the phase shift would be different. I chose the cosine curve because .3 was closer to the y-axis, and the cosine curve starts with y=1 at x = 0, so my shift would be smaller.

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2. yzax dwme xqkw brmexchi zlwsva jexs tvrheqpk

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