While flying a kite, 400 ft of string was let out. The string ,when pulled tight, formed a 40 angle with the ground .What was th
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Answer:
a. 18.5 degrees
b. 58.5 degrees
c. 24 hours
d. T(t) = 58.5 + 18.5cos(πt/12)
e. predicted: 53.7°, about 1.7° high
Step-by-step explanation:
No doubt your curriculum materials have examples of problems of this sort. You would do well to follow an example.
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There are several ways to go about making a sinusoidal model of this data, depending on the error you are willing to tolerate. The data can be modeled exactly using the result of a discrete Fourier Transform, but we suspect that is pretty far beyond the intended scope of the problem.
The easiest model to make is one that matches the extremes of the data, so that is what we'll describe here.
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a. The amplitude of the sinusoidal function can be computed as half the difference between the maximum (77) and minimum (40). So, the amplitude is ...
A = (77 -40)/2 = 37.2 = 18.5
The amplitude is 18.5 (degrees).
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b. The vertical shift can be computed as the average of the maximum and minimum:
D = (77 +40)/2 = 117/2 = 58.5
The vertical shift is 58.5 (degrees).
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c. The data is presumed to be periodic with a period of 24 hours.
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d. The temperature at t=0 is a maximum, so can be modeled best using a cosine function (which has a maximum when t=0). Using these parameters, the model is ...
T(t) = D + Acos(2π/period·t)
T(t) = 58.5 + 18.5cos(πt/12)
If you insist on a sine (not cosine) model, the sine function is the same as the cosine function when it is shifted left by 1/4 period.
T(t) = 58.5 +18.5sin(π(t+6)/12)
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e. 10 a.m. corresponds to 7 hours before t=0, so is t = -7 or +17.
T(17) = 58.5 +18.5cos(17π/12) ≈ 58.5 +18.5(-0.2588) ≈ 53.7
The predicted temperature at 10 a.m. is 53.7 °F. The actual value was 52 °F, so the prediction was a little high.
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The graph shows a plot of the given data points and the sinusoidal model we created. (It only matches the given data at the two extreme values used to create the model.)
