Question

Suppose that the monthly average precipitation at a particular location varies sinusoidally between a maximum of 5 inches in March and a minimum of 2 inches in September. Let t = 1 correspond to January and t = 12 correspond to December. (a) Use $f(t) = a \cos (b(t - c)) + d$ to model these conditions. (b) Evaluate $f(3)$ and $f(9)$. (a) Estimate the values for the constants a, b, c, and d to substitute in the given model. $f(t) = \boxed{} \cos \left(\boxed{} (t - \boxed{})\right) + \boxed{}$ (Simplify your answers. Type exact answers, using $\pi$ as needed.) 

          Suppose that the monthly average precipitation at a particular location varies sinusoidally between a maximum of 5 inches in March and a minimum of 2 inches in September. Let t = 1 correspond to January and t = 12 correspond to December.
(a) Use $f(t) = a \cos (b(t - c)) + d$ to model these conditions.
(b) Evaluate $f(3)$ and $f(9)$.
(a) Estimate the values for the constants a, b, c, and d to substitute in the given model.
$f(t) = \boxed{} \cos \left(\boxed{} (t - \boxed{})\right) + \boxed{}$
(Simplify your answers. Type exact answers, using $\pi$ as needed.)
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Suppose that the monthly average precipitation at a particular location varies sinusoidally between a maximum of 5 inches in March and a minimum of 2 inches in September. Let t = 1 correspond to January and t = 12 correspond to December. (a) Use f(t) = a cos (b(t - c)) + d to model these conditions. (b) Evaluate f(3) and f(9). (a) Estimate the values for the constants a, b, c, and d to substitute in the given model. f(t) = cos( (t - )) + (Simplify your answers. Type exact answers, using π as needed.)

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Suppose that the monthly average precipitation at a particular location varies sinusoidally between a maximum of 5 inches in March and a minimum of 2 inches in September. Let t=1 correspond to January and t=12 correspond to December. (a) Use f(t)=acos(b(t-c))+d to model these conditions. (b) Evaluate f(3) and f(9). (a) Estimate the values for the constants a, b, c, and d to substitute in the given model. (Simplify your answers. Type exact answers, using π as needed.) Suppose that the monthly average precipitation at a particular location varies sinusoidally between a maximum of 5 inches in March and a minimum of 2 inches in September. Let t=1 correspond to January and t=12 correspond to December. (a) Use f(t)=acos(b(t-c))+d to model these conditions. (b) Evaluate f(3) and f(9). (a) Estimate the values for the constants a, b, c, and d to substitute in the given model. (Simplify your answers. Type exact answers, using π as needed.)
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Transcript

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00:01 Okay, in this question we have been given this precipitation function, which is the average monthly precipitation in inches for a city.
00:12 Now what we want to do is begin by finding the maximum and minimum precipitation and when it occurs.
00:18 And what's nice is that we don't have to use calculus techniques to do this if you recognise that sine is biggest at 1 and smallest at minus 1.
00:28 So the maximum is going to occur when this sine function, when say sine of x equals 1.
00:37 And sine of x equals 1 when x is pi over 2, 5 pi over 2, and so on.
00:44 You can just keep adding 2 pi.
00:48 So sine of x equals 1 at these points and x in this case is 0 .59t plus 3 .94.
00:57 So what happens when 0 .59t plus 3 .94 equals pi over 2? what value of t do we get? and you should get a value of t of minus 4 .016.
01:12 And we're going to disregard this because that's negative time, we aren't interested in that.
01:17 So let's look at the next solution.
01:18 What happens when 0 .59t plus 3 .94 equals 5 pi over 2? what value of t does that correspond to? and you should find that you get a value of t of 6 .633867 and you can round that however you like.
01:38 Okay so the maximum occurs when t takes on this value.
01:42 And what's the precipitation when t takes on that value? let's call it p max.
01:48 Okay well we don't actually have to put in this value of t into the original equation because we know already that sine is 1 there.
01:56 So p max is just 1 .09 times 1 plus 1 .63.
02:02 So if you type that in, 1 .09 plus 1 .63, you should get 2 .72.
02:15 Okay so that is your maximum and now let's consider your minimum.
02:20 Now we're going to use the same idea here, remembering that sine of x has a minimum at minus 1.
02:27 And that happens when x is 3 pi over 2 and also when x is 7 pi over 2...
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