Seasonal Sales The monthly sales (in millions of units) of snow blowers can be modeled by S=15+6

Seasonal Sales The monthly sales (in millions of units) of...

Key Concepts

Sinusoidal Modeling

Sinusoidal modeling involves using sine and cosine functions to represent periodic phenomena. This approach is useful for capturing data that exhibits regular fluctuations over time, such as seasonal variations, by incorporating amplitude, frequency, and phase shift parameters that adjust the behavior of the sinusoidal curve.

Average Value of a Function

The average value of a function over a given interval is found by integrating the function over that interval and then dividing by the length of the interval. This concept is a fundamental tool in calculus that helps in summarizing the overall behavior of functions, particularly when dealing with fluctuating or periodic data.

Definite Integration

Definite integration is the process of calculating the integral of a function over a specific interval. It is a key technique in calculus used to compute total accumulation, average values, and areas under curves, and plays an essential role when analyzing functions that model real-world phenomena.

Properties of Trigonometric Functions

Trigonometric functions, such as sine and cosine, have well-known properties including periodicity, symmetry, and average values over complete cycles. These properties are often exploited in problem solving to simplify calculations, particularly when integrating over intervals that encompass whole periods of the function.

Transcript

00:01 So in this problem, we are given an equation that demonstrates the monthly sales of snowblowers.

00:07 And this question in part a is asking how many snowblowers are sold during an entire year.

00:13 So we need to add up all of the snowblowers sold in january and february and march and april all the way through december.

00:19 So the best way to do that would be to find the anti -derivative, aka the area under the curve, aka the sum.

00:26 So we're going to go ahead and evaluate this equation.

00:29 I split it into two anti -derivatives because there's a plus sign between the 15 and then the trache function.

00:34 And then we're going to evaluate this antiderivative using the first fundamental theorem of calculus between january and december.

00:42 So let's go ahead and get started.

00:43 The anti -derivative of 15 is just 15.

00:47 Oops, they don't actually need to write this symbol anymore.

00:50 Is 15t? and like i said, that's going to be evaluated between 0 and 12, which they are telling us in the point.

01:00 Problem.

01:02 Plus, now when it comes to this trig function, this is sign of this whole thing in here.

01:12 And so if we were to take the derivative of this, it would require chain rules.

01:17 So now when we're taking the antiderivative, we need to use use substitution because we know what the antiderivative of like sine of x is, but it gets a lot more complicated when you have sine of 2x, or in this case, sine of pi over six times the quantity x minus eight.

01:33 And so in order to deal with that more complicated step, we go ahead and use what we call use substitution.

01:40 So i'm going to come over here and make that long thing equal to you.

01:51 So i can come down here and rewrite this.

01:54 This is now 6 sine of u d t.

02:03 So i've taken care of the fact that now i know what the antiderivative of sign of u is, but i'm still multiplying by d t, and so my variables at the moment don't match.

02:16 So i need to rearrange some things over on the right here to get d t in terms of d u instead.

02:24 I need my variables to match.

02:26 So i'm going to go ahead and take the derivative of u, which is d u over d t, and then this is really pi over 6t, minus 8.

02:41 Pi over 6.

02:43 So the derivative of pi over 6 t is just pi over 6 and then the derivative of 8 pi over 6 that's a constant term so the derivative of that is 0.

02:54 Like i said i want to replace the d t with a du so i need a equation that starts with d t equals so i can do a direct substitution rearranging this equation up here i can cross multiply and i get d t is equal to 6 over pi du.

03:17 So on this line, i'm going to be rewriting this 15t a couple more times.

03:32 Dt is the same thing as 6 over pi du, so i can do a direct replacement here.

03:41 And now i have my trig in terms of something i know how to take the anti -derivative of, and i have matching variables.

03:49 So i can go ahead and do the anti -derivative now.

03:58 I'm going to bring the coefficients out in front, and so 6 times 6 over pi is 36 over pi.

04:07 The anti -derivative of sine is negative cosine, so i'm going to make that a negative.

04:16 And again, this is being evaluated between 0 and 12.

04:21 Now that 0 and the 12 is in terms of t.

04:24 It is not in terms of u.

04:26 So we need to plug this value back in.

04:52 Now everything is matching as it should be.

04:55 Let's go ahead and finish the algebra.

04:57 I'm going to use the first fundamental theorem of calculus, which means i plug in the larger number, in this case 12, and then subtract off the smaller number.

05:06 So we can do 15 times 12, minus 15 times 0, minus.

05:26 Oops.

05:27 I'm going to write that 36 over pi outside...

Please give Ace some feedback