Refer to Exercises 81 and 82 . A dermatologist recommends protection from the sun when the inten

Refer to Exercises 81 and 82 . A dermatologist recommends...

Key Concepts

Mathematical Modeling of Natural Phenomena

This concept involves representing real-world processes, such as the variation of sunlight intensity throughout a day, using mathematical functions. In many cases, periodic functions (often sinusoids) are used to mimic the rise and fall of the sun’s intensity over time, allowing predictions and calculations about the behavior of the system under different conditions.

Threshold Analysis

Threshold analysis is the process of determining the intervals during which a modeled function exceeds or falls below a certain value. In applications like this one, a specific percentage of the maximum value is set as a threshold to decide when certain actions, such as applying sun protection, are needed. This requires solving inequalities involving the function to identify the corresponding time intervals.

Numerical Approximation Methods

When exact analytic solutions are difficult or impossible to obtain, numerical approximation techniques are used to estimate solutions to equations. These methods help in approximating the times at which a function exceeds a given threshold, by either graphically analyzing the function or employing methods like iteration or interpolation.

Comparative Analysis of Functional Models

This concept involves comparing how different modeled scenarios, such as varying weather conditions (e.g., clear versus cloudy days), affect the behavior of the function. By analyzing the differences in model outputs, one can assess how changes in parameters or environmental conditions lead to different durations for which a specified threshold is exceeded.

Transcript

00:04 All right.

00:05 So this particular problem actually refers to a couple previous problems.

00:10 So i have written down some information that we need.

00:15 Okay.

00:16 We have the intensity of the sun on a clear day.

00:21 Okay.

00:21 That's this top equation.

00:24 And then the intensity of the sun on a cloudy day, which is that second equation.

00:30 So i stands for the intensity of the sun.

00:33 I sub -ed m is the maximum intensity of the day.

00:39 Okay.

00:40 T stands for the number of hours.

00:43 And d stands for the number of hours of daylight in the day.

00:47 Okay.

00:49 So basically what we're looking at here is dermatologists recommend that we need protection from the sun when the intensity exceeds 75 % of the maximum.

01:04 Intensity over, oh, i did something.

01:16 When the intensity exceeds 75 % of the maximum of intensity, okay, over the 12 hours of the daylight.

01:26 Okay.

01:26 So we're going to approximate the number of hours for which we need protection on a cloudy day as well as on a clear day.

01:38 Okay, well, clear day first and then a cloudy.

01:41 So i'm going to go ahead and start with the clear day, and i'm going to do the clear day in blue just so that it looks different than when i get to the cloudy.

01:51 Okay, so i'm going to start by substituting these values that we know into the equation.

02:03 So 0 .75 of the maximum is going to be equal to the maximum intensity times the sine cubed of pi times t over 12.

02:26 Okay.

02:27 So now i'm just going to solve this.

02:31 So because i have this, i am, if i divide both sides by the maximum intensity, i'm going to be.

02:37 Be left with 0 .75 equals the sine cubed of pi times t over 12.

02:48 All right.

02:50 So i will take the sine in, or sorry, i will take the cube root of both sides to start.

03:00 So i get the cube root of 0 .75 plus the sign of.

03:11 Pi times 2 over 12.

03:13 And i really like to try to get my answers as close to exact as possible.

03:20 And so for that reason, i am going to go ahead and just take this on inverse of both sides instead of approximate this radical.

03:34 You can approximate it.

03:36 If you do, i would suggest going out to like 12 or.

03:42 Or it's 12, four decimal places is usually sufficient.

03:47 Okay, so i'll do the sine inverse of the cube root of 0 .75.

03:53 And then my last step will be to multiply both sides by pi over 12.

04:00 Or sorry, by 12 over pi...

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