Assignment
University
Rice UniversityCourse
Preparing for the AP Calculus AB ExamPages
2
Academic year
2022
Deskov
Views
30
Related Rates. Problem Example 1 Let us now consider the situation where a man walks at a constant rate of speed either toward or away from a light source. The single light source is a certain height above ground. The problem is to determine how fast the shadow grows as a person moves away from the light source. The distance from a light source is not an important variable in determining the intensity of light on an object. In this drawing, the artist has created a ground and a man who is 6 feet tall. The light source is 20 feet above the ground. The light source is 14 feet above the man's head. He is walking toward it at five feet per second. The shadow of his head is projected onto the ground 5 feet ahead of him by the angle of light. The man is 5 feet away from the light source. If he were walking toward it at 5 feet per second, the shadow on his head would fall 5 feet behind him, and therefore be 14 feet above his head. And so the reason I said the distance from the light source doesn't matter is because this distance is changing. It is a variable. So, you should label this 10 as some variable that we'll just call x. We'll call the length of the shadow y, and x will represent the distance from the light source. The length of the shadow cast by an object (y) is determined by its distance from a light source (x). As we explained earlier, we are taking the derivative of the length of the shadow to find the rate of change of that length. However, since the pythagorean theorem is so cumbersome to apply, we will instead examine triangles with sides of known lengths. Simply, we need to work with more than 1 triangle. In such a situation – with 3 of them.
There are three triangles in this problem. The sides of these triangles are all proportional to each other, allowing you to create proportional triangles based on the whole large triangle and one of the smaller ones. You could look at the entirety of this problem, but it's also possible to set up a proportion with two different triangles. In which case, you can set up the proportion 20/x x plus y is equal to another triangle, which you choose. You may choose to draw the figure from the top of the head to the light source. In that case, you would have a height of 14 and a base of x. Or you might choose to draw the triangle created by the person and his shadow. In that case, the height is 6 and the base is y. I prefer to use these two right here when I create my problem. The derivative can be taken at this point, but I want to clean up the notation a little bit first. We have two variables, 14 and x. By doing a little multiplication, 14y is equal to 6x. That looks a lot better. Now, I will go ahead and take the derivative with respect to time. Thus, we need to identify two items. The rate of change of the man's shadow is dx dt. Since the rate of change of the man is known, dx dt must be equal to that. The man is walking at a rate of 5 feet per second. So the differential equation for the distance traveled, dx dt, is 5. Thus, integrating both sides of the equation yields 14 dy dt = 30. The solution to this equation is dy dt = 30/14, which reduces to 15/7 when divided by 2. Shadow questions are a great way to help students remember the proportions of a right triangle.
Problem Example 1. Related Rates
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