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A 6 foot tall man is walking towards a 24 -foot lamppost at the rate of 8 feet per second. At what rate is the tip of his shadow moving towards the lamppost when he is 5 feet from the lamppost?

$32 / 3$ ft/sec

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

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University of Michigan - Ann Arbor

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Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

04:54

A 6-foot-tall man walks aw…

05:26

A man standing 3 feet from…

05:41

A moving shadow $A$ man 6 …

02:02

Moving Shadow A man 6 ft t…

04:17

Moving shadow A 5-foot-tal…

06:28

A 6 -ft tall man is walki…

03:46

A man 6 feet tall walks at…

02:12

A man $6 \mathrm{ft}$ tall…

05:22

A streetlight is on top of…

for this problem. We have a lamppost now. It's always important to start by drawing pictures with related rates. If you can. Doesn't have to be a great picture. As you will see, I'm not a good artist, but it's enough to see how things relate. So we have a lamppost and it's casting shadows on the ground. And we have a man who's walking. He is casting a shadow behind him. Okay, let's see what we know. What we're trying to find will put as much information on our picture as we can were told that the lamppost is 24 ft tall and the Manus 6 ft tall. Those numbers are not changing. The lamppost isn't growing and shrinking, so I can put those constants right on my picture Now. There are some things, though, that are changing. I have a distance. He's moving toward the lamppost so I can call that X, and I'm looking at how his tip this shadow is moving toward the light so I can call this piece here. Why? And that's changing as well as he moves, the tip of his shadow is moving. So this piece where his shadow is I could just call that. Why, minus X. Okay, what numbers do I know? I know that he's walking toward the lamppost. That tells me DX DT is shrinking. It's a negative change. Negative 8 ft per second and I want to know at what right? The tip of his shadow is moving. So I want to know how Why is changing when he is 5 ft from the lamppost. So that's my unknown. And when X is 5 ft 5 ft away, what are we going to dio? Okay, so we have our picture. We've written down the information. Now it's time to go about actually solving the problem. First, I need an equation that relates thes different variables to each other. Well, one thing I do have here I have some nested right triangles, thes air, similar triangles so I can compare the sides to each other. For example, I could take the bigger one here. I could say that the height compared to the length for the big triangle iss Samos the height compared to the length for the second triangle I cross multiply. We get that and I'm going to simplify this a little. I'm gonna divide both sides by six. So this gives me for why minus four X equals Y or three y equals four x. Okay, Now let's take our derivative. Okay. On the left hand side, I will have three d Y d t on the right. I will have four DX DT. Now let's substitute in our particular pieces of information. I know three. You know that we're given that D Y d t is my unknown DX DT is negative eight. So when I go to solve this, I get negative 32/3. So that tells me this negative tells me that it's decreasing, which makes sense as the man moves toward the lamppost. The tip of his shadow is going to be moving toward the lamppost as well, and it's moving at a rate of negative 32 3rd feet per second.

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