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For the case described in the previous exercise, find the rate at which the length of the man's shadow is changing. Is it getting longer or shorter?

$$8 / 3 \mathrm{ft} / \mathrm{sec}, \text { shorter }$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Oregon State University

Baylor University

Idaho State University

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.

03:14

A man of height $1.8 \math…

05:41

A man of height 1.8 meters…

07:44

Shadow Length $A$ man 6 ft…

02:12

A man $6 \mathrm{ft}$ tall…

02:02

Moving Shadow A man 6 ft t…

for this problem. We once again are looking at our lamppost. So let's draw our picture. Here's our lamppost and here's our man again. Just like last problem again. This is referencing 23. So when I say last problem, that's the one I'm looking at. Just like the last problem. The light post is 24 ft tall and the Manus 6 ft tall. Those numbers are not changing their constant, so I could put them right on my picture. What are changing are the numbers on the ground because he's walking toward the lamppost. So that's changing X, and it's changing. Why? So what do we know about these rates? Well, we know that he's walking toward the lamppost, so that means DX DT is shrinking, so it's gonna be a negative number. He's walking at a rate of 8 ft per second, and I want to know, um, the rate at which the length of the man shadow was changing. Well, his shadow is why there's his shadow. So I want to know how Why is changing. There's my unknown. That's what I'm looking for. And I want this, um, just in general. At what? Yep, At What rate is his shadow changing now, just like in the last example if we need it. We were looking at the point when he's 5 ft from the lamppost. Okay, so if we need it, that's where we're looking at. So what Equation puts X and Y together? I need to know how they relate to each other. Well, thes are similar triangles, so I can say that the height of the Big Triangle, compared to the height of the small triangle, is the length of the Big Triangle X plus. Why, compared to the length of the small triangle now, it's tempting to say that's X over Y. But remember, you want the whole length, so it's X plus y. I could do a little bit of simplifying here. Just keep my number small like what? Six. Into both of those. So when I do my cross multiplication, I get four. Y equals X plus y. Or I could say this is three y equals X. So there's my equation. That's how these relate to each other. So how do we, um, solve it from here? Well, we have our equation. The next thing we do is take the derivative. The derivative of the left is three d y d x. I'm sorry d Y d t helps fight right that correctly d Y d t. And that's going to equal the derivative of what's on the right, which is just DX DT. Now let's solve What do we know? Well, I know that d Y d t is what I'm looking for. DX DT is negative eight, which means I can solve for D y d t as negative eight thirds so d Y d t. The length of his shadow is decreasing. That's what that negative means at eight thirds feet per second.

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