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The area of a square is increasing at the rate of 1 square inch per minute. At what rate is the: (a) side of the square changing; (b) diagonal changing; when the side is 6 inches long?

(a) $1 / 12$ in $/ \min$(b) $\frac{\sqrt{2}}{12} \approx 0.1179$ in $/ \mathrm{min}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

Campbell University

Oregon 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.

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The sides of a square decr…

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Each side of a square is i…

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for this problem. We're considering a square. And like all related rate problems, it's always a good idea to start with the diagram if you can. So I have a square and I'm going to call the sides of my square X And I know some rates. I know some values, anything that's constant I can put on my picture. If it's a rate or a varying number, I'm gonna put it beneath. So what do we know? The area is increasing. Well, that's the change of area. With respect to time, it's increasing at the rate of one square inch permit. It's an increase. This is a positive one now. First thing I want to know is I want to know what the rate is at. The side of the square is changing. So that's my unknown. How is the side of the square changing when, um, the side is six inches long? Okay, Now, do I put that on our square or not? And the answer is no, I'm not gonna put I'm gonna put that over here. That my side's gonna be six inches. The sides are changing, so I don't wanna put, uh, one instance of that Teoh on the picture. I want to show it with a variable to show that that really is a varying amount. If it was a constant number, I put it on. So how do we go about doing this? Our first step, I need a function, something that relates area and the sides Well, Fortunately, the area of a square is X squared. It's a nice, simple formula relating the areas to the sides. Things has two variables, so I should have a rate to go with each one of these. I know d a d t I'm trying to find the x d t. We're good to go now that I have my formula. Let's take the derivative that gives me d a d t equals two x dx DT. And now we can substitute and solve. I know d a. D T equals one X is six dx DT is what we're trying to find. I want to divide by 12. I end up with 1/12 and the units here have been working in is inches per minute. So that's how fast the side is changing at that particular moment with respect to time. Now, let's look at the same square again. Axes going thio equal six at the instant that I'm looking at it. But I'm looking at something else Now. I want to know what the rate is of the diagonal changing. So let's draw in a diagon off. We'll call it Capital D. So I want to know here. How fast is that Diagonal changing with respect to time. Okay, so how do we go about finding this? Well, we need a new formula. So we're back to step one because the formula we used last time had nothing to do with distance. But I do know how distance and the size of related that's a right triangle. So I can say that X squared plus X squared is D squared or two X squared equals d squared. And I can use that to help me come up with my area because my area is X squared. Well, X squared is d squared, divided by two. So now I have a formula that's not area inside but relating area and diagonals. Now let's take the derivative of this new formula. I get d a d t. I get. I bring down that to it cancels with the two in the denominator that just gives me a d times d d d t Now that I've taken the derivative weaken substitute so d a D t is still one inch per minute Now what's d? Well, I know that X is six. So here, um if I simplify this down just a little bit, I take the square root of both sides. I get the square root of two times X equals d. I know Xa six So d is six times the square root of two. So when I go to plug that in, I'm gonna have six times the square root of two d d d t. Okay, divide both sides, and I don't like to have that radical in the denominator, So I'm just gonna multiply top and bottom, and that's going to give me square root of 2/12, and we again, we are in inches per minute. So that is D D d t at that particular point

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