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The legs of an isosceles right triangle increase in length at a rate of $2 \mathrm{m} / \mathrm{s}$a. At what rate is the area of the triangle changing when the legs are 2 m long?b. At what rate is the area of the triangle changing when the hypotenuse is $1 \mathrm{m}$ long?c. At what rate is the length of the hypotenuse changing?

A. $=4 \mathrm{m}^{2} / \mathrm{s}$ B. $=\sqrt{2} \mathrm{m}^{2} / \mathrm{s}$ C. $=2 \sqrt{2} \mathrm{m} / \mathrm{s}$

Calculus 1 / AB

Chapter 3

Derivatives

Section 10

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here we have And I saw sleaze. Right? Triangle. Okay, so, beings, it's I saw Seles A and B are equal and we're told that the side lengths or the legs are changing at a rate of two meters per second. So I'm just gonna select leg be and label that as DBT T is equal to two meters per second. Okay, What we're trying to find out here is how fast is our area changing When that leg is two meters long. So we need the formula for the area of a triangle So a equals 1/2 base times height. Well, I don't have a height here. What? I know that my base is be and my height is a Okay, so now I know that a is equal to be so I'm gonna substitute of be in for a and that's going to give me area is equal to 1/2 b squared to take the derivative of both sides with respect to t. And then I'm gonna substitute in. What I know so be is to DBT tea is too. So our area is changing at a rate of four meter squared per second when our side length is two meters. Next thing we want to find out is how fast is our area changing when C is equal to one so I can go ahead and start with this formula. So d a d t. Is equal to what is be. You have to figure that out. We know that C squared is equal to a squared plus B squared. And since a is equal to B, C squared is equal to two b squared. Okay, so C squared is one. So 1/2 is equal to B squared B is equal to one over route to No, we usually rationalize, But for right now, we're just gonna leave that because we're not at the end of our problem yet. So we're gonna multiply. This is gonna be to over route to We can go ahead and rationalize that to root Teoh over to de a t T is equal to fruit till and that is meters squared per second. Lastly, what we want to know is how fast is our high pot news changing when C is equal to one. So we're gonna go ahead and use our Pythagorean theorem formula that we already figured out C squared is equal to two b squared and we're gonna take the derivative of that. So to C dcd t is equal to four b d b d t. We're gonna plug in some things here. We have two times while we no see is one dcd tea is what we're trying to find. Four. We also know that B is one over route to and DBT tea is too. We're gonna go ahead and move up here a little bit, so I have to dcd t is equal to eight over route to I'm gonna divide out the to D C D T is equal to eight over to root two When we reduce and rationalize that we end up with d C D T is equal to to root two meters per second

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