the-altitude-of-a-triangle-is-increasing-at-a-rate-of-1-cmmin-while-the-area-of-the-triangle-is-incr

Name: the altitude of a triangle is increasing at a rate of 1 cmmin while the area of the triangle is incr
Uploaded: 2018-03-02
Duration: 2 min 7 s
Channel: Numerade
Description: The altitude of a triangle is increasing at a rate of $ 1 cm/min $ while the area of the triangle is increasing at a rate of $ 2 cm^2/min. $ At what rate is the base of the triangle changing when the altitude is $ 10 cm $ and the area is $ 100 cm^2? $

Question

The altitude of a triangle is increasing at a rate of $ 1 cm/min $ while the area of the triangle is increasing at a rate of $ 2 cm^2/min. $ At what rate is the base of the triangle changing when the altitude is $ 10 cm $ and the area is $ 100 cm^2? $

Wen Zheng · Accepted Answer

it&#x27;s a problem. This attitude off a triangle is increasing at reach of one centimeter from minute whilst area off The train, though, is increasing at a reach of two centimeter square per minute at whatever it is the base of the Tran Go changing when altitude is ten centimeters And this area is what handlers and meteor squire So first we have area of a triangle is equal to one half times the base eh? On dams had each and different state each side respect to tea We have He asked that he it&#x27;s unconscionable. We&#x27;LL have terms. Hee hee has a judge Us bitch p And the one altitude is equal to ten area. Is it one hundred? We have a handler is equal to one half have stayed house ten in this case you see twenty Then we have you ass titties. You got to choose to its half ham&#x27;s Hey, did he h And last. The city is one on a twenty. So we have a key. Is the control or minus twenty over hands with this is negative one point six This is the base of the triangle is decreasing at a rate of one point six centimetres per minute

Suman Saurav Thakur · Answer

oh this is a question in which will be using the concept of increasing and decreasing our application of derivative. We should say if this is triangle abc, this is the altitude and this is the base so Area is 1/2 and two. Base into the altitude question number one. Now various things we have been given ah that altitude is increasing at a rate of one centimeter per minute. So D. H by DT is one cm/min. An area is increasing at the rate of two centimeters Squire per minute. So we need to find the raid outfits. The basis is changing when altitude is 10 centimeter and base area and the area is 100 centimeters square. So let us differentiate equation number eight with respect to T so dear by duty is half since be an edge all our variables. So you&#x27;ll be using the concept of the differentiation of the product. So be the H by DT place DB by DT into at Disability is 2 1 x two B. What should be base should be taken as okay. At what rate is the base of the triangle is changing when altitude? Okay no problem. Be into DHB oddity we have one place delivery T. We need to find out and altitude is 10 cm we have to find the value of B when area is and the centimeter and ah Altitude is 10 cm 8800 centimeter square half into base into altitude 10. So the system. So BB. is 20 cm so we&#x27;ll be plugging in value of 20 over here. So this will be to equal to one x 2, 22 and 20 plus DB by detained 2 10. So this will be four Equal to 20 plus and D. V by DT So D V. But it will be cool too, -16 x 10 -8 x five. So the rate of change of the base will be minus 85 centimeter permit on minus one point six centimeter per minute. Now this negative science shows that the base is decreasing with time. Thank you.

Anthony Han · Answer

So here we get an example of a related rates problem. So we have an equilateral triangle where the three sidelines are equal. So for an equilateral triangle we have 60°, 60° and 60° where all the angles are congruent. So we know that the area of a triangle is one half base times height. Let&#x27;s call the length of one side of our equilateral triangle. S. So this part, if we split the triangle in half, This region would be s divided by two. Well this region here is as radical three divided by two. Since we have a special triangle with angles of 33, and 93s. And as a result, area equals 1/2 base which is S. And heights S. Radical three divided by two. And this would be equivalent to Radical three divide by four S. Squared. So this would be the area of our triangle. So we&#x27;re given information that dS over with respect to time Is equivalent to 10 cm permanent. And we&#x27;re also giving information that S. is 30 cm. So for related to rates we have to take the derivative of both sides with respect to time. We have to be careful here since we have to use chain rule. So this would be the result of differentiation. So we know that this is equivalent to radical three times two Times 10 times 30. So this is equivalent to 150 radical three centimetres squared permanence. And this gives our final answer

Amrita Bhasin · Answer

We know the formula for the area of a triangle with based B and high age is a equals 1/2 the age differentiating using the product rule. If I split this up into two components 1/2 B and H using the product rule day over DT, I got 1/2 times D B that indicates base times age plus B times D H you can probably in furnace indicates height changing height so h and B just using those with derivatives. You know dp over. Dt is negative. One d h over dt is five. B is 10 and ages 22 So plugging him, we end up with D A over DT equals 14 centimeters per minute.

J Hardin · Answer

the area of this triangle may be written as one half X Y sign data. Now they&#x27;re telling you that the area of the triangle remains constant with respect to time. So this tells us that if we take the derivative of A with respect to time will get zero. On the other hand, you can use the chain rule here to rewrite the right hand side. Now, since this term in blue is the one we&#x27;re trying to solve for we should go ahead and isolate this equation for the blue term. Yeah, So here, you see, you have three partial derivatives that need to be computers. So let&#x27;s go ahead and find those and the data. So all three of these partial derivatives were all found by using the original area equation. So now going into the next page will replace these terms here in the apprentices with the partial derivatives we just found and notice Also, you&#x27;LL be able to cancel those one half so going on to the next page. Come on, Come. There&#x27;s our numerator and there&#x27;s our denominator. And remember we canceled off the one half. So now we&#x27;re ready. Tto plug in all the information, so this is given from the problem. Status pie over. Six. Why is thirty x is twenty dx over d. T was positive. Three. The idea was minus two. So plugging this all in to our formula we have thirty times have that. That was because of the sign of pie or six. Sometimes three twenty, another one half And then this time of minus two, then we have X And why in the bottom that&#x27;s thirty and a twenty and then co sign a pie over six. Well, three over too. And this all can be simplified. But if you want approximation. And so here we say that data is decreasing at a rate of point zero five radiance per second, and there&#x27;s our final answer.

Daphne Pusey · Answer

here we have And I saw sleaze. Right? Triangle. Okay, so, beings, it&#x27;s I saw Seles A and B are equal and we&#x27;re told that the side lengths or the legs are changing at a rate of two meters per second. So I&#x27;m just gonna select leg be and label that as DBT T is equal to two meters per second. Okay, What we&#x27;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&#x27;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&#x27;m gonna substitute of be in for a and that&#x27;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&#x27;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&#x27;re just gonna leave that because we&#x27;re not at the end of our problem yet. So we&#x27;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&#x27;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&#x27;re gonna take the derivative of that. So to C dcd t is equal to four b d b d t. We&#x27;re gonna plug in some things here. We have two times while we no see is one dcd tea is what we&#x27;re trying to find. Four. We also know that B is one over route to and DBT tea is too. We&#x27;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&#x27;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

Wen Zheng · Answer

in order to figure out the rate. The first thing we know is that h is altitude and abuse base. Looking at the formula 1/2 base times height, we can plug in, let me know 100 centimeters squared his area. I only know H is 10 centimetres. Giving us B is 20 centimeters. Therefore, into the equation D A over DT is 1/2 HDB over DT. So it&#x27;s 10 times D B over DT plus B times D H over DT 20 times one was just 20. Simplify this. We got a DP over. DT is negative 1.6 centimeters her minutes. So clearly the rate is decreasing since native.

Problem

The altitude of a triangle is increasing at a rat…

Problem 21 Easy Difficulty

The altitude of a triangle is increasing at a rate of $ 1 cm/min $ while the area of the triangle is increasing at a rate of $ 2 cm^2/min. $ At what rate is the base of the triangle changing when the altitude is $ 10 cm $ and the area is $ 100 cm^2? $

Answer

$-1.6 \mathrm{cm} / \mathrm{min}$

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$-1.6 \mathrm{cm} / \mathrm{min}$

Calculus

Grace H.

Anna Marie V.

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Kristen K.

Precalculus Review - Intro

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James StewartCalculus

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

James Stewart

Calculus