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Problem 8 Medium Difficulty

The area of a triangle with sides of length $ a $ and $ b $ and contained angle $ \theta $ is
$ A = \frac {1}{2}ab \sin \theta $
(a) If $ a = 2 cm, b = 3 cm, $ and $ \theta $ increases at a rate of $ 0.2 rad/min, $ how fast is the area increasing when $ \theta = \pi/3? $
(b) If $ a = 2 cm, b $ increases at a rate of $ 1.5 cm/min, $ and $ \theta $ increases at rate of $ 0.2 rad/min, $ how fast is the area increasing when $ b = 3 cm $ and $ \theta = \pi/3? $
(c) If $ a $ increases at a rate of $ 1.5 cm, b $ increases at a rate of $ 1.5 cm/min, $ and $ \theta $ increases at a rate of $ 0.2 rad/min, $ how fast is the area increasing when $ a = 2 cm, b = 3 cm, $ and $ \theta = \pi/3? $

Answer

a) 0.3 $\mathrm{cm}^{2} / \mathrm{min}$
b) 1.6 $\mathrm{cm}^{2} / \mathrm{min}$
c) 4.85 $\mathrm{cm}^{2} / \mathrm{min}$

Discussion

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MA

Mona A.

July 2, 2021

suppose y=root 2x+1

Video Transcript

Okay. Okay, we have this triangle. I'm going to draw it down here. Hope that's not a triangle. Okay, here we go. And let's say this side is A and this site is B. And then theta is in between them. Okay, In the first question part A A and B are fixed. They are two and 3 and there are not getting any bigger. And D Theta DT, r Theta is changing the rate that data changes is 0.2 radiance per minute. So since it's a raid, its name is D. Theta DT. Yeah. Okay. So this would be see right here And then a was two and B was three can update is getting bigger. Okay, so in a little while it's this big and then in another little while it's this big. Then in another little while it's this big. But A and B are staying the same length. But you can see that the area is changing. Okay, so the question is, how fast is the area change? Okay, so we have a cause one half a B signed data And a was to be was three And d. Theta DT was .2. Okay, so A and B are fixed. They're not going to change. So let's put them in before we take the derivative one half times two times three times to sign up data. So a is three sine theta. Okay, now the two things that are changing here, R Theta and A. So when we take their derivatives, they get A D T stuck on it. So here we go derivative of A D A D T. Derivative of three. Sine theta is three. Cosign Theta times the derivative of theta which is D theta DT. I think I missed something here. Also find the ATT1 thing. It is five or 3 fine. B A T. T. Okay. So it's not saying that data is pi over three all the time. It says at one of these instances over here, data is pi over three. And at that instant, how fast is the area changing? Okay, so just plug them in three Times The coastline of pirate. three times D. Theta DT which was 0.2 543. That's 60°. That square 23 to 1. So the coastline is one half. So three times 1/2 times two. So three three what? Well A and B. Are centimeters so it's centimeters squared. And D Theta DT was in ratings per minute. So it's centimeter squared per minute. Okay? Now and the next one B is changing and feta is changing. So the rate that B is changing is called D. B D. T. And the rate that that is changing is called D. Theta DT but A is fixed this time. Okay, So here's the picture. Okay. Is to and then here's B. And here's data. So here's our other side. All right, so that is getting bigger. It is getting bigger. See it getting bigger at the same time. Be is getting bigger. So now we have this and then tha tha gets even bigger and bigger gets even bigger and now we have this and then they gets even bigger and be gets even bigger and now we have this. Okay, so two things are changing this time be and theta, so A is fixed at two B. Is changing at a rate of, I forgot um 1.5 centimeters per minute and that is changing at a rate of .2 radiance per minute. And so we're trying to find D. D. C. To D A D. T. The rate the area's changing Win B is three and theater is pi over three because 3 85 or three. So it's like this is happening and we take its picture right at the instant when B is three and theta pi over three, what's the rate the area is changing right there? Okay, same formula A equals 1/2 a b. sine. Theta. We can plug in equals two but we can't plug in B equals three. Cosby is changing. So you can't put anything in below this line until after you take the derivative. Okay, so right before you say fine you can't plug any of that other stuff in but we can't play the two in before we start. So we get one half times two times B times signed data. So that's be santa. Okay, now this time three things were changing. Remember B is changing data is changing and the area is changing. So over here on the right, we're gonna have to use the product group. Alright, here we go. The derivative of A. Which is D. A. D. T. Equals the first B. Times the derivative of the second which is cosign Theta times the derivative of theta plus the second to sync data times the derivative of the first, which is D. B. D. T. Okay, the first B derivative of the second cosign Theta. D. Theta DT plus the second santa times the derivative of the first DVD T. Okay, so now just plug stuff in. So b. is three Date is Pi over three. D. Theta DT is .2 state is five or three and D. V. D. T. Was 1.5 543. Remember 60°2321. So here we get three times 2 Times The Co Sign, which is 1/2. Let's say we're here. We get 1.5 Times The Sine which is the square root of three over to. So it's three plus .75 times the square to three. I don't know where my calculator is, which is something and again, cm squared per minute because it's area changing. All right. Now in the 3rd 1 A. Is changing B is changing data is changing. So they're all getting bigger and they just keep getting bigger and then let's find the A. D. T. All right. So, we have D. A. D. T. Was something D. B. D. T. Was something D. Theta DT was something And we want to find the A. D. T. Wind. They have to tell us what A. Is what B. Is and what they do is can remember nothing below this line. Can go into the equation until after you take the derivative. So the A. D. T. Was 1.51 point 5.2. Okay. 1.5. 1.52. And I bet this is 2, 3 and five or 3. Okay, so big A. Is one half a B. Sign theater. Okay. This time we have four things changing. Sorry is changing. It is changing state is changing therefore the area is changing. So when we take the derivative over here, on the right hand side, we got three pieces. So we're going to have to do a fancy fancy product rule. So, let me show you an easy way if why is the function F. G. H. So three functions distributive is take the derivative of F. Times G. H. Plus take the derivative of G times FH plus the derivative of H. Times the other two. So the derivative of each one times the other two. That way we don't have to do anything fancy. Alright, here we go. D. A. D. T. Equals. Okay, the one half is just gonna be out in the front there. Okay, the derivative of the first thing that's D. A. D. T. Times the other two plus the derivative of the middle thing. That's DBT Times the other two plus the derivative of the third thing which is coastline, tha tha d theta DT Times The Other two Things. Okay in this rule work, if you have four things or five things are six things you take the derivative of one times the other ones plus the derivative of the next one times the other one. So you never have to use the product rule more than once. All right now we're gonna do is plug some stuff in. Okay? We got 1.5 times three times a sign of data which remember was squared 3/2 plus 1.5 times two times the square to 3/2 Plus the co sign which is 1/2 times point to Times two Times 3. All right, let's see if I can do this without a calculator. No. No. Okay. So just calculate all of this. And then put it put I mean put it in your calculator and calculate and out will come the answer. I can set it up a little bit more. Um Three times 1.5. That's 4.5 divided by two. That's 2.25 square 23 plus. This one is 1.5 square +23 Plus This one is um Uh those cancelled three times .2. So .6. So one half of 3.75 times that square 23 plus 230.6. So I have those together divide by two and that will be D a d t again centimeter squared per minute. Hope that helps.