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The length width %, and heignt h of a rectangular box (wth change with time: Ar certain instant the dimensions are =2m, W = Smh =1m, and and increasing at rate of 4...

Question

The length width %, and heignt h of a rectangular box (wth change with time: Ar certain instant the dimensions are =2m, W = Smh =1m, and and increasing at rate of 4 ns while h decreasing at a rate of 8 mrs: Die lengte breedte w, en hoogte van rerjcekige boks (met deksel) verander met tyd. Op sekere oomolik is die cimensies? 2 m, W = 5 n h =1m,en/en Loeneem teen tempovan m/s rerwyl afneem iccn rempovan nrs Suppose S the surtace area of the box At the given instant we have: Veroncerstel $ die bui

The length width %, and heignt h of a rectangular box (wth change with time: Ar certain instant the dimensions are =2m, W = Smh =1m, and and increasing at rate of 4 ns while h decreasing at a rate of 8 mrs: Die lengte breedte w, en hoogte van rerjcekige boks (met deksel) verander met tyd. Op sekere oomolik is die cimensies? 2 m, W = 5 n h =1m,en/en Loeneem teen tempovan m/s rerwyl afneem iccn rempovan nrs Suppose S the surtace area of the box At the given instant we have: Veroncerstel $ die buiteoppervlak van die boks Op die gegewe pcmhlik het ons: The rate at wnich the surface area ofthe box changing at that instant is Die tempo wraarteen de buiteoppervlak van die boks verander OpJ dlaardie oomblik is;



Answers

The length $\ell,$ width $w,$ and height $h$ of a box change with
time. At a certain instant the dimensions are $\ell=1 \mathrm{m}$ and
$w=h=2 \mathrm{m},$ and $\ell$ and $w$ are increasing at a rate of 2 $\mathrm{m} / \mathrm{s}$
while $h$ is decreasing at a rate of 3 $\mathrm{m} / \mathrm{s}$ . At that instant find the
rates at which the following quantities are changing.
$\begin{array}{l}{\text { (a) The volume }} \\ {\text { (b) The surface area }} \\ {\text { (c) The length of a diagonal }}\end{array}$

All right, so and this question were given the length with height a rectangular prism also given young by DT just two meters per second. And both of the change with her perspective time which is two meters per second and the change in height with respect to time which is negative three meters per second. Basically, it's when a negative three meters per second. That basically means that the height is decreasing at a rate of three meters per second. All right now were cold part a to determine d v by d. T. So we know that he is like friends with terms height. So the derivative volume depends on length within height which in turn depend on time. So we can write this because we have to use the chain room so that we can find the derivative of the respected so d V by d. T is equal to d v by d l times d l Biting to you plus d v by d. W times dw by DT plus DV by th times Huge So we already know what So the lbgt is given the w by DT years given into huge writing, they're all given in the course. But now what we need to determine is you be by D L D V by D. W on TV. But well, that's not difficult. We just take the derivative of blank friends with times height with respect, the link with and height respectively. So, do you be like yo is just, uh, times height? Because the derivative of length will be one So two times two which is for the derivative of volume, with perspective with well, the during averted the derivative of deadly used wants us just length times one times height, just just length, times, height, which is one time to do it too. And then finally the derivative of value would respect. The height is just held of a year or two times one. Well, great. Now we've got everything. So we plug everything back into TV times DT. And then what we get is six cubic meters per second. Zahra unit for volume is cubic meters cube and the unit for times just seconds. All right now in part B were asked to determine the derivative ball surface every with respect to time. Well, any their service areas, just two times like things with most length as I plus high times so we can find We know that the surface area is a function of length, width and height, which in turn are a function of time. So when we want to find a derogative off the air by DT gonna have to use the change, so be able to use equal to be by the l a Times the lbgt was the a bi dw times W by DT and the eight times the age times you to buy So deal by DT again, this is given in the question. This is also given in the question of this. What? So what we need to calculate is the year by year old a Brady WMD. Well, do you? Every deal is just, um, two times work. That's height, because we're finding the derivative of a right here with respectively. And then ah, with and height are given in the question. So this is it was that the eight times DW So the derivative of felt times W's just Bell and a derivative als l times each is zero and the derivative of each times that will use just so what we're left with its two times like times height which is and then finally we do the same thing. We take the derivative all day with respect age. There's just two times, like, plus with which is two times one plus two, which is equal to six. So great. Now we got everything. So what we do is we plug it back into our equation for D A body which we use a chain saw. So we get is 10 meters square for a second, cause the unit off a is meter square to the unit of tea is seconds. All right, so that we need to find the rate of change. Is that the agonal with respect to time. So the formula to determine the day agonal is he squared is equal toe elsewhere but a W squared Plus Now we take the derivative of the left hand side. So we're gonna have to use the jingles or doing the derivative with respect to time. Well, this is the derivative of thes squared is just to 20 times the derivative of the divided by and that is gonna equal to know we take the derivative of the right hand side. So that's just dfd. Would respect that? The derivative of dealing, respected at all times the derivative of Albert Respect the times derivative of our respective t plus the a d were respect Duty by DW Times www by DT was d D by D h times each so we know what you'll buy in two years. We know what w already 18 years. We know what the each party so now but we need to determine is the of the word respect l the derivative of the with respect of l. You're a bit of a deal with respectability of derivative of So we take the derivative of the right hand side with respect to l. It's just too well, just two times one or just again the derivative of the right hand side with respect to W just to w. So when you're playing in two for W, you're gonna get two times two for and finally there is a derivative of the new would respect of ancient just tweet or two times two, which is for, and we can are. We can also determine what ideas because we're given length with. So we plug that in the square, root off one square, plus two square foot to square. There's square it or nine or just three. All right. Now we plug everything back into original equation. Treat around using the chain room. So two times during his six his D D by D T times two times two plus two times for four times to what's four times negative. Three. So on the right hand side, we're gonna get four plus eight minus 12 which is dearer. Now we divide by six on both sides, where we get a zero divided by six or just zero.

Hi there. So in this problem, let's write down what we know. So we know at a certain time length l equals one and W and H are both too. We also know that the length and the width Ellen W are increasing at a rate of two meters per second. So in other words, D l d t equals two. The w whoops. D t also equals two and h, the height is decreasing at a rate of three per second. That means D H D t is negative. Three. Okay, so that's all the information that we have. And we're supposed to find the rates at which three quantities air changing. So for a party, it's the Valium. So for party, you want to find DVD T. Okay, well, to do that, let's remember the formula for the volume. So well, I'm of the boxes just length times with time site. So times tell you times h okay. And what we in order to get our DVD t Let's follow the chain rule for the chain Rule says we will need the partial of D. A partial of the respect Thio the little many times d l d t plus partial of the respected W time W d t Close partial of V with respect to h times, DHD tiu Okay, so these will be pretty quick The derivative of partial of V. With respect to l. A gambler's gonna look up here for all these partials. L is considered to be the variable than the derivative is just w times h and we know d l. The tea is too partial of V with respect to w again just looking up here it'll be l Times, age and D W T t is also too finally partial of the with respect. H l Time W and DHD tea was given to us is minus three. Now it is putting in our values for W and H and l will get her answer. So c w nature both to two times. Two times 28 l and H one times two is two times two is four L times W one times 22 times negative. Three is negative. Six A plus for his 12 minus six is, uh, six, and we should get our units correct. So everything is in meters and seconds of volume is measured in cubic meters, of course. So six cubic meters per second. It's increasing at that rate. So that's the answer for part A. OK, so apart be we will do much the same thing. Let's have to race this year. Uh, this time, instead of volume, we want help quickly. The surface area is changing. So as before, we will have to begin with a formula for surface area. So that's change this. So now we want we'll use a for area you want d a t t. Okay. In the surface area of a box or rectangular prism, um is well, we can think of it, but it's going to be two times and just think about their six faces to the box. But really, there's just two copies of each of the three different possible pairs, right? So the base is like L W on one of the walls would be, say, W h and either wall would be Which one haven't we done yet? L h to that surface area and we'll do the exact same thing to get d a t t. We're just going to need a partial of a with respect to now times the LD teams plus partial of A with respect to w times w d t close Partial A with respect thio h times D h d t. Okay, uh, for the partials will have to look at our formula for surface area. Appear so with respect to l think term this first term Here, give us a derivative of to w Okay, And this term here will give us a derivative of to age. D l E t again is given to us as two. With respect to w, this will feel much the same. First term gives us a primitive of two l Second term gives us a derivative of two h third term zero. Since there's no w in it, D W t t is given to us as two. And finally partial of eight. Respect to H. Let's see nothing in the first term the kids to w from the second term, and we get to L. A from the third term all times D h d t, which is negative three and now looking in our current values W h you're both too. Four plus four is eight times to that. Should give us 16 from the first term. Ellis Juan ages to get two plus four is six times two is 12. And for this 3rd 1 W is too. That's four. That was one. That six again. Times negative. Three is negative. 18. So 16 plus 12 28 28 minus 18 is 10. And this time our units since the area are units will be, uh, square meters her second. So that's your answer for B. Okay, what race? This? And now, finally, we're gonna do the exact same thing for part C. Except, uh, this time, instead of volume or area, we're going to be calculating how quickly the length of the diagonal is changing. This is a little more interesting. This is part. See? And let's just call it Diagonal Capital D. So you remember what it is. So we're gonna want how quickly is the diagonal changing with respect to time? Okay, um, there's several ways to think of this. Hopefully you think about me, miss, um, if you've got any vector calculus already or think about three dimensional a distance between three points, Really, this is just, uh should feel like the Pythagorean theorem. So the diagonal is from one corner of the box to the opposite corner or whatever this distances and the formula for that again, this is just like a three dimensional version of the Pythagorean theorem. It is Ah, square root of l squared plus W squared plus h squared. Just, uh, just as the diagonal of a rectangle would have been the square root of the length plus with squirt. So we just add one more dimension here. So that's the formula for the diagonal of the box. Okay, Which a CZ always for calculus will think of this as 1/2 power to make our derivatives easier. Okay, so for the last time, dd dd t will equal Well, we're gonna need a partial of D With respect. Thio l times d l d t Partial deep respect to W times wd teen plus partial d Who expected h time th d t Okay, partial deal respected w with respect to l. The good news is, once we calculate, this is gonna be the same for the other two variables. Just because everything is symmetrical in this formula. So off we go, we'll get this. So we have to be 1/2 times, everything in there still power rule. So this has been pure power well, so far now where we ought to be a little careful chain rule tells us we need to multiply this by the derivative of that quantity inside. Now that derivative with respect to El Ella's, they're only variable. We just get two times out. So that is a partial of D with your spectacle. And now for a deal, the tea and that's given to us earlier in the problem. That's too good. So again, the good news is we could do the exact same thing for W. The derivative look the same in Elite 1/2 l squared plus W skirt. What's H squared? Minus 1/2 with this time times to W and D. W. D. T was given us, too, and same thing over here. 1/2 elsewhere Tell you Squared was H squared to the minus 1/2 times to H this time, and D H D T is given to us as negative three. Okay, now, just time to plug in our values. So let's see elsewhere. But W. Scarpa's H squared l is one. W is two ages, too, so if we square all these will get one plus four plus four, which is nine. Right? So I'm thinking how this negative 1/2 power works. That's the same thing at the square root on the bottom. So, uh, this entire part becomes right in the parentheses. This helping was one over the square root of nine. Which, of course, will be three in the next step of this 1/2 of that to cancel out. So we just get l times two. But L A is one. So we get everything is just times too. So when everything is done, we should get to over the square to nine, which, of course, will be 2/3 at the end. Over here. Let's see what we get. Well, again, the good news is, every time we do this, we're gonna get the same thing. We just did this calculation. So this is one over a squirt of nine again this 1/2 and that too, will cancel out all we'll be left this foot with his w times two and W is to set three times too. Will give us four on top. Finally, once again, this part we know already It's just one over the square root of nine The 1/2 from the to cancel We're left with each times Negative three h is too so two times negative three is negative six. So we're left with 2/3 plus 4/3 minus 6/3. But if you look at the numerator Sze so is that the nominated are all the same to pose for a six minus 60 So after half of that and we end up with the final answer of zero and we should think of our units, this is the length of the diagonal. So this is just meters and our time this seconds zero meters per second. Uh, in other words, at this moment, given the and what we have, even though all of sides or changing lengths that diagonal at this moment is actually staying constant, it's not growing or shrinking. Hopefully that helped

Hi, everyone. So we're gonna be solving problem 42 from chapter 14 Section four on the chain rule. So the problem gives us our variables for rectangular box. We have eight millimeter b equals two meters C equals three meters and we also have the derivatives of these and respected times. So we have here thes two equal one and equals negative three. And given all this information, we want to find the derivative of the volume in respect to time a derivative of the surface area, and respect the time and the derivative of l and respected time where l we have defined as our boxes interior, diagonal. And we want to use this derivative to find if it is a positive or negative. So we can, uh, find if it is increasing or decreasing. So to start, we're going to start by solving for the derivative of the volume and respected time. So we're gonna define our volume first and that's gonna be a B. C. Is it's dependent on a B and C, and from here, we're going to use our chain rule to solve for the derivative of the volume in respect to time. So using the chain rule. We get DV over D a times d A over DT plus TV over db times db over DT plus do be over d c times d c over d t So we already have some of these already given Um, we know that's d A over DT is equal to one meter per second and we know that d over DT is He wouldn't want meter per second and we know that d c over D t is equal to negative three meters per second. So now we have to solve for these three. So to do that, we'll do that on the side here. So Devi over d day, we're going to be using our original volume equation and we're gonna get Do you see TV over? DB is a c devi over d. C is a beat and we can just plug that in because we already know that a is one of these two seas three oo meters, so we could just calculate for that BC times, and once you plug R B A, B and C, we should get three meters cube per second and that is our answer and moving on to the surface area. Do you want to first define our surface area, which will call us and we know that surface area is That equation is to length with plus two length height plus two with height. So we have our variables as a B and C, so we'll pull again accordingly. Now same steps is before we're going to use our chain role. So we have DS over D A Times Day over duty, Yes, over db time zb over duty plus DS over D C temps D C over t t again. We already have these three guys to find. So let's calculate for our menacing variables Sorry, That should be Do you be? And yes, you see, So of course, using our original service area equation, we get, uh, d s or D A is equal to DB Plus because our to be plus to see yes over D. B is equal to two A plus two c and D s already see is to a plus to be we plug all of this into our equation. We have to be close to see times one meter per second plus to a pause to see comes one meter per second and to a was to be times negative, three meters per second. We plug in our A, B and C, which is 12 and three, respectively. And it should be cool. Zero meter squared per second. Great. Moving on to the last part of the problem. We have to find the derivative of the internal diagonal, which we have to find a spell. And this is gonna be a little more work because you need to find a formula for l depending on ABC. We're essentially doing the same steps. So to start funding are equated for l. We have to apply the three d printer degree in Europe. So without going into too much detail, read what we know is we have to find this side, which I have, uh, marked in green and using the Pythagorean theorem. We find that this side is the square root of a squared plus B squared. So now using this, we can find the information for l. So what we get is l squared is equal to a squared plus B squared plus C squared. And now we can isolate l and just put all of that under a square root. And we have our equation for help. So now we have to find the derivative of L in respect to time. And once again, we're gonna use our chief no accident dio over d a times d a over DT plus dio over db times to be over duty and deal over D's tens d c over d t Once again, we're gonna have to find our derivatives of l in respect to a B and C So we'll do that over here, and we're gonna find those to be a over the square root of a squared plus B squared closely square. And actually, we're gonna see that this, um, is actually the same thing as L. So for the next one, I'm just gonna, uh, substitute else it up. So our drew of L over d B is b over out, and then we have see, overall, it's not plugging everything in, including our information that really no, that's given in our problem. We have a over O times one meter per second plus B over L times, one litre per second, plus see over, oh times negative. Three commuters, her second and what that is all get equal. Once we plug, everything in is gonna be negative. Six over the square of 14 which is about negative 1.6 meters per second. And what we see here is that it is negative. So we know that it is decreasing. And that is our answer to the last time the problem.

Hi, everyone. So we're gonna be solving problem 42 from chapter 14 Section four on the chain rule. So the problem gives us our variables for rectangular box. We have eight millimeter b equals two meters C equals three meters and we also have the derivatives of these and respected times. So we have here thes two equal one and equals negative three. And given all this information, we want to find the derivative of the volume in respect to time a derivative of the surface area, and respect the time and the derivative of l and respected time where l we have defined as our boxes interior, diagonal. And we want to use this derivative to find if it is a positive or negative. So we can, uh, find if it is increasing or decreasing. So to start, we're going to start by solving for the derivative of the volume and respected time. So we're gonna define our volume first and that's gonna be a B. C. Is it's dependent on a B and C, and from here, we're going to use our chain rule to solve for the derivative of the volume in respect to time. So using the chain rule. We get DV over D a times d A over DT plus TV over db times db over DT plus do be over d c times d c over d t So we already have some of these already given Um, we know that's d A over DT is equal to one meter per second and we know that d over DT is He wouldn't want meter per second and we know that d c over D t is equal to negative three meters per second. So now we have to solve for these three. So to do that, we'll do that on the side here. So Devi over d day, we're going to be using our original volume equation and we're gonna get Do you see TV over? DB is a c devi over d. C is a beat and we can just plug that in because we already know that a is one of these two seas three oo meters, so we could just calculate for that BC times, and once you plug R B A, B and C, we should get three meters cube per second and that is our answer and moving on to the surface area. Do you want to first define our surface area, which will call us and we know that surface area is That equation is to length with plus two length height plus two with height. So we have our variables as a B and C, so we'll pull again accordingly. Now same steps is before we're going to use our chain role. So we have DS over D A Times Day over duty, Yes, over db time zb over duty plus DS over D C temps D C over t t again. We already have these three guys to find. So let's calculate for our menacing variables Sorry, That should be Do you be? And yes, you see, So of course, using our original service area equation, we get, uh, d s or D A is equal to DB Plus because our to be plus to see yes over D. B is equal to two A plus two c and D s already see is to a plus to be we plug all of this into our equation. We have to be close to see times one meter per second plus to a pause to see comes one meter per second and to a was to be times negative, three meters per second. We plug in our A, B and C, which is 12 and three, respectively. And it should be cool. Zero meter squared per second. Great. Moving on to the last part of the problem. We have to find the derivative of the internal diagonal, which we have to find a spell. And this is gonna be a little more work because you need to find a formula for l depending on ABC. We're essentially doing the same steps. So to start funding are equated for l. We have to apply the three d printer degree in Europe. So without going into too much detail, read what we know is we have to find this side, which I have, uh, marked in green and using the Pythagorean theorem. We find that this side is the square root of a squared plus B squared. So now using this, we can find the information for l. So what we get is l squared is equal to a squared plus B squared plus C squared. And now we can isolate l and just put all of that under a square root. And we have our equation for help. So now we have to find the derivative of L in respect to time. And once again, we're gonna use our chief no accident dio over d a times d a over DT plus dio over db times to be over duty and deal over D's tens d c over d t Once again, we're gonna have to find our derivatives of l in respect to a B and C So we'll do that over here, and we're gonna find those to be a over the square root of a squared plus B squared closely square. And actually, we're gonna see that this, um, is actually the same thing as L. So for the next one, I'm just gonna, uh, substitute else it up. So our drew of L over d B is b over out, and then we have see, overall, it's not plugging everything in, including our information that really no, that's given in our problem. We have a over O times one meter per second plus B over L times, one litre per second, plus see over, oh times negative. Three commuters, her second and what that is all get equal. Once we plug, everything in is gonna be negative. Six over the square of 14 which is about negative 1.6 meters per second. And what we see here is that it is negative. So we know that it is decreasing. And that is our answer to the last time the problem.


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The decond ibor d 3 houseabowaruet len|pan AHor IcIe quire d [0 270-kg retigerator sacond story letel? Express Your Jnswt Mo fuifcant Floutr und include tne appropilein uninKaltteUmitsSubniXuquutrnser
The decond ibor d 3 house abow aruet len| pan A Hor Ic Ie quire d [0 270-kg retigerator sacond story letel? Express Your Jnswt Mo fuifcant Floutr und include tne appropilein unin Kaltte Umits Subni Xuquutrnser...
5 answers
Rroor All answer are wrong2. At what condition, the electric field of the dipole is vanished:(a) At double of the distance between the dipole (6) At point is far away from the center of dipole At half of the distance between the dipole It cannot be known (e All answers are wrong
rroor All answer are wrong 2. At what condition, the electric field of the dipole is vanished: (a) At double of the distance between the dipole (6) At point is far away from the center of dipole At half of the distance between the dipole It cannot be known (e All answers are wrong...
5 answers
Write the equation of the ellipse which satisfies the givenconditionsa. Foci (0, ±8) and major semi axis 17b. Vertices (2,6) and (2,-4), b = 4
Write the equation of the ellipse which satisfies the given conditions a. Foci (0, ±8) and major semi axis 17 b. Vertices (2,6) and (2,-4), b = 4...
4 answers
Cross a pink 4 o'clock flower with another pink flower. Pink isheterozygous. Give the phenotypic ratio. (incomplete dominance)
cross a pink 4 o'clock flower with another pink flower. Pink is heterozygous. Give the phenotypic ratio. (incomplete dominance)...
5 answers
Part 1: For the coefficient matrix below; determine the solution set for the homogeneous system AND describe the solution set geometrically (line; plane; hyperplane through point _ ) [2 3 5 20022-3Part 2: For the same coefficient matrix, assume Ax = b has the particular solution Xp given by the vector below. Write the full solution set for Ax = b AND describe the solution set geometrically (line, plane, hyperplane through point )42 Xp = 03
Part 1: For the coefficient matrix below; determine the solution set for the homogeneous system AND describe the solution set geometrically (line; plane; hyperplane through point _ ) [2 3 5 2 0 0 2 2 -3 Part 2: For the same coefficient matrix, assume Ax = b has the particular solution Xp given by th...

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