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anonymous
 3 years ago
the dimensions of a box are 10" by 12" by 16". if the shorter sides are decreasing at the rate of .2"/min. and the longest side is increasing at .3"ming. how fast is the volume changing?
anonymous
 3 years ago
the dimensions of a box are 10" by 12" by 16". if the shorter sides are decreasing at the rate of .2"/min. and the longest side is increasing at .3"ming. how fast is the volume changing?

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zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1355878324269:dwHmm this one is a little tricky, because we're going to have the product rule with 3 terms. Does the picture make sense? The prime terms are the RATE OF CHANGE of the sides. You can think of them as dx/dt if that's more clear to you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You got it now take it apart and solve

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do you set it up to solve it though?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The question asks for volume changing so you take your formula V=LWH and d/dt both sides. Are you familiar with implicit diferentiation? d/dt(V=LWH) dV/dt = LWH' + LW'H + L'WH the last bit is just from the product rule. You want to find dVdt, you have L, W, H, and the derivative of each. so you just have to solve

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If youneed anymore help with these types of problems let me know

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Were you able to figure this one out erica? :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no I haven't yet, I understand that it is lookin for the volume but the implicit rule still gets me, I was trying to figure it out but I got confused

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Do you understand the product rule with 3 variables? It's a bit tricky, it'll follow the same rule though,\[\large (uv)'=u'v+uv'\]\[\large (uvw)'=u'vw+uv'w+uvw'\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large V=xyz\]\[\large V'=x'yz+xy'z+xyz'\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes but how do you take the derivatives of those sides when there isn't a variable, for example like 3x^2 becomes 6x

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Oh you mean since they gave us the side values? So it's confusing you because you're thinking of the sides as constants?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0lolol i always find that funny "erica is typing a reply.... .... ... . .. . .. . ...... . ...*3 minutes later* " > "yes :3"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry haha I was trying to think about what it was that you pointed of my mistake

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what are the numbers that are being derived in this case?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0also how do you use the given rates to solve the problem?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0The problem is, we don't want to think of the sides as CONSTANT. The sides are CHANGING. We can write Volume as a function of the 3 other sides, x, y and z. The fact that the x, y and z have particular values at one point in time doesn't mean they simply equal those values. It's just has to do with the time that we looked at them. Maybe that's making it more confusing.. hmm it's difficult for me to word it nicely. The point is, our x, y and z are NOT constant. They are variables. They equal their face value, x=x and so on... So when we take the derivative, the normal rules will apply. If we had,\[\large y=x\]If we took the derivative with respect to TIMEEEE, we would get,\[\large y'=x'\]Does that much make sense? Or are the primes confusing you? This is the same thing as,\[\large \frac{dy}{dt}=\frac{dx}{dt}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Is the implicit things confusing you? You might wanna think back to taking derivatives implicitly. \[\large x^2+xy=0\]Remember how we would solve this implicitly? We would take the derivative of both sides normally, and then whenever we took the derivative of y, a y' would pop out. The reason an x' never showed up is because we were differentiating WITH RESPECT TO X. The same thing is happening in our Volume problem except now we're differentiating WITH RESPECT TO TIME, so EVERY variable will produce a prime.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Uh oh, did your brain esplode? :c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes it did lol I read through it and tried to think about the process.. is it alright it you could solve it out for me showing the steps and then I'll go back and explain how I see that its been solved, like I'll try to explain how you solved it? I feel like that would help me understand better

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Yah if you see it step by step it might help.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you :3 I really appreciate you taking your time to help me learn this~

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1355899110491:dwSo we have all of the sides labeled. We have the "rate of change" of the sides labeled. Think of these as 6 variables that we will LATER plug in.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know that we need the Area formula A=L*W*H

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Volume, yes :) I labeled them as x y and z, I hope that isn't confusing. Same as L W H. \[\large V=xyz\]We're trying to solve for V', the rate at which V is changing.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0To find dV/dt, we'll take the derivative of both sides, with respect to t. MAKING SURE TO NOT PLUG IN ANY NUMBERS BEFORE WE DIFFERENTIATE.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so we need the derivative, do we plug in the number yet? or it's still later on

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0_ lol read the sentence in all caps.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol sorry it showed up after I pressed enter o.o

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0So if we take the derivative, we'll have to apply the product rule,

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large \frac{d}{dt}V=\frac{d}{dt}(xyz)\] \[\large \frac{dV}{dt}=\frac{dx}{dt}yz+x\frac{dy}{dt}z+xy\frac{dz}{dt}\]Product rule :O

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Personally I think a messy formula like that is easier to read with the prime notation, but whatever works for you :)

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Uh oh :( Silence... So this is the tough part huh?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0personally I don't understand how I'm maintaining a 76.8% in calc it's getting more difficult for me T___T

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Yah tough class c: so much fun though! lol

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Do you plan on taking calc 2 at any point? c: that class is harrrrd

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no thank you lol maybe if it's alright for you to keep continuing and Ill still keep trying to piece it together?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0OpenStudy froze again.. sigh have to type all that over again :( man that's so frustrating.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dang that bites, I think it's frustrating enough trying to help me o.o"

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0We took the derivative of V,\[\large V=xyz\]\[\large V'=x'yz+xy'z+xyz'\]Now from this point, we want to plug in our variables from before.\[\large x=10, \quad\;\;\;\qquad y=12, \quad\;\;\;\qquad z=16\]\[\large x'=0.2, \qquad y'=0.2, \qquad z'=0.3\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0And then we will get our solution for V'.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[V'=[0.2(12)(16)]+[10(.2)(16)]+[10(12)(.3)]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then input this into the calc ill do it :3

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0That's what I got also.. Hmm that seems kinda large.. but maybe that's ok :O the base is decreasing in size, getting smaller and smaller, but the box is getting taller and taller. So I would have thought it was decreasing very slowly. I guess I was wrong :D hah

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would this be the final answer?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0If the concept is really confusing, maybe just try to focus on a few of the steps. It's really just a matter of labeling things correctly, taking a derivative, and then plugging things in.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what? I feel like there's supposed to be more to it. O.O ok so now i understand the product rule better the more parts there are to it than the first stays and you take the derivative of the next part and so on right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I understand how to do this problem better (:

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0k good c: i hope so hehe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there is less confusion :D

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0If you saw a problem like this on a test, I think you would be more likely to see a square changing, (2 variables), rather than a box (3 variables). But it's the same idea, so you gotta try to get comfortable with all these prime terms floating around :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you for helping me this really did help, basically the concept is to know where to label the picture, derive the volume formula, and plug in using product rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0does the same idea apply to a cone? just a different volume formula?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Yah the same idea applies to a cone and many other shapes. I dunno what type of text book you use, but you might come across a problem where a person is blowing up a balloon. They'll provide you with the rate at which the radius of the balloon is increasing, and want to know how much the rate of change of the volume. I remember seeing a problem where sand is being poored into a pile that forms a cone shape. If the height is changing at this rate, how fast is the volume changing. Yah these problems come up in all shapes and sizes. So try to get comfortable with how this might apply to a circle or sphere as well as square objects :D It's really just about getting the correct formula to differentiate.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you so much :3 there are more problems similar to this so I'll be practicing on it oh hey, even though the word problems in calculus are some what related to real life situations but do you ever really use them?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Related rates are fairly basic, you probably won't use them much. If you took a course on physics or differential equations, you might see how these things become important. Especially in fields of engineering ~ Building bridges, skyscrapers. Understanding how electricity flows and stuff. I dunno... I haven't gotten very far yet, so I don't really know much about applications :D hehe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Alrighty, I'm really glad you helped me through this. I'm scared to take the AP test for it in a couple of months, there's so much to know where to apply certain processes.
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