Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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?
 one year ago
 one year 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?
 one year ago
 one year ago

This Question is Closed

zepdrixBest ResponseYou've already chosen the best response.0
dw: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.
 one year ago

bear4343Best ResponseYou've already chosen the best response.0
You got it now take it apart and solve
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
how do you set it up to solve it though?
 one year ago

bear4343Best ResponseYou've already chosen the best response.0
The 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
 one year ago

bear4343Best ResponseYou've already chosen the best response.0
If youneed anymore help with these types of problems let me know
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Were you able to figure this one out erica? :D
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
no 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
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.0
\[\large V=xyz\]\[\large V'=x'yz+xy'z+xyz'\]
 one year ago

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

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

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

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

ericatrangBest ResponseYou've already chosen the best response.0
what are the numbers that are being derived in this case?
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
also how do you use the given rates to solve the problem?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
The 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,\[\large \frac{dy}{dt}=\frac{dx}{dt}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Is 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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Uh oh, did your brain esplode? :c
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
yes 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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Yah if you see it step by step it might help.
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.0
dw: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.
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
I know that we need the Area formula A=L*W*H
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Volume, 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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
To 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.
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.0
_ lol read the sentence in all caps.
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
lol sorry it showed up after I pressed enter o.o
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
So if we take the derivative, we'll have to apply the product rule,
 one year ago

zepdrixBest 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
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.0
Uh oh :( Silence... So this is the tough part huh?
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.0
Yah tough class c: so much fun though! lol
 one year ago

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

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

zepdrixBest ResponseYou've already chosen the best response.0
OpenStudy froze again.. sigh have to type all that over again :( man that's so frustrating.
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
dang that bites, I think it's frustrating enough trying to help me o.o"
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
We 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.\[\large x=10, \quad\;\;\;\qquad y=12, \quad\;\;\;\qquad z=16\]\[\large x'=0.2, \qquad y'=0.2, \qquad z'=0.3\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
And then we will get our solution for V'.
 one year ago

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

ericatrangBest ResponseYou've already chosen the best response.0
and then input this into the calc ill do it :3
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
That'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
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
would this be the final answer?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
If 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.
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
what? 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?
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
I understand how to do this problem better (:
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
k good c: i hope so hehe
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
there is less confusion :D
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
If 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
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
thank 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
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
does the same idea apply to a cone? just a different volume formula?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Yah 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.
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
Thank 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?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Related 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
 one year ago

ericatrangBest ResponseYou've already chosen the best response.0
Alrighty, 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.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.