Question

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Answer 1

has anyone answered your problem yet?

Answer 2

No! No one has...I was kinda of waiting on you to help me...

Answer 3

ok so the volume formula, what do you have for that?
V=?

Answer 4

The volume formula is V=4/3(pi)r^3

Answer 5

oh wait that's for the sphere. lol sorry

Answer 6

no this is a box, a rectangular box has
volume=lwh
notice this box has a square base so what is the formula here?

Answer 7

yes you caught that yours is for a sphere

Answer 8

is it hxw

Answer 9

I mean h x l

Answer 10

well remember we had befor \[V=lwh\]in this case l=w, so we can sub that into our formula and it becomes\[V=x^2h=500 \]we know the volume because it is given in the problem
so we want to rearrange this for the variable we want to solve for the derivative of. Which variable is that?
Can you set up that equation?

Answer 11

sorry \[V=w^2h\]

Answer 12

same thing

Answer 13

ok! so we want to find the height, right?

Answer 14

actually change in height, but solving for height first and then taking the derivative is how you get that. So what have you got as far as that goes?

Answer 15

h=500/w^2

Answer 16

Nice, and we want the derivative of that with respect to time on both sides. what do you have for that?

Answer 17

would i do dt(h)=500/w^2(dt)

Answer 18

no we want the derivative
if this derivative is giving you problems rewrite the \[h={500\over x^2}\] as \[h=500x^{-2}\]

Answer 19

ohhhh ok

Answer 20

can you do this derivative with respect to time? (it requires implicit differentiation)

Answer 21

So would that be -1000w????

Answer 22

so close, but you forogt the exponent on the x and you need to remember that this derivative is not w/respect to x, but to t, so we need to use the chain rule:\[{d \over dt}f[g(t)]=f'[g(t)]g'(t)\]so here we get\[{dh \over dt}={-1000}w^{-1}{dw \over dt}={-1000}w^{-1}{dw \over dt}={-100\over w}{dw \over dt}\]please try to see the errors you made.
Now you can just plug in the numbers. What do you get for that?

Answer 23

Oh I'm sorry I see I misread the question and sat solve it properly on the other post, disregard most of the above, I thought it gave the rate change of the SIDE of the base, not the area, so my work is invalid. My mistake

Answer 24

Lol it's fine!!!!So can we do it over?

Answer 25

You mean you want to do it over with me again? I can do it right this time if so.

Answer 26

Yes, because I you're making me actually do the work so that I can learn. I need to ask questions and see how to actually work it out

Answer 27

Cool, ok so let me try to combine my approach with satellite's.

I gave you \[V=w^2h=500\]which you understood, so how can we rewrite this formula in terms of the area of the base? What can we substitute?

Answer 28

Can we substitue the w?

Answer 29

Because we're looking for the rate of change in the height?

Answer 30

Yes, well we want area in there so we need a representation of the area if the base.

A side of our square base is w, so what is the area of that base in terms of w?

After finding that and substituting A into our formula, what does it become?

Answer 31

A^2h=500?

Answer 32

Not quit, watch:

area of base=length of base times width of base

so A=lw

in this case, because it is a square l=w so our formula becomes

A=w^2

subbing that into

w^2h=500 yields...?
sorry that took a while my connection seems bad:(

Answer 33

ohhh because we already have the volume? Is that why we can substitute A for V? And I'm not quite sure where to go from there...

Answer 34

No it's fine! Mine i bad too.

Answer 35

No we cannot sub A for V, we can sub A for w^2. What do you get subbing A for w^2 into
w^2h=500
?

Answer 36

A^2h=500, which would be h=500/A^2?

Answer 37

no we get \[Ah=500\]we don't need to include the square because we subbed A=x^2, not A=x
so solving for h, which I know you can do so I won't test you, gives\[h={500\over A}\]please try to think about the logic behind that, it is fairly important. So what do you get for the derivative w/respect to time for that
(remember that the chain rule, which I gave above, is necessary because this is respect to time, not w)

Answer 38

?

Answer 39

hint: remember this is the same as \[h=500A^{-1}\]

Answer 40

so would it be dh/dt=500A^-1 dA/dt???? And i'm not sure after that

Answer 41

I just copied what you typed from before

Answer 42

check that derivative again. the chain rule part is right, but the derivative of 500A^-1 is wrong, you left it the same.

Answer 43

would it be -500 ?

Answer 44

You're still there?

Answer 45

no careful with the power rule
\[{d \over dt}500A^{-1}=-500A^{-2}={-500\over A^2}{dA \over dt}\]now we just need to find A and we can solve our problem. since the question tells us the length of the side at this moment, and the rate change of the area,
1)what is A ?
2)what is dA/dt ?
3)what is the answer ?

again sorry for the connection business...

Answer 46

Wait here does the A^2 come from?

Answer 47

the derivative of A^-1 is -A^-2, which you hopefully remember can be written as -1/A^2

Answer 48

Would A be 18? and the dA/dt become 3

Answer 49

dA/dt=3, that's correct
but it says the length of ONE SIDE (w) is 18, so what is the area of the base? remember

A=w^2 for a square base

then, what do you get for an answer?

Answer 50

-.0142....

Answer 51

Exactly! well, to nitpick -0.1428..., so you should round it to -0.0143 but you got it. I hope that helped.

I must say it seems you need to practice some basics, like the power rule, the chain rule, and why that substitution gives Ah=500, not A^2h=500.

If you have any questions let me know!

Answer 52

Lol well thank you so much. And I understand the Chain rule and the power rule. I just don't know how to apply to this optimization problem

Answer 53

I do have another question about expressing a relationship between small changes and correspondence...

Answer 54

umm... not sure what you mean by those terms but you can try me :)

Answer 55

Ok this is the question: Express the relationship between a small change in x and the corresponding change in y=ln(5+x^3) in the form of dy=f'(x)dx

Answer 56

I have no clue how to do this and I am horrible at logs

Answer 57

Well, no log rules required here. The question is about differentials, which basically amounts to taking a derivative and leaving a "dx" at the end.

That's what they means by a small change, that's what a differential represents.

So all we need to do is take the derivative of y=ln(5+x^3)
and scribble a dx next to it on the right.

Can you take that derivative? it requires

1)The chain rule
2)The derivative of the form ln(x)

Show me what you get, and don't forget to write dx next to it!

Answer 58

d/dx\[dy= \ln(5+x^3) dx\]

Answer 59

minus the d/dx, i meant to delete that

Answer 60

would the (5+x^3) become 3x^2(5+x^3)?

Answer 61

Well, no log rules required here. The question is about differentials, which basically amounts to taking a derivative and leaving a "dx" at the end.

That's what they means by a small change, that's what a differential represents.

So all we need to do is take the derivative of y=ln(5+x^3)
and scribble a dx next to it on the right.

Can you take that derivative? it requires

1)The chain rule
2)The derivative of the form ln(x)

Show me what you get, and don't forget to write dx next to it!

Answer 62

You forgot about the natural log part! can you do the derivative of\[ y=\ln x\]?

Answer 63

I hope I don't come off as mean here, but this is why I say you may need to reevaluate some basics. try again and remember to take the derivative of \[ \ln(5+x^3) \]1)derivative of natural log
2)chain rule
3)power rule
Apply those steps in that order and you should get your answer.

Answer 64

So the derivative of y=ln x is y=1/x

Answer 65

good! so using that, the chain rule, and the part you did correctly, plus the dx, gives...

Answer 66

But I'm not good at logarithmic functions. and no you're fine. You're seeing my faultws and errors and where I need to study more at

Answer 67

I meant faults*

Answer 68

would that become 1/5+x^3 ?

Answer 69

Well, no log rules required here. The question is about differentials, which basically amounts to taking a derivative and leaving a "dx" at the end.

That's what they means by a small change, that's what a differential represents.

So all we need to do is take the derivative of y=ln(5+x^3)
and scribble a dx next to it on the right.

Can you take that derivative? it requires

1)The chain rule
2)The derivative of the form ln(x)

Show me what you get, and don't forget to write dx next to it!

Answer 70

Well try to see that not being "good at logarithmic functions" doesn't matter much, it's just a rule to remember

\[{d \over dx}\ln x={1\over x}\]I'm afraid the above is still not right. This is why I say you should reexamine the chain rule.

I'm afraid it's quite late where I am so I'm afraid I'm just going to give the answer at this point? I think with the info I gave you you should be able to see how I got it. \[dy={3x^2dx \over5+x^2}\]Notice the chain rule left the whole 5+x^2 in the denominator. Think about the formula
\[{d \over dx}f[g(x)]=f'[g(x)]g'(x)\]

Answer 71

Goodnight and good luck, I'm sure you'll get the hang of it :)

Answer 72

Thank you!!!