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anonymous
 one year ago
Calculus Help!
anonymous
 one year ago
Calculus Help!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Rate of Change: Find the instantaneous rate of change of the surface area S = 6x^2 of a cube with respect to the edge length x at x = a.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1instantaneous rate of change, also known as the derivative in this case it will be with respect to side length x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1need to take \[\frac{ d }{ dx }\] of both sides

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the derivative with respect to x, the side length, or sometimes it could be maybe, d/dt , with respect to time, or anything else

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1wait, what have you learned so far? just limits?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ya I only learned that.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so you have to use the limit definition of the instantaneous rate of change then i guess

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would you do that?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the [ f(x)+delta x)  f(x)] / delta x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Isn't that \[\frac{ f(x+h)f(x) }{ h }\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that the same thing?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{\Delta x \rightarrow 0}\frac{ f(x+\Delta x )f(x)}{ \Delta x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so what would be h and what would be x?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1Trust me , the class will get easier after this , well less work at least

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ya I wish I had a good teacher though. She doesn't even teach!!

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1use khan academy, lots of vids there, or MIT opencourse ware... even better if you have time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0She can't "explain the method". She can solve it but she can't explain it. I don't see why she is teaching.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ya I was able to understand most of it. I just couldn't find the ones talking about Rate of Change.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1rate of change is just like from algebra rise over run  a slope

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1y has a rate of change for each x change

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry but that was just my opinion about her.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So plug in x in 6x^2?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1but the derivative you are gonna learn will show you what the slope is of a tangent line to a curve

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1there is only 1 point how can that have a slope?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1that is where the limit we are doing here comes from... as the change in x shrinks infinitley small, then you can have an 'instantaneous' rate of change...the rate of change of a single point? sounds messed up , but that is the power of calc

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1that is all the derivative is, then you will learn like 10 shortcuts and ways to find derivatives for like 2 months

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's fine. What would I do with a if x=a?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1Rate of Change: Find the instantaneous rate of change of the surface area S = 6x^2 of a cube with respect to the edge length x at x = a.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the function is S(x) surface area as a function of length of side

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so this is a dynamic system of a box growing or shrinking, they want to know the rate the surface area is changing the moment the side length x is a (x=a)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so in the limit definiton use S(a) = 6a^2 and plug all that in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But then what would a equal?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1a is just a constant arbitrary value

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0S(a) =6a^2 is what then?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so that is the surface area value when x=a...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1plug s(a) in for the f(x) in that limit definition of inst. rate of change

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where would S(a) be plugged into? f(x+h) or f(x) or h?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{h \rightarrow 0}\frac{ S(x+h)  S(x) }{ h }=\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1actually just keep it as the general S(x) function for now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so what is s(x+h) equal?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1s(x) = 6x^2 replace all x with (x+h) s(x+h) = 6(x+h)^2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so you get \[\lim_{h \rightarrow 0}\frac{ 6(x+h)^2  6x^2 }{ h }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so you have to do the algebra now and see if you can take the limit... as h goes to zero... the denominator goes to zero... not good

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1expand and combine the top part

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{h \rightarrow 0}\frac{ 6(x+h)^2  6x^2 }{ h }~~=~~\lim_{h \rightarrow 0}\frac{ 12hx+6h^2}{ h }\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1 (x+h)^2 = x^2 + 2xh + h^2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1you see the x^2 term canceled out nicely, 6x^2  6x^2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1If i remember, that is usually the case, the squared term cancels, so you can now factor an h out that will cancel with the h in the denominator...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer is 12x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{h \rightarrow 0}(12x + 6h)\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1positive 12 yeah, put h=0 in

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1we can call that function the instantaneous rate of change of the surface area with respect to the side length x, for any length x.... name it S prime (x) ... S ' (x) = 12x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1pretty much... So overall this is a box that is changing size at a constant linear rate (since 12x is a linear function)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1and now all they want you to do is calculate it when x=a S(x) = 6x^2 S ' (x) = 12x > S ' (a) = 12*a

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1let me show you something real fast that will help

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1we can start calling the 'instantaneous rate of change now' the derivative... notice... S(x) = 6x^2 S ' (x) = 12x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1The first shortcut to finding the derivative is the simple power rule... for exponents like S(x) the rule is simply....

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the derivative of any term like polynomial terms \[\huge a*x^n\] is found by this simple formula...multiply by the exponent, then decrease that exponent by 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay. So how would the derivitive help in this?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the surface area function was S(x) = 6x^2 after all that work, we figured the derivative was S ' (x) = 12x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1all you have to do is... follow that power formula... 6*2*x^(21)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow!!! That is seriously so simple with that method!!

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1haha, yeah, but the limit stuff helps to understand better what the derivative is...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1so the derivative of f(x) = 10x^3 + 12x^2 + 3

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1derivative of a constant number is zero

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the limit thing we did will get the same answer with all the algebra...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1aight, got to go.. goodluck

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1ill add you as friend and can help ya out whenever i am on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That would be great! I'll do the same. :)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1no prob, i would rather recall and relearn calc again than do another system of linear equations .. hah bye
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