anonymous
  • anonymous
Calculus Help!
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Rate 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
  • DanJS
instantaneous rate of change, also known as the derivative in this case it will be with respect to side length x
DanJS
  • DanJS
need to take \[\frac{ d }{ dx }\] of both sides

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anonymous
  • anonymous
Wait what is d/dx?
DanJS
  • DanJS
the derivative with respect to x, the side length, or sometimes it could be maybe, d/dt , with respect to time, or anything else
DanJS
  • DanJS
wait, what have you learned so far? just limits?
anonymous
  • anonymous
Ya I only learned that.
DanJS
  • DanJS
so you have to use the limit definition of the instantaneous rate of change then i guess
anonymous
  • anonymous
How would you do that?
DanJS
  • DanJS
the [ f(x)+delta x) - f(x)] / delta x
anonymous
  • anonymous
Isn't that \[\frac{ f(x+h)-f(x) }{ h }\]?
anonymous
  • anonymous
Is that the same thing?
DanJS
  • DanJS
\[\lim_{\Delta x \rightarrow 0}\frac{ f(x+\Delta x )-f(x)}{ \Delta x }\]
DanJS
  • DanJS
yeah
anonymous
  • anonymous
Okay so what would be h and what would be x?
DanJS
  • DanJS
Trust me , the class will get easier after this , well less work at least
anonymous
  • anonymous
Does f(x)=6x^2?
anonymous
  • anonymous
Ya I wish I had a good teacher though. She doesn't even teach!!
DanJS
  • DanJS
use khan academy, lots of vids there, or MIT opencourse ware... even better if you have time
anonymous
  • anonymous
She can't "explain the method". She can solve it but she can't explain it. I don't see why she is teaching.
anonymous
  • anonymous
Ya I was able to understand most of it. I just couldn't find the ones talking about Rate of Change.
DanJS
  • DanJS
rate of change is just like from algebra rise over run -- a slope
DanJS
  • DanJS
y has a rate of change for each x change
anonymous
  • anonymous
Sorry but that was just my opinion about her.
anonymous
  • anonymous
So plug in x in 6x^2?
anonymous
  • anonymous
x=0
anonymous
  • anonymous
So y=0?
DanJS
  • DanJS
but the derivative you are gonna learn will show you what the slope is of a tangent line to a curve
DanJS
  • DanJS
there is only 1 point how can that have a slope?
DanJS
  • DanJS
that 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
  • DanJS
that is all the derivative is, then you will learn like 10 shortcuts and ways to find derivatives for like 2 months
DanJS
  • DanJS
sorry i got off track
anonymous
  • anonymous
It's fine. What would I do with a if x=a?
DanJS
  • DanJS
Rate 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
  • DanJS
the function is S(x) surface area as a function of length of side
anonymous
  • anonymous
Okay
DanJS
  • DanJS
so 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
  • DanJS
so in the limit definiton use S(a) = 6a^2 and plug all that in
anonymous
  • anonymous
But then what would a equal?
DanJS
  • DanJS
a is just a constant arbitrary value
anonymous
  • anonymous
S(a) =6a^2 is what then?
DanJS
  • DanJS
so that is the surface area value when x=a...
anonymous
  • anonymous
Is that it?
DanJS
  • DanJS
plug s(a) in for the f(x) in that limit definition of inst. rate of change
anonymous
  • anonymous
Where would S(a) be plugged into? f(x+h) or f(x) or h?
DanJS
  • DanJS
\[\lim_{h \rightarrow 0}\frac{ S(x+h) - S(x) }{ h }=\]
DanJS
  • DanJS
actually just keep it as the general S(x) function for now
anonymous
  • anonymous
Okay so what is s(x+h) equal?
DanJS
  • DanJS
s(x) = 6x^2 replace all x with (x+h) s(x+h) = 6(x+h)^2
DanJS
  • DanJS
so you get \[\lim_{h \rightarrow 0}\frac{ 6(x+h)^2 - 6x^2 }{ h }\]
DanJS
  • DanJS
you get it all so far?
anonymous
  • anonymous
Ya I got that part.
DanJS
  • DanJS
so 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
  • DanJS
expand and combine the top part
anonymous
  • anonymous
Okay hold on.
anonymous
  • anonymous
12x+6h
DanJS
  • DanJS
\[\lim_{h \rightarrow 0}\frac{ 6(x+h)^2 - 6x^2 }{ h }~~=~~\lim_{h \rightarrow 0}\frac{ 12hx+6h^2}{ h }\]
DanJS
  • DanJS
--- (x+h)^2 = x^2 + 2xh + h^2
DanJS
  • DanJS
you see the x^2 term canceled out nicely, 6x^2 - 6x^2
DanJS
  • DanJS
got to hurry a bit,
anonymous
  • anonymous
Ya I got that.
DanJS
  • DanJS
If 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
  • anonymous
So the answer is -12x
DanJS
  • DanJS
\[\lim_{h \rightarrow 0}(12x + 6h)\]
DanJS
  • DanJS
positive 12 yeah, put h=0 in
DanJS
  • DanJS
we 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
anonymous
  • anonymous
So that's it?
DanJS
  • DanJS
pretty much... So overall this is a box that is changing size at a constant linear rate (since 12x is a linear function)
DanJS
  • DanJS
and now all they want you to do is calculate it when x=a S(x) = 6x^2 S ' (x) = 12x ---> S ' (a) = 12*a
anonymous
  • anonymous
Okay thanks!
DanJS
  • DanJS
let me show you something real fast that will help
DanJS
  • DanJS
we can start calling the 'instantaneous rate of change now' the derivative... notice... S(x) = 6x^2 S ' (x) = 12x
DanJS
  • DanJS
The first shortcut to finding the derivative is the simple power rule... for exponents like S(x) the rule is simply....
DanJS
  • DanJS
the 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
  • anonymous
Okay. So how would the derivitive help in this?
DanJS
  • DanJS
\[\huge n*a*x ^{n-1}\]
DanJS
  • DanJS
the surface area function was S(x) = 6x^2 after all that work, we figured the derivative was S ' (x) = 12x
DanJS
  • DanJS
all you have to do is... follow that power formula... 6*2*x^(2-1)
anonymous
  • anonymous
Wow!!! That is seriously so simple with that method!!
DanJS
  • DanJS
or 12x
DanJS
  • DanJS
haha, yeah, but the limit stuff helps to understand better what the derivative is...
DanJS
  • DanJS
so the derivative of f(x) = 10x^3 + 12x^2 + 3
DanJS
  • DanJS
f '(x) = 30x^2 + 24x
DanJS
  • DanJS
derivative of a constant number is zero
DanJS
  • DanJS
the limit thing we did will get the same answer with all the algebra...
DanJS
  • DanJS
aight, got to go.. goodluck
anonymous
  • anonymous
Thanks!
DanJS
  • DanJS
ill add you as friend and can help ya out whenever i am on
anonymous
  • anonymous
That would be great! I'll do the same. :)
DanJS
  • DanJS
no prob, i would rather recall and relearn calc again than do another system of linear equations .. hah bye
anonymous
  • anonymous
Alright bye!

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