Calculus Help!

- anonymous

Calculus Help!

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- 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

instantaneous rate of change, also known as the derivative
in this case it will be with respect to side length x

- DanJS

need to take
\[\frac{ d }{ dx }\] of both sides

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## More answers

- anonymous

Wait what is d/dx?

- 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

wait, what have you learned so far? just limits?

- anonymous

Ya I only learned that.

- DanJS

so you have to use the limit definition of the instantaneous rate of change then i guess

- anonymous

How would you do that?

- DanJS

the [ f(x)+delta x) - f(x)] / delta x

- anonymous

Isn't that \[\frac{ f(x+h)-f(x) }{ h }\]?

- anonymous

Is that the same thing?

- DanJS

\[\lim_{\Delta x \rightarrow 0}\frac{ f(x+\Delta x )-f(x)}{ \Delta x }\]

- DanJS

yeah

- anonymous

Okay so what would be h and what would be x?

- DanJS

Trust me , the class will get easier after this , well less work at least

- anonymous

Does f(x)=6x^2?

- anonymous

Ya I wish I had a good teacher though. She doesn't even teach!!

- DanJS

use khan academy, lots of vids there, or MIT opencourse ware... even better if you have time

- 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

Ya I was able to understand most of it. I just couldn't find the ones talking about Rate of Change.

- DanJS

rate of change is just like from algebra rise over run -- a slope

- DanJS

y has a rate of change for each x change

- anonymous

Sorry but that was just my opinion about her.

- anonymous

So plug in x in 6x^2?

- anonymous

x=0

- anonymous

So y=0?

- DanJS

but the derivative you are gonna learn will show you what the slope is of a tangent line to a curve

- DanJS

there is only 1 point how can that have a slope?

- 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

that is all the derivative is, then you will learn like 10 shortcuts and ways to find derivatives for like 2 months

- DanJS

sorry i got off track

- anonymous

It's fine. What would I do with a if x=a?

- DanJS

- DanJS

the function is S(x) surface area as a function of length of side

- anonymous

Okay

- 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

so in the limit definiton use S(a) = 6a^2 and plug all that in

- anonymous

But then what would a equal?

- DanJS

a is just a constant arbitrary value

- anonymous

S(a) =6a^2 is what then?

- DanJS

so that is the surface area value when x=a...

- anonymous

Is that it?

- DanJS

plug s(a) in for the f(x) in that limit definition of inst. rate of change

- anonymous

Where would S(a) be plugged into? f(x+h) or f(x) or h?

- DanJS

\[\lim_{h \rightarrow 0}\frac{ S(x+h) - S(x) }{ h }=\]

- DanJS

actually just keep it as the general S(x) function for now

- anonymous

Okay so what is s(x+h) equal?

- DanJS

s(x) = 6x^2 replace all x with (x+h)
s(x+h) = 6(x+h)^2

- DanJS

so you get
\[\lim_{h \rightarrow 0}\frac{ 6(x+h)^2 - 6x^2 }{ h }\]

- DanJS

you get it all so far?

- anonymous

Ya I got that part.

- 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

expand and combine the top part

- anonymous

Okay hold on.

- anonymous

12x+6h

- DanJS

\[\lim_{h \rightarrow 0}\frac{ 6(x+h)^2 - 6x^2 }{ h }~~=~~\lim_{h \rightarrow 0}\frac{ 12hx+6h^2}{ h }\]

- DanJS

--- (x+h)^2 = x^2 + 2xh + h^2

- DanJS

you see the x^2 term canceled out nicely, 6x^2 - 6x^2

- DanJS

got to hurry a bit,

- anonymous

Ya I got that.

- 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

So the answer is -12x

- DanJS

\[\lim_{h \rightarrow 0}(12x + 6h)\]

- DanJS

positive 12 yeah, put h=0 in

- 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

So that's it?

- 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

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

Okay thanks!

- DanJS

let me show you something real fast that will help

- DanJS

we can start calling the 'instantaneous rate of change now' the derivative...
notice...
S(x) = 6x^2
S ' (x) = 12x

- DanJS

The first shortcut to finding the derivative is the simple power rule... for exponents like S(x) the rule is simply....

- 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

Okay. So how would the derivitive help in this?

- DanJS

\[\huge n*a*x ^{n-1}\]

- DanJS

the surface area function was
S(x) = 6x^2
after all that work, we figured the derivative was
S ' (x) = 12x

- DanJS

all you have to do is... follow that power formula...
6*2*x^(2-1)

- anonymous

Wow!!! That is seriously so simple with that method!!

- DanJS

or 12x

- DanJS

haha, yeah, but the limit stuff helps to understand better what the derivative is...

- DanJS

so the derivative of
f(x) = 10x^3 + 12x^2 + 3

- DanJS

f '(x) = 30x^2 + 24x

- DanJS

derivative of a constant number is zero

- DanJS

the limit thing we did will get the same answer with all the algebra...

- DanJS

aight, got to go.. goodluck

- anonymous

Thanks!

- DanJS

ill add you as friend and can help ya out whenever i am on

- anonymous

That would be great! I'll do the same. :)

- DanJS

no prob, i would rather recall and relearn calc again than do another system of linear equations .. hah bye

- anonymous

Alright bye!

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