Question

Calculus Help!

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

Wait what is d/dx?

Answer 5

the derivative with respect to x, the side length, or sometimes it could be maybe, d/dt , with respect to time, or anything else

Answer 6

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

Answer 7

Ya I only learned that.

Answer 8

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

Answer 9

How would you do that?

Answer 10

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

Answer 11

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

Answer 12

Is that the same thing?

Answer 13

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

Answer 14

yeah

Answer 15

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

Answer 16

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

Answer 17

Does f(x)=6x^2?

Answer 18

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

Answer 19

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

Answer 20

She can't "explain the method". She can solve it but she can't explain it. I don't see why she is teaching.

Answer 21

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

Answer 22

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

Answer 23

y has a rate of change for each x change

Answer 24

Sorry but that was just my opinion about her.

Answer 25

So plug in x in 6x^2?

Answer 26

x=0

Answer 27

So y=0?

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

sorry i got off track

Answer 33

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

Answer 34

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.

Answer 35

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

Answer 36

Okay

Answer 37

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)

Answer 38

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

Answer 39

But then what would a equal?

Answer 40

a is just a constant arbitrary value

Answer 41

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

Answer 42

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

Answer 43

Is that it?

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

s(x+h) = 6(x+h)^2

Answer 50

so you get

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

Answer 51

you get it all so far?

Answer 52

Ya I got that part.

Answer 53

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

Answer 54

expand and combine the top part

Answer 55

Okay hold on.

Answer 56

12x+6h

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

got to hurry a bit,

Answer 61

Ya I got that.

Answer 62

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...

Answer 63

So the answer is -12x

Answer 64

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

Answer 65

positive 12 yeah, put h=0 in

Answer 66

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

Answer 67

So that's it?

Answer 68

pretty much...

So overall this is a box that is changing size at a constant linear rate (since 12x is a linear function)

Answer 69

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

Answer 70

Okay thanks!

Answer 71

let me show you something real fast that will help

Answer 72

we can start calling the 'instantaneous rate of change now' the derivative...

notice...

S(x) = 6x^2
S ' (x) = 12x

Answer 73

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

Answer 74

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

Answer 75

Okay. So how would the derivitive help in this?

Answer 76

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

Answer 77

the surface area function was

S(x) = 6x^2

after all that work, we figured the derivative was

S ' (x) = 12x

Answer 78

all you have to do is... follow that power formula...

6*2*x^(2-1)

Answer 79

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

Answer 80

or 12x

Answer 81

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

Answer 82

so the derivative of

f(x) = 10x^3 + 12x^2 + 3

Answer 83

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

Answer 84

derivative of a constant number is zero

Answer 85

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

Answer 86

aight, got to go.. goodluck

Answer 87

Thanks!

Answer 88

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

Answer 89

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

Answer 90

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

Answer 91

Alright bye!