anonymous one year ago Calculus Help!

1. 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.

2. DanJS

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

3. DanJS

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

4. anonymous

Wait what is d/dx?

5. 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

6. DanJS

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

7. anonymous

Ya I only learned that.

8. DanJS

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

9. anonymous

How would you do that?

10. DanJS

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

11. anonymous

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

12. anonymous

Is that the same thing?

13. DanJS

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

14. DanJS

yeah

15. anonymous

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

16. DanJS

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

17. anonymous

Does f(x)=6x^2?

18. anonymous

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

19. DanJS

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

20. 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.

21. anonymous

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

22. DanJS

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

23. DanJS

y has a rate of change for each x change

24. anonymous

Sorry but that was just my opinion about her.

25. anonymous

So plug in x in 6x^2?

26. anonymous

x=0

27. anonymous

So y=0?

28. DanJS

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

29. DanJS

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

30. 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

31. DanJS

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

32. DanJS

sorry i got off track

33. anonymous

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

34. 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.

35. DanJS

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

36. anonymous

Okay

37. 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)

38. DanJS

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

39. anonymous

But then what would a equal?

40. DanJS

a is just a constant arbitrary value

41. anonymous

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

42. DanJS

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

43. anonymous

Is that it?

44. DanJS

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

45. anonymous

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

46. DanJS

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

47. DanJS

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

48. anonymous

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

49. DanJS

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

50. DanJS

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

51. DanJS

you get it all so far?

52. anonymous

Ya I got that part.

53. 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

54. DanJS

expand and combine the top part

55. anonymous

Okay hold on.

56. anonymous

12x+6h

57. DanJS

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

58. DanJS

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

59. DanJS

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

60. DanJS

got to hurry a bit,

61. anonymous

Ya I got that.

62. 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...

63. anonymous

64. DanJS

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

65. DanJS

positive 12 yeah, put h=0 in

66. 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

67. anonymous

So that's it?

68. DanJS

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

69. 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

70. anonymous

Okay thanks!

71. DanJS

let me show you something real fast that will help

72. DanJS

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

73. DanJS

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

74. 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

75. anonymous

Okay. So how would the derivitive help in this?

76. DanJS

$\huge n*a*x ^{n-1}$

77. DanJS

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

78. DanJS

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

79. anonymous

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

80. DanJS

or 12x

81. DanJS

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

82. DanJS

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

83. DanJS

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

84. DanJS

derivative of a constant number is zero

85. DanJS

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

86. DanJS

aight, got to go.. goodluck

87. anonymous

Thanks!

88. DanJS

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

89. anonymous

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

90. DanJS

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

91. anonymous

Alright bye!