## anonymous 5 years ago The mechanics at Lincoln Automotive are reboring a 6-in deep cylinder to fit a new piston. The machine they are using increases the cylinders radius one-thousandth of an inch every 4min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800in

1. anonymous

okay, you can write it down like this in every 4 mins ---> radius increase = 1000in in every x mins? ----> radius increase R= D/2 =1.504 after finding the answer in mins, convert to seconds :) give it a try now :)

2. anonymous

cross multiply

3. anonymous

4. anonymous

they want how much is the volume increasing right?

5. anonymous

Its okay :)

6. anonymous

at what rate is the volume increasing, supposed to use derivitives (this is calculus)

7. anonymous

then find the derivative of Volumeof Cylinder

8. anonymous

you have r' = 1000

9. anonymous

all you have to do is find h from the original equation which is Vol = 1/3 pi r^2 h using r = 1.504

10. anonymous

$Vol = 1/3 \pi r^2h , Vol' = 2/3 \pi rr'h'$ I guess

11. anonymous

you have r' = 1000 r= 1.504 you'll have to find h'

12. anonymous

am I making any sense lol?

13. anonymous

Not really...

14. anonymous

How did you get r' ?

15. anonymous

jmcwhs2012, if you allow, may I say a few words

16. anonymous

it's given, it says that the radius is increasing 1000 in every 4 mins so r' = 1000in/min

17. anonymous

18. anonymous

sure:)

19. anonymous

"one-thousandth of an inch"

20. anonymous

Okay, as far as I understand, the rate at which the radius is increasing is equal to the radius bored in a second

21. anonymous

you have the original radius, you can computer r' I guess it's 1000(1.504)

22. anonymous

So we can write dr/dt= (2.54/100000)/(60*4)

23. anonymous

compute*

24. anonymous

Now all we need to find is dv/dt

25. anonymous

I dont understand where you got dr/dt?

26. anonymous

So we can use the result (dr/dt)x(dv/dr)

27. anonymous

(dr/dt)x(dv/dr)=dv/dt

28. anonymous

@jmcwhs2012 the thing is coming like this 2.54 cm is equal to one inch,

29. anonymous

Why do we have to use cm?

30. anonymous

I later convert it into m

31. anonymous

You see for that I divide it by 100000

32. anonymous

Ok let me explain the whole thing

33. anonymous

The machine is working at the rate of 1/1000 inch per 60*4 secs

34. anonymous

is that clear?

35. anonymous

wait @iam , he wants to find V'

36. anonymous

Thats what I am doing, if you all allow me

37. anonymous

lol, sorry, proceed :)

38. anonymous

The machine is working at the rate of 1/1000 inch per 60*4 secs, is that thing clear?

39. anonymous

yes, got you.

40. anonymous

Which means 2.54/100000 m per 240 secs

41. anonymous

So now we can find dr/dt from there

42. anonymous

okay

43. anonymous

But we need to find dv/dt

44. anonymous

so, let's find it

45. anonymous

So we can use the formula dv/dt=(dr/dt)x(dv/dr)

46. anonymous

Is this step clear

47. anonymous

honestly, no, why are you multiplying them?

48. anonymous

wait! chain rule?

49. anonymous

Write that thing on a piece of paper, and try to cancel the commong terms, you will get why I am doing that

50. anonymous

Yes chain rule

51. anonymous

lol, got you! okay then what?

52. anonymous

So we have already found dr/dt which is 2.54/100000/240

53. anonymous

Now we need to find dv/dr

54. anonymous

alright

55. anonymous

For that we will have to differenciate 6 x pi x r^2 with respect to r

56. anonymous

why 6 pi r^2?

57. anonymous

Oops I am sorry, it would be 6*2.54*pi*r^2

58. anonymous

6 is the length of the cylinder given on the problem

59. anonymous

I don't get where you got 6?

60. anonymous

oh, the height

61. anonymous

yes

62. anonymous

So now just we will have to multiply both the things

63. anonymous

but isn't the vol = 1/3 pi r^2 h?

64. anonymous

Thats for a cone

65. anonymous

which will be 2 pi r^2?

66. anonymous

lol...sorry ^^"

67. anonymous

So we just need to multiply both the things

68. anonymous

then differentiate with respect to r :)

69. anonymous

But unfortunately the person who gave this problem is not much interested

70. anonymous

No no no need to differenciate again

71. anonymous

lol, he'll come back later :)

72. anonymous

Its over

73. anonymous

so V = 15.34 pi r^2?

74. anonymous

yes

75. anonymous

what is r? 1.504?

76. anonymous

I didn't get you

77. anonymous

r is a variable

78. anonymous

wouldn't you derive to V' then substitute r = 1.504 and r' in the equation and then u are done?

79. anonymous

No not 1.504, we will have to substitute 1.900

80. anonymous

then by then you have found V' which is the volume rate

81. anonymous

I didn't get you

82. anonymous

alright, 1.900 and r' the one you have found which is dr/dt, then you're done

83. anonymous

right? :)

84. anonymous

Actually I am not getting you at all

85. anonymous

{(2.54/100000)/240}*2.54*6*pi*2*1.900=dv/dt Thats it and thats all

86. anonymous

so the given is: r = 1.900 r' = dr/dt = 2.54 so V' = 15.34(2) pi rr' = 30.68 pi (1.900)(2.54) = 465.15 m / secs So the volume is increasing = 465.15m/sec right?

87. anonymous

simply ^_^

88. anonymous

dr/dt = 2.54 this is not right

89. anonymous

so then it's 2.54 /1000/240?

90. anonymous

= 1.05 x10^-5

91. anonymous

dr/dt ={(2.54/100000)/240} dv/dr= 2.54*6*pi*2*1.900

92. anonymous

so: V' = dv/dr = 2.54*12 pi *1.900 = 181.94 m/secs