## anonymous one year ago university 1st year math question.

1. welshfella

i looked up torricelli's law and only found a formula for a cylindrical tank..

2. anonymous

@ganeshie8

3. anonymous

@IrishBoy123

4. anonymous

@ganeshie8

5. anonymous

i am finding it very difficult

6. ganeshie8

@ParthKohli try this

7. ParthKohli

are you mathmath333?

8. anonymous

what ?

9. ganeshie8

No, he is not

10. ParthKohli

I haven't studied fluids yet, but I looked at the Wikipedia page and it sorta makes sense.

11. ParthKohli

This is what we're really looking at though. https://en.wikipedia.org/wiki/Torricelli%27s_law#Application_for_time_to_empty_the_container

12. ganeshie8

same here, it seems torcelli's law gives the speed of fluid of height $$h$$, leaving tank from bottom : $v = \sqrt{2gh}$

13. anonymous

noo $v=\sqrt{2gh}$

14. ytrewqmiswi

What does that k represents?

15. anonymous

i was thinking of an approach where i find A(h) and use integration

16. ParthKohli

Is that the area of cross-section as a function of height? And are you sure this is the equation?$A(h) \frac{dh}{dt} = -k \sqrt{h}$

17. anonymous

yes 100% because we havent used other v=sqrt2gh yet and my teacher wouldnt expect us to use that

18. anonymous

the only problem i am having is how to find A(h)

19. ytrewqmiswi

That wuld mean that the area of cross section varies with height will that make any sense?

20. ParthKohli

But in your equation, $$\sqrt{2g}$$ or $$g$$ won't appear. We can find the area of the cross-section as a function of height.

21. ParthKohli

If my eyes work, we're looking at a rectangular cross-section, right?

22. anonymous

yes

23. anonymous

and yes i wan to find the area as a function of height plz

24. ParthKohli

Width increases in a linear manner from 4 to 5. At h = 0, it is 4 and at h = 4, it is 5. Thus in general it should be $$w(h) = h/4 + 4$$. Meanwhile the length remains 10 so $$l(h) = 10$$. Multiply...

25. anonymous

$\frac{ 10h }{ 8 }$

26. ParthKohli

noooo I didn't write h/8 h/4+4 means (1/4)*h + 4

27. anonymous

$\frac{ 10(h+16) }{ 4 }$

28. ganeshie8

40----> 50 A(y) = 40 + (y/4)*10 does that work ?

29. ParthKohli

looks good

30. ParthKohli

it only works because one of them remains constant can't do the same thing when both width and length increase which is why it's a good practice to do them separately

31. ganeshie8

yeah 10cm is fixed, only the other dimension is changing, so it is linear

32. anonymous

just a question srry interpreting ur thought process how did u get width as (1/4) times h +4

33. ganeshie8

$A(h)\frac{ dh }{ dt }=-k \sqrt{h}$ plugin $$A(h)$$, the de becomes $(40+(\frac{h}{4})*10)\frac{ dh }{ dt }=-k \sqrt{h}$

34. ytrewqmiswi

the liquid escapes with a velocity =sqrt(2gh) which varies with the level of liquid. Hence, we have to use integral. Let y be the height of the liquid at an instant. This height changes by dy in time dt. Volume of water leaving out per secound=-Ady.dt at the hole volume escaping per second is av=a[sqrt(2gy)] a[sqrt(2gy)] =-Ady/dt ---->$\int\limits_{H}^{0}\frac{ -dy }{ \sqrt{y} } =\frac{ a \sqrt{2g} }{ A }\int\limits_{t}^{0}dt$ $t=\frac{ A }{ a \sqrt{2g}}2\sqrt{H}$ a<<A? what did i do o.o

35. ganeshie8

maybe think of it in reverse : A(h) = 40 + (h/4)*10 plugin h=0, what do you get ? plugin h=4, what do you get ?

36. ParthKohli

|dw:1443179501695:dw|

37. ytrewqmiswi

well...that thing can be used to find time tho..?

38. anonymous

@ytrewqmiswi we cant use v=sqrt2gh because well i want the area as a function of h

39. ytrewqmiswi

A = cross sectional area of tank H=height till water is filled a=cross sectional area of hole

40. ParthKohli

but didn't you assume A to be constant?

41. ytrewqmiswi

yes i took it constant

42. ParthKohli

yes that solution matches ours so far

43. anonymous

good because i am not sure about my final answer it doesnt make sense so i need to be assured

44. ParthKohli

you shouldn't have any problem with integration

45. anonymous

noo

46. anonymous

but would love it if you solve it step by step plzzz

47. anonymous

it will make my life so easy

48. ganeshie8

you don't have teamviwer ?

49. anonymous

again what is teamviewer?

50. anonymous

do i have to buy it ?

51. ganeshie8

no, it is a free screen sharing app

52. anonymous

for the solution u an see i am stuck in the end

53. anonymous

ahm do i have to get it for this question

54. ParthKohli

$A(h) \frac{dh}{dt} = -k\sqrt{h}$We've got all messy differentials here. What we do is we take all the $$h$$-terms to to one side and $$t$$-terms to the other. Then integrate both sides with proper limits.

55. anonymous

ok

56. ParthKohli

$\frac{A(h)}{\sqrt h} dh=-kdt$

57. anonymous

$\frac{ A(h) }{ \sqrt{h} } dh=-k dt$

58. ParthKohli

Great. Now when $$t= 0$$, $$h=4$$. When $$t = t_0$$, $$h = 0$$. So you integrate both sides with those limits corresponding to each other.

59. anonymous

how?

60. ParthKohli

OK, first put the value of $$A(h)$$ in the equation and simplify. You don't have to type it out here. Keep doing that on paper. I'll guide you through each step.

61. anonymous

which A(h) value must i put?

62. ParthKohli

The one we found out earlier?

63. ParthKohli

We found the value of area of the cross-section in terms of height earlier, yes?

64. anonymous

$\frac{ 10(h+16) }{ 4 }\ /\sqrt{h} =-kdt$

65. ParthKohli

Good. I know you're having trouble with typing everything out. So don't. Just keep doing it on paper alright?

66. anonymous

no i need to show u , i can be writting jubrish

67. ganeshie8

//ahm do i have to get it for this question Absolutely not, but considering the fact that you already have a detailed solution which you don't understand it, the jibber jabber that I do here only confuses you more. If you have teamviewer, we could use voice also and engage more efficiently.

68. anonymous

$\frac{ h+16 }{ \sqrt{h}} dh = - \frac{ 2 }{ 10 } \times 0.03 dt$

69. anonymous

@ganeshie8 i am on computer , will that work on this and how do i add uz

70. ganeshie8

Hey it is really not required, let try and finish it off here

71. anonymous

ok

72. ganeshie8

have you deleted the attachment ?

73. anonymous

noo

74. ganeshie8

i couldn't find the solution above

75. ganeshie8

76. anonymous

ok u see i did the A(h) wrong then towards the end i thought when t=0 , h=3

77. anonymous

my solution is wrong

78. anonymous

i did A(h) wrong

79. ganeshie8

your A(h) looks fine to me

80. anonymous

and i dont seem to get what that refers to as 3 quarter filled

81. anonymous

so plz ignore the solution

82. ganeshie8

dude you're good, don't you see what you have is same as A(h) = 40 + (h/4)*10

83. anonymous

ok so my solution makes sense all of it ok what about part c

84. anonymous

i just gave up on that

85. anonymous

i mean iii

86. anonymous

and also i dunno how to go furthur from here

87. ganeshie8

so whats you final answer for part i

88. anonymous

i dunno what is h and t if i dunno t how am i suppose to find h

89. anonymous

90. ganeshie8

Notice that the depth of water, h, is 0 when the tank is empty. so, for part iii, i think you simply plugin h=0 and solve $$t$$

91. anonymous

is this what u saying?

92. ganeshie8

I am asking you to simply replace $$h$$ by $$0$$ because, empty tank means that the depth of water is 0

93. anonymous

ohhh

94. ganeshie8

$\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89$ $0=-0.012t+58.89$ solve $$t$$ and you're done!

95. anonymous

t=4907.5

96. anonymous

just one question did uz consider the three quarter filled thing

97. ganeshie8

Oops! we missed that, my mistake, sorry, lets start over..

98. anonymous

ok so part i is finding an expression we have done it

99. anonymous

ur being sarcastic right

100. ganeshie8

I'm never sarcastic

101. ganeshie8

we need to start over, our work for all parts is wrong because we assumed that the tank was full

102. anonymous

ok

103. ganeshie8

gimme some time, il type in the complete solution

104. anonymous

ok

105. Jhannybean

nice comforter I mean. Lol.

106. ganeshie8

Hey, it turns out our previous solution works out perfectly fine because we have worked it from bottom.

107. ganeshie8

No need to redo anything. we are good.

108. ganeshie8

@ParthKohli please double check if psble

109. anonymous

srry last thing what about part b

110. anonymous

as u said make h=0 and solve for T

111. anonymous

thnx guys cheers i got it

112. ganeshie8
113. ganeshie8

deleted, let me know if you want me delete any other stuff

114. anonymous

no its ok thak uso much

115. ganeshie8

np

116. anonymous

how can i manupilate above equation so the subject is h because thats what it is ultimately asking

117. anonymous

@dan815

118. anonymous

@pooja195

119. anonymous

but the question is asking me to make h the subject

120. dan815

ok

121. dan815

it says you can use wolfrad alpha so use that

122. anonymous

noo i cant because its for graph only h vs t

123. dan815

yeah?

124. anonymous

when u read the question b (i) last line determine the depth of the milk at any instant after the drain has opened

125. anonymous

so we need to find expression where h is the subject and i did use wolfram alpha it didnt work there were always two h

126. anonymous

did u understand i am after i want an expression where h = .....

127. dan815

$\frac{ 2 }{ 3 }\sqrt{h} (h+48)=-0.012t+58.89$

128. dan815

you just specify in wolfram plot h vs t

129. anonymous

look i am not talking about ii use wolfram alpha to obtain plot h verus t

130. dan815

what do u want then

131. anonymous

i am talking about question above (i)

132. dan815

oh i thought u are done that

133. anonymous

nooo

134. dan815

do u want me to solve b(i)

135. anonymous

how do i make h the subject

136. anonymous

in this expression

137. anonymous

how do i get rid of sqrt h

138. anonymous

so i have an expression like h =..... when i manipulate above eq

139. dan815

it's not going to be pretty

140. dan815

yes.. this 0.03 for k means nothing to me

141. anonymous

k is just viscosity its a constant

142. dan815

doesnt help

143. anonymous

see the solution u will realise

144. dan815

k im assuming is a function of the radius of the opening

145. dan815

yep see that equation u are using there is an application of the bernoullis principle, where that k in there would be some factor with the radius of the opening at the bottom accounted in for it

146. anonymous

yea can u see any way where h is the subject

147. dan815

what do u mean h is the subject

148. anonymous

ok h is the depth

149. anonymous

at 4 sec

150. anonymous

h= depth in equation and t = time

151. dan815

what you have is good enough

152. dan815

i dont see where it says to make h the subject in your question

153. anonymous

ok at 10 sec what would be the depth of tank can u give me answer for that by substituting 4 into the final eq

154. anonymous

substituing 10*

155. dan815

you gotta use nonlinear methods lol

156. anonymous

how would i use non linear method

157. dan815

just use matlab or something for now

158. anonymous

ok thnx for ur help btw before u go mind deleting this whole question plz

159. dan815

i have no powers

160. dan815

you know as far as applications and solutions go solving for t=f(h) is good enough

161. anonymous

yea i hope so

162. dan815

as you can graph that, and rotating it returns h=f(t)

163. anonymous

yea true

164. dan815

if you really must solve for h i can show u a way

165. anonymous

can u plz delete the expression above

166. dan815

it's not pretty yo uwill need to use complex numbers

167. dan815

and it might be too early for you right now in first year

168. anonymous

no dont i have solved its not pretttyyy

169. dan815

okay