## anonymous 5 years ago given y>0 and dy/dx=3x^2+4x/y. if the point (1,sqrt{10}) is on the graph relating x and y, then what is y when x=0?

1. TuringTest

no chance that's supposed to be dy/dx=(3x^2+4x)/y is it ?

2. myininaya

it is

3. myininaya

he didn't like my answer maybe you can try turing

4. TuringTest

Ah, I see... You should trust myininaya spikey

5. anonymous

lol i didnt understand it :(

6. myininaya

but no he might understand you better

7. myininaya
8. anonymous

dy/dx=(3x^2+4x)/y yes thats correct

9. myininaya

this is what i did turing if you want to look

10. anonymous

yep thats the other one :) do you wanna work off here or of the other one.

11. TuringTest

ok... this is a very basic separation of variables problem, in which we can treat dy/dx like a regular fraction... dy/dx=(3x^2+4x)/y ydy=3x^2+4xdx now integrate both sides... but according to the other post you don't seem to know how integrate yet :/

12. TuringTest

you don't know how to integrate that* correct?

13. anonymous

hmm im not sure. if you mean using anti derivaties than sort of

14. anonymous

$\int\limits_{?}^{?}$ but i havent used this yet

15. TuringTest

yes exactly antiderivatives

16. myininaya

$\int\limits_{}^{}x^n dx=\frac{x^{n+1}}{n+1}+C, n \neq -1$

17. myininaya

or you could say the antiderivative of x^n = that

18. anonymous

ahh yes that sort of makes sense.

19. anonymous

i understand that c is a constant.

20. TuringTest

check that the derivative of that gives x^n

21. myininaya

right

22. anonymous

the derivative of ydy=(3x^2+4x)dx?

23. TuringTest

antiderivative of that

24. myininaya

no we want to integrate both sides or we can use the term antiderivative

25. myininaya

like turing said

26. anonymous

ahh okay.

27. TuringTest

so what is the antiderivative of ydy according to the formula myin posted?

28. anonymous

x^3

29. anonymous

for 3x^2

30. anonymous

i dont know for 4x

31. TuringTest

yes but that is not ydy... like I said, I can't use latex, myin will have to show you

32. myininaya

$\int\limits_{}^{}y dy =\int\limits_{}^{}y^1 dy=\frac{y^{1+1}}{1+1}+c_1$

33. myininaya

see n was 1 here

34. anonymous

okay im struggling alot right now because i havent had enough practice with this.

35. myininaya

its cool

36. myininaya

struggling happens but the main is don't give up

37. anonymous

okay so the rule is for 3x^2 x^n+1

38. anonymous

what formula for the 4x?

39. anonymous

the anti derivative of dx is x?

40. TuringTest

4x=4x^(1) so we can use the same formula integral of (x^n)dx=x^(n+1)/(n+1)

41. myininaya

$\int\limits_{}^{}3x^2 dx=3 \int\limits_{}^{}x^2 dx=3 \cdot \frac{x^{2+1}}{2+1}+c_2$

42. myininaya

$\int\limits_{}^{}4x dx=4 \int\limits_{}^{}x dx=4\int\limits_{}^{}x^1 dx=4 \cdot \frac{x^{1+1}}{1+1}+c_3$

43. myininaya

see i'm using that same formula every time

44. anonymous

oh okay so i have to do it for each.

45. myininaya

right

46. myininaya

$\int\limits_{}^{}(f(x)+g(x))dx=\int\limits_{}^{}f(x) dx+\int\limits_{}^{}g(x)dx$

47. myininaya

so we have $\frac{y^2}{2}+c_1=3 \cdot \frac{x^{2+1}}{2+1}+c_2+4 \cdot \frac{x^{1+1}}{1+1}+c_3$

48. anonymous

so for the formula ydy you used the formula.

49. anonymous

okay i understand how u got that equation.

50. myininaya

or you could write instead $\frac{y^2}{2}=3 \cdot \frac{x^{2+1}}{2+1}+4 \cdot \frac{x^{1+1}}{1+1}+C$ since the sum of some constants is still a constant

51. anonymous

yes

52. myininaya

ok great! lets make this prettier

53. myininaya

$\frac{y^2}{2}=x^3+2 x^2+C$

54. myininaya

is that okay?

55. anonymous

yes my simplification came out the same

56. myininaya

ok we can also multiply two on both sides

57. myininaya

$y^2=2x^3+4x^2+C$

58. myininaya

2C is still constant so I left it as C

59. anonymous

yes

60. myininaya

you can write 2C if you feel more comfortable with that

61. anonymous

and now square both sides?

62. myininaya

square root of both sides

63. myininaya

we don't need to keep the negative value since your directions say y>0

64. anonymous

where do we have a negative value?

65. myininaya

take square root of both sides you get plus or minus

66. anonymous

thats right!

67. myininaya

we only need the plus since y>0

68. myininaya

so what I'm saying is that we have $y=\sqrt{2x^3+4x^2+C}$

69. anonymous

yes.

70. myininaya

you were given a point on this curve

71. myininaya

$(1,\sqrt{10} )$

72. myininaya

x=1 and y=sqrt(10)

73. anonymous

yes i plug it in for x and y?

74. myininaya

so we can use this to find C

75. myininaya

$\sqrt{10}=\sqrt{2 (1)^3+4(1)^2+C}$ => $10=2(1)^3+4(1)^2+C$

76. myininaya

$10=2+4+C$

77. myininaya

=>C=4

78. myininaya

so we have $y=\sqrt{2x^3+4x^2+4}$

79. myininaya

you wanted to know y when x=0, right?

80. anonymous

yes

81. myininaya

so how do you think we do that?

82. myininaya

i brb i think it will be pretty easy for turing to help without latex on this last part that you have to do

83. anonymous

okay turing i get now :D

84. anonymous

but i need help finishing

85. myininaya

ok i'm back

86. myininaya

i had to get my glasses

87. myininaya

so i can be fully nerd

88. anonymous

lol

89. myininaya

so we have $y=\sqrt{2x^3+4x^2+4}$

90. anonymous

yes

91. myininaya

it says what is y if x=0

92. myininaya

$y=\sqrt{2(0)^3+4(0)^2+4}$

93. myininaya

i replaced x with 0 so I can see what y is when x is 0

94. myininaya

$y=\sqrt{0+0+4}=\sqrt{4}=2$

95. myininaya

This says when x is 0, y is 2

96. anonymous

got it! :D thank you!!!! sorry i didnt really try hard enough before you were a great help!

97. myininaya

It is okay. I didn't think any offense to anything you did. Sometimes you may get someone who can explain it better. I know that I'm probably not the best explaining some things.

98. myininaya

Or you know like I way you prefer.

99. anonymous

lol yeah :D thanks again.

100. myininaya

np