- anonymous

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?

- katieb

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- TuringTest

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

- myininaya

it is

- myininaya

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

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## More answers

- TuringTest

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

- anonymous

lol i didnt understand it :(

- myininaya

but no he might understand you better

- anonymous

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

- myininaya

this is what i did turing if you want to look

- anonymous

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

- 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 :/

- TuringTest

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

- anonymous

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

- anonymous

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

- TuringTest

yes exactly antiderivatives

- myininaya

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

- myininaya

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

- anonymous

ahh yes that sort of makes sense.

- anonymous

i understand that c is a constant.

- TuringTest

check that the derivative of that gives x^n

- myininaya

right

- anonymous

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

- TuringTest

antiderivative of that

- myininaya

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

- myininaya

like turing said

- anonymous

ahh okay.

- TuringTest

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

- anonymous

x^3

- anonymous

for 3x^2

- anonymous

i dont know for 4x

- TuringTest

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

- myininaya

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

- myininaya

see n was 1 here

- anonymous

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

- myininaya

its cool

- myininaya

struggling happens
but the main is don't give up

- anonymous

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

- anonymous

what formula for the 4x?

- anonymous

the anti derivative of dx is x?

- TuringTest

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

- myininaya

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

- myininaya

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

- myininaya

see i'm using that same formula every time

- anonymous

oh okay so i have to do it for each.

- myininaya

right

- myininaya

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

- myininaya

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

- anonymous

so for the formula ydy you used the formula.

- anonymous

okay i understand how u got that equation.

- 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

- anonymous

yes

- myininaya

ok great! lets make this prettier

- myininaya

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

- myininaya

is that okay?

- anonymous

yes my simplification came out the same

- myininaya

ok we can also multiply two on both sides

- myininaya

\[y^2=2x^3+4x^2+C\]

- myininaya

2C is still constant so I left it as C

- anonymous

yes

- myininaya

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

- anonymous

and now square both sides?

- myininaya

square root of both sides

- myininaya

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

- anonymous

where do we have a negative value?

- myininaya

take square root of both sides
you get plus or minus

- anonymous

thats right!

- myininaya

we only need the plus since y>0

- myininaya

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

- anonymous

yes.

- myininaya

you were given a point on this curve

- myininaya

\[ (1,\sqrt{10} )\]

- myininaya

x=1 and y=sqrt(10)

- anonymous

yes i plug it in for x and y?

- myininaya

so we can use this to find C

- myininaya

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

- myininaya

\[10=2+4+C\]

- myininaya

=>C=4

- myininaya

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

- myininaya

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

- anonymous

yes

- myininaya

so how do you think we do that?

- 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

- anonymous

okay turing i get now :D

- anonymous

but i need help finishing

- myininaya

ok i'm back

- myininaya

i had to get my glasses

- myininaya

so i can be fully nerd

- anonymous

lol

- myininaya

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

- anonymous

yes

- myininaya

it says what is y if x=0

- myininaya

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

- myininaya

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

- myininaya

\[y=\sqrt{0+0+4}=\sqrt{4}=2\]

- myininaya

This says when x is 0, y is 2

- anonymous

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

- 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.

- myininaya

Or you know like I way you prefer.

- anonymous

lol yeah :D thanks again.

- myininaya

np

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