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tmn50
Group Title
let's say we have a function F(y(t),t)=y*G(t)
the partial derevative of F with respect to t
is y*dG/dt or dy/dt*G+dG/dt*y
I would really apreciate an explanation
 one year ago
 one year ago
tmn50 Group Title
let's say we have a function F(y(t),t)=y*G(t) the partial derevative of F with respect to t is y*dG/dt or dy/dt*G+dG/dt*y I would really apreciate an explanation
 one year ago
 one year ago

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Mashy Group TitleBest ResponseYou've already chosen the best response.1
since both are function of t!
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
\[F(y(t), t) = y \times G(t)\] ?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ ∂G }{ ∂t }\]
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
if both G and Y are function of t.. then you have to use the product rule!
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
mashy our teacher used the first one in determining the solution of a differential equation of first order that's what makes me crazy
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
no he mentioned its a function of time see the question!!
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
y is a function of time
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
Then Y must be independent of t... !!
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
stop yelling @ me!! Ur hurting my feelings, bro :'(
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
awww.. come here you !!
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
wow i get a medol and abbot gets none.. yay yay :P
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
who the hell gave u a medal i think your answer is wrong man here we're talking about partial derevative which is diffrent from derevating everything with respect to t
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Is it the partial derivative of \[ F(u, t) \]Or of \[ F(y(t),t) \]?
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
the second one bro partial derevative with respect to t
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Well it was ambiguous, and \(F(u, t) \neq F(y(t),t)\)
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
if both functions are time dependent.. then really there is no difference between partial and normal derivatives.. HELL why would you even DO a partial derivative doesn't even make any sense :/
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
_ it makes sense when u'r solving a diferential equation of first order first you find F(y,t) then you derevative partially with respect to y or t and force it to be equal to the equation you have left so yeh u need a partial derevative
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
give me a differential equation!!
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Like @Mashy is saying, \(F(y(t),t)\) doesn't have a partial derivative because it is not a multiple variable function. It's the OUTPUT of \(F(u, t)\) which happens to be a single variable function.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
well, that escalated quickly.
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
M(y,t)dt+N(y,t)dy=0 where M=df/dt and N=df/dy
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Whoa, now we're doing ODE's?
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
thats called as EXACT FORM right?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Correct, Mashy :)
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
wio i didnt catch up with what you said i mean here we got 2 variable well not two variables y is a function of time and hell
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
well then in that.. case.. i have forgetting how to do it :D
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
This is escalating too damn quickly.
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
it's like i have holes in mathematics so when i advance i fall into some of them
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
holes in mathematics.. lol funny :D
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Can you give us \(M, N\)?
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
he already mentioned.. df/dy and df/dt!
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Well, you want to find the PD for either M or N first. I think it's ∂M/∂x and ∂N/∂y ?
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
it doesn't matter what M and N are the only restriction is that M and N are the partial derevatives of a certain function F(y,t)
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
wow i did this 4 years back!! ..
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
so i don't remember much :/
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Well, you have to check that they are exact first for this to work.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
Otherwise, you are wasting time trying to do furhter work than is necessary.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
\[ \begin{array}{rcl} M(y,t)dt+N(y,t)dy &=& 0 \\ M(y,t)dt &=& N(y,t)dy \end{array} \] I would try integrating \(M\) with respect to \(t\) and then differentiating with respect to \(y\).
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
wait can i post a picture ?
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
yup i can wait
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
really?!? of yourself?? we are in the middle of math here!
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
here check tht
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
i'm too handsome :p
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
so i have heard :P
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
i mean .. i did.. just now :P.. from you!!
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
so any idea about what i posted?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
What it looks like you're trying to do in that step is find a integrating factor to make the solution exact?
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
wait wait wait we found the integrating factor we're in the middle of finding F
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
see where's the arrow pointing
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.1
I'm not quite sure what you did there, I'm used to doing it a different method..
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
tht's the teacher's doing not me :3
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
oh well i'm off now i'll check again after half an hour or so thanks ppl
 one year ago

tmn50 Group TitleBest ResponseYou've already chosen the best response.0
i'm back people any news,
 one year ago
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