Asked by Barry
Please help don't understand how to solve it.
3xy-6x+2y^2
dx/dt=-6
x=2
y=-3
thank you
3xy-6x+2y^2
dx/dt=-6
x=2
y=-3
thank you
Answers
Answered by
Steve
Neither do I. You haven't asked a question.
If you want dy/dt at (2,-3) then you just need to do
3y + 3x dy/dt - 6 + 4y dy/dt
but you can't evaluate anything, because you have no equation.
Why not try posting the entire question, not just part of it?
If you want dy/dt at (2,-3) then you just need to do
3y + 3x dy/dt - 6 + 4y dy/dt
but you can't evaluate anything, because you have no equation.
Why not try posting the entire question, not just part of it?
Answered by
Barry
Assume x and y are functions of t, Evaluate dy/dt,
3xy-6x+2y^2
dx/dt=-6
x=2
y=-3
dy/dt
3xy-6x+2y^2
dx/dt=-6
x=2
y=-3
dy/dt
Answered by
Steve
Still no good, but let's assume you have
3xy-6x+2y^2 = 0
Then
3y dx/dt + 3x dy/dt - 6 dx/dt + 4y dy/dt = 0
(3x+4y) dy/dt = (6-3y) dx/dt
dy/dt = 3(2-y)(dx/dt)/(3x+4y)
dy/dt = 3(5)(-6)/(6-12) = 15
But that's so only if my original equation is close.
3xy-6x+2y^2 = 0
Then
3y dx/dt + 3x dy/dt - 6 dx/dt + 4y dy/dt = 0
(3x+4y) dy/dt = (6-3y) dx/dt
dy/dt = 3(2-y)(dx/dt)/(3x+4y)
dy/dt = 3(5)(-6)/(6-12) = 15
But that's so only if my original equation is close.
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