y^2 = 2 + xy; at t = 5, y = 3, dy/dt = 6; find dx/dt

Math wiz ya rite 09

Junior Member
Joined
Aug 27, 2006
Messages
136
Let x and y be functions of time t that are related by the equation y^2 = 2 +xy. At time t=5, the value of y is 3 and dy/dt =6. Find the value of dx/dt at time =5.

Note: The derivative of y^2 = 2 + xy = y / (2y - x)
 
Re: calc2

Hello, Math wiz ya rite 09!

I explained this at another site . . .


Let \(\displaystyle x\) and \(\displaystyle y\) be functions of time \(\displaystyle t\)
that are related by the equation: \(\displaystyle \, y^2 \:=\: 2\, +\,xy\)
At time \(\displaystyle t\,=\,5:\;y\,=\,3\) and \(\displaystyle \frac{dy}{dt}\,=\,6\)

Find the value of \(\displaystyle \frac{dx}{dt}\) at \(\displaystyle t\,=\,5.\)

When \(\displaystyle t\,=\,5,\:y\,=\,3\;\;\Rightarrow\;\;3^2\:=\:2\,+\,3x\;\;\Rightarrow\;\;x\,=\,\frac{7}{3}\)


Differentiate the equation with respect to \(\displaystyle t:\;\;\L 2y\cdot\frac{dy}{dt}\;=\;x\cdot\frac{dy}{dt}\,+\,y\cdot\frac{dx}{dt}\)

And we know: \(\displaystyle \:x\,=\,\frac{7}{3},\;y\,=\,3,\;\frac{dy}{dt}\,=\,6\) .. . . Go fot it!

 
Top