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

[Math wiz ya rite 09]

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)

 

[soroban]

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!

 

[stapel]

Multi-post.

 

[pka]

Multi-post.

That is truly funny.