Respuesta :
Solution:
Given the equation:
[tex]\begin{gathered} x^2-3xy=9y\text{ --- equation 1} \\ \text{where} \\ \frac{dy}{dt}=7 \\ x=3,\text{ y=4} \end{gathered}[/tex]Required: solve for dx/dt.
[tex]\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}\text{ ---- equation 2}[/tex]Step 1: Solve for dy/dx
From equation 1,
[tex]\begin{gathered} x^2-3xy=9y \\ \frac{d(x^2-3xy)}{dx}=\frac{d(9y)}{dx} \\ 2x-3\mleft(y+x\frac{d}{dx}\mleft(y\mright)\mright)=9\frac{dy}{dx} \\ \Rightarrow9\frac{dy}{dx}+3x\frac{dy}{dx}=2x-3y \\ \therefore\frac{dy}{dx}=\frac{2x-3y}{9+3x} \end{gathered}[/tex]Step 2: Evaluate dy/dx.
Recall that x=3, y=4.
Thus,
[tex]\begin{gathered} \frac{dy}{dx}=\frac{2(3)-3(4)}{9+3(3)} \\ =\frac{6-12}{9+9}=-\frac{6}{18} \\ \Rightarrow\frac{dy}{dx}=-\frac{1}{3} \end{gathered}[/tex]Step 3: solve for dx/dt.
From equation 2,
[tex]\begin{gathered} \frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx} \\ \end{gathered}[/tex]Recall that
[tex]\begin{gathered} \frac{dy}{dx}=-\frac{1}{3} \\ \frac{dy}{dt}=7 \end{gathered}[/tex]Thus,
[tex]\begin{gathered} \frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx} \\ \Rightarrow-\frac{1}{3}=7\times\frac{dt}{dx} \\ -\frac{1}{3}=7\frac{dt}{dx} \\ \Rightarrow21dt=-dx \\ \text{hence,} \\ \frac{dx}{dt}=-21 \end{gathered}[/tex]Hence,
[tex]\frac{dx}{dt}=-21[/tex]