Question

Please help: Given x=3, y=4, dy/dt=7, solve for dx/dt.x^2-3xy=9y

Answer 1

Given the equation:

`\begin{gathered} x^2-3xy=9y\text{ --- equation 1} \\ \text{where} \\ (dy)/(dt)=7 \\ x=3,\text{ y=4} \end{gathered}`

Required: solve for dx/dt.

`(dy)/(dx)=(dy)/(dt)*(dt)/(dx)\text{ ---- equation 2}`

Step 1: Solve for dy/dx

From equation 1,

`\begin{gathered} x^2-3xy=9y \\ (d(x^2-3xy))/(dx)=(d(9y))/(dx) \\ 2x-3\mleft(y+x(d)/(dx)\mleft(y\mright)\mright)=9(dy)/(dx) \\ \Rightarrow9(dy)/(dx)+3x(dy)/(dx)=2x-3y \\ \therefore(dy)/(dx)=(2x-3y)/(9+3x) \end{gathered}`

Step 2: Evaluate dy/dx.

Recall that x=3, y=4.

Thus,

`\begin{gathered} (dy)/(dx)=(2(3)-3(4))/(9+3(3)) \\ =(6-12)/(9+9)=-(6)/(18) \\ \Rightarrow(dy)/(dx)=-(1)/(3) \end{gathered}`

Step 3: solve for dx/dt.

From equation 2,

`\begin{gathered} (dy)/(dx)=(dy)/(dt)*(dt)/(dx) \\ \end{gathered}`

Recall that

`\begin{gathered} (dy)/(dx)=-(1)/(3) \\ (dy)/(dt)=7 \end{gathered}`

Thus,

`\begin{gathered} (dy)/(dx)=(dy)/(dt)*(dt)/(dx) \\ \Rightarrow-(1)/(3)=7*(dt)/(dx) \\ -(1)/(3)=7(dt)/(dx) \\ \Rightarrow21dt=-dx \\ \text{hence,} \\ (dx)/(dt)=-21 \end{gathered}`

Hence,

`(dx)/(dt)=-21`