suppose that x=x(t) and y=y(t) are both functions of t if y^2 +xy-3x=1 and dy/dt = -2 when x=3 and y =2 what is dx/dt?

Question

suppose that x=x(t) and y=y(t) are both functions of t

if y^2 +xy-3x=1 and dy/dt = -2 when x=3 and y =2 what is dx/dt?

Answer 1

The key to this is that, per the chain rule, dy/dt = (dy/dx)•(dx/dt) and therefore dx/dt = (dy/dt)/(dy/dx)

We are give dy/dt but, to use this equation, we need dy/dx. But we can find dy/dx through implicit differentiation:

y² + xy- 3x = 1

taking the derivative of both sides with respect to x we get:

2y•dy/dx + [x'y + xy'] - 3 = 0

2y•dy/dx + y + x•dy/dx = 3

dy/dx[2y + x] = 3 - y

dy/dx = (3 - y)/(2y + x)

at x = 3 and y = 2 we get dy/dx(3,2) = (3-2)/(2•2+3) = 1/7

So since dx/dt = (dy/dt)/(dy/dx) then dx/dt = (-2)/(1/7) = -14