Assume that z = f(x, y)^2, x = g(t), y = h(t), f_y(1, 0) = 1, f_x(1, 0) = -1, f(1, 0) = 2, g(3) = 1, h(3) = 0, g'(3) = -3, and h'(3) = 4. Find dz/dt at t=3.

Steps to Solve

1. Differentiate z with respect to t

   We start with the relationship ( z = f(x, y)^2 ). To find ( \frac{dz}{dt} ), we apply the chain rule:

   $$ \frac{dz}{dt} = 2 f(x,y) \frac{df}{dt} $$

   where ( \frac{df}{dt} ) needs to be calculated using the chain rule again.

2. Apply the chain rule to find ( \frac{df}{dt} )

   Using the chain rule for ( f ), we have:

   $$ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} $$

   Here, ( x = g(t) ) and ( y = h(t) ), so ( \frac{dx}{dt} = g'(t) ) and ( \frac{dy}{dt} = h'(t) ).

3. Substitute the values into ( \frac{df}{dt} )

   At ( t = 3 ):

   • ( g(3) = 1 )
   • ( h(3) = 0 )
   • ( g'(3) = -3 )
   • ( h'(3) = 4 )
   • ( f_x(1, 0) = -1 )
   • ( f_y(1, 0) = 1 )
   • ( f(1, 0) = 2 )

   Substitute these values:

   $$ \frac{df}{dt} = f_x(1, 0) g'(3) + f_y(1, 0) h'(3) $$

   Plugging in:

   $$ \frac{df}{dt} = (-1)(-3) + (1)(4) = 3 + 4 = 7 $$

4. Calculate ( \frac{dz}{dt} )

   Now substitute ( f(1, 0) = 2 ) and ( \frac{df}{dt} = 7 ) back into the expression for ( \frac{dz}{dt} ):

   $$ \frac{dz}{dt} = 2 f(1, 0) \frac{df}{dt} = 2(2)(7) = 4 \times 7 = 28 $$