A system of equations is given by:
F1(x,y,a,b) = x² + bxy + y² - a – 2 = 0
F2(x,y,a,b) = x² + y ² - b² + 2a + 3 = 0
Where x and y are endogenous variables while a and b are exogenous variables. Compute the differentials δx/δb and δy/δa at the point (x,y,a,b) = (0,1,3,4).
To compute the differentials δx/δb and δy/δa at the point (x,y,a,b) = (0,1,3,4), we first need to find the partial derivatives of the given system of equations F1 and F2 with respect to the variables x, y, a, and b. Then we can substitute the values of x, y, a, and b into these partial derivatives to find the differentials.
Let's determine the partial derivatives of F1 and F2:
Partial derivative of F1 with respect to x (denoted as ∂F1/∂x):
∂F1/∂x = 2x + by
Partial derivative of F1 with respect to y (denoted as ∂F1/∂y):
∂F1/∂y = bx + 2y
Partial derivative of F1 with respect to a (denoted as ∂F1/∂a):
∂F1/∂a = -1
Partial derivative of F1 with respect to b (denoted as ∂F1/∂b):
∂F1/∂b = xy
Partial derivative of F2 with respect to x (denoted as ∂F2/∂x):
∂F2/∂x = 2x
Partial derivative of F2 with respect to y (denoted as ∂F2/∂y):
∂F2/∂y = 2y
Partial derivative of F2 with respect to a (denoted as ∂F2/∂a):
∂F2/∂a = 2
Partial derivative of F2 with respect to b (denoted as ∂F2/∂b):
∂F2/∂b = 0
Now, substitute the values (x,y,a,b) = (0,1,3,4) into these partial derivatives to find the differentials:
δx/δb = (∂F1/∂x * δb) + (∂F1/∂y * δy) + (∂F1/∂a * δa) + (∂F1/∂b * δb)
= (2(0) + 1(4)) + (4(1)) + (-1(0)) + (0(δb))
= 4 + 4 + 0 + 0
= 8
δy/δa = (∂F2/∂x * δx) + (∂F2/∂y * δy) + (∂F2/∂a * δa) + (∂F2/∂b * δb)
= (2(0) + 2(1)) + (2(1)) + (2(δa)) + (0(4))
= 2 + 2 + 2δa + 0
= 4 + 2δa
Therefore, at the point (x,y,a,b) = (0,1,3,4), δx/δb = 8 and δy/δa = 4 + 2δa.