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import sympy as sp
# Test the corrected numerical examples separately
def test_corrected_examples():
x, y = sp.symbols('x y')
print("Testing Corrected Numerical Examples:")
print("=" * 50)
# Case 1 Example: F(x,y) = [2, 3y²]
print("Case 1: F(x,y) = [2, 3y²]")
Vx = 2
Vy = 3 * y**2
# Manual calculation: φ = ∫2 dx + ∫3y² dy = 2x + y³
phi_manual = 2*x + y**3
# Verify
print(f"∂φ/∂x = {sp.diff(phi_manual, x)} = Vx? {sp.diff(phi_manual, x) == Vx}")
print(f"∂φ/∂y = {sp.diff(phi_manual, y)} = Vy? {sp.diff(phi_manual, y) == Vy}")
print()
# Case 2 Example: F(x,y) = [2x + 3y + 1, 3x + 4y + 2]
print("Case 2: F(x,y) = [2x + 3y + 1, 3x + 4y + 2]")
Vx2 = 2*x + 3*y + 1
Vy2 = 3*x + 4*y + 2
# Check gradient condition: ∂P/∂y = 3, ∂Q/∂x = 3 → equal, so gradient field exists
print(f"Gradient condition: ∂P/∂y = {sp.diff(Vx2, y)}, ∂Q/∂x = {sp.diff(Vy2, x)}")
print(f"Field is gradient? {sp.diff(Vx2, y) == sp.diff(Vy2, x)}")
# Find potential using our method
phi_x = sp.integrate(Vx2, x) # ∫(2x + 3y + 1)dx = x² + 3xy + x
remaining = Vy2 - sp.diff(phi_x, y) # (3x + 4y + 2) - 3x = 4y + 2
phi_y = sp.integrate(remaining, y) # ∫(4y + 2)dy = 2y² + 2y
phi_calculated = phi_x + phi_y # x² + 3xy + x + 2y² + 2y
print(f"Calculated potential: φ = {phi_calculated}")
print(f"∂φ/∂x = {sp.diff(phi_calculated, x)} = Vx? {sp.diff(phi_calculated, x) == Vx2}")
print(f"∂φ/∂y = {sp.diff(phi_calculated, y)} = Vy? {sp.diff(phi_calculated, y) == Vy2}")
# Run the test
test_corrected_examples() |