Linearization and Partial Derivatives

The video tutorial explains how to find critical points and perform linearization on a given system of equations. It starts by defining the functions f(x, y) and g(x, y) and solving for critical points by setting both functions to zero. The tutorial then demonstrates the linearization process by introducing new variables and calculating the Jacobian matrix. Finally, it compares the slope fields of the original system and its linearization, highlighting the approximation around the critical point.

What is the function f(x, y) in the given system?

• f(x, y) = x + y^2

• f(x, y) = x^2 + y^2

• f(x, y) = x + y + y^2

• f(x, y) = x + y

What is the function g(x, y) in the given system?

• g(x, y) = y^2

• g(x, y) = x

• g(x, y) = x + y

• g(x, y) = y

What is the critical point of the system?

• (0, 0)

• (1, 0)

• (1, 1)

• (0, 1)

What variable substitution is used for linearization?

• U = x - x0, V = y - y0

• U = x * x0, V = y * y0

• U = x / x0, V = y / y0

• U = x + x0, V = y + y0

What is the partial derivative of f with respect to x?

• 2y

• x + y

• 0

• 1

What is the partial derivative of f with respect to y?

• 1

• 2x

• y^2

• 1 + 2y

What is the partial derivative of g with respect to x?

• 0

• 1

• y

• x

What is the partial derivative of g with respect to y?

• 0

• 1

• x

• y

What is the linearization of U' in terms of U and V?

• U' = U / V

• U' = U - V

• U' = U + V

• U' = U * V

How does the linearization compare to the original system's slope field?

• It is unrelated.

• It is a good approximation near the critical point.

• It is an exact match.

• It is a poor approximation.