Limit, continuity and differentiability of functions of two variables, chain rule, change of variables

Correct Answer: c) Does not exist

Explanation:

To check the limit \(\lim_{(x, y) \to (0, 0)} f(x, y)\), we evaluate along different paths:

1. **Path \(y = x^2\)**: \[ f(x, x^2) = \frac{x^2x^2}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac{1}{2} \]

2. **Path \(y = 0\)**: \[ f(x, 0) = \frac{x^2 \cdot 0}{x^4 + 0} = 0 \]

3. **Path \(x = 0\)**: \[ f(0, y) = \frac{0 \cdot y}{0 + y^2} = 0 \]

Since the function approaches different values along different paths, the limit does not exist.

Therefore, the correct answer is:

Option (c).

Correct Answer: c) lim_{(x,y)→(a,b)} f(x,y) = f(a,b)

Explanation:

For a function \( f(x, y) \) to be continuous at \( (a, b) \), three conditions must be satisfied:

1. \( f(a, b) \) must exist.
2. \( \lim_{(x, y) \to (a, b)} f(x, y) \) must exist.
3. \( \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) \).

The existence of partial derivatives is not a requirement for continuity. Therefore, the correct answer is:

Option (c).

Correct Answer: c) The tangent plane approximation is valid

Explanation:

For a function \( f(x, y) \) to be differentiable at a point \( (a, b) \), the most important condition is that the tangent plane approximation is valid. This means the function can be closely approximated by a linear function near that point.

While continuity and the existence of partial derivatives are necessary for differentiability, they are not sufficient on their own. The differentiability condition ensures that the error in the linear approximation approaches zero faster than the distance to the point.

Hence, the correct answer is:

Option (c).

Correct Answer: b) \( \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \)

Explanation:

Using the chain rule for multivariable functions, we write:

\[ \frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \]

This accounts for how both \( x \) and \( y \) change with respect to \( t \).

Hence, the correct answer is:

Option (b).

Correct Answer: b) \( \frac{\partial(u, v)}{\partial(x, y)} \)

Explanation:

The Jacobian determinant is given by the determinant of the matrix of first-order partial derivatives:

\[ J = \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \]

This matrix represents how the new variables \( u \) and \( v \) change with respect to \( x \) and \( y \).

Hence, the correct answer is:

Option (b).

Correct Answer: b) \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \)

Explanation:

According to Clairaut's Theorem (or the symmetry of second derivatives), if a function \( f(x, y) \) has continuous second-order partial derivatives, then the mixed partial derivatives are equal:

\[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]

This property holds true under the condition of continuity for the second-order partial derivatives.

Hence, the correct answer is:

Option (b).

Correct Answer: a) \( \nabla f \cdot \mathbf{u} \)

Explanation:

The directional derivative of a function \( f(x, y) \) in the direction of a unit vector \( \mathbf{u} = (a, b) \) is given by the dot product of the gradient of the function with the vector \( \mathbf{u} \):

\[ D_{\mathbf{u}} f(x, y) = \nabla f(x, y) \cdot \mathbf{u} = \frac{\partial f}{\partial x} a + \frac{\partial f}{\partial y} b \]

Note that the gradient vector \( \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Hence, the correct answer is:

Option (a).

Correct Answer: a) \( -\frac{F_x}{F_z} \)

Explanation:

Using implicit differentiation for \( F(x, y, z) = 0 \), we apply the total derivative:

\[ F_x + F_y \frac{\partial y}{\partial x} + F_z \frac{\partial z}{\partial x} = 0 \]

Since \( y \) is independent of \( x \), we have:

\[ F_x + F_z \frac{\partial z}{\partial x} = 0 \]

Rearrange to solve for \( \frac{\partial z}{\partial x} \):

\[ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \]

Hence, the correct answer is:

Option (a).

Correct Answer: a) \( z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \)

Explanation:

The general form of the tangent plane equation to \( z = f(x, y) \) at the point \( (a, b, f(a, b)) \) is given by:

\[ z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \]

Where:

• \( f_x(a, b) \) is the partial derivative of \( f \) with respect to \( x \) evaluated at \( (a, b) \).
• \( f_y(a, b) \) is the partial derivative of \( f \) with respect to \( y \) evaluated at \( (a, b) \).

Therefore, the correct answer is: Option (a).

Correct Answer: a) \( r \)

Explanation:

The Jacobian determinant for the transformation from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) is given by:

\[ J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} \]

Using \( x = r \cos \theta \) and \( y = r \sin \theta \), we calculate the partial derivatives:

• \( \frac{\partial x}{\partial r} = \cos \theta \)
• \( \frac{\partial x}{\partial \theta} = -r \sin \theta \)
• \( \frac{\partial y}{\partial r} = \sin \theta \)
• \( \frac{\partial y}{\partial \theta} = r \cos \theta \)

Now, the determinant becomes:

\[ J = \begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} = (\cos \theta \cdot r \cos \theta) - (-r \sin \theta \cdot \sin \theta) = r (\cos^2 \theta + \sin^2 \theta) = r \]

Therefore, the correct answer is: Option (a).