NEOCODE

UNIT 5 PART 2

1. If \( f(x,y) \) is a homogeneous function of degree \( n \), then according to Euler's theorem:

Correct Answer: a) \( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f \)

Explanation:

Euler's theorem for homogeneous functions states that if \( f(x, y) \) is a homogeneous function of degree \( n \), then:

\[ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y) \]

This result is a direct consequence of applying the definition of homogeneity to the function and differentiating using the chain rule.

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

2. For the transformation \( u = x^2 - y^2 \), \( v = 2xy \), the Jacobian determinant \( \frac{\partial(u,v)}{\partial(x,y)} \) is:

Correct Answer: a) \( 4(x^2 + y^2) \)

Explanation:

The Jacobian determinant for the transformation is calculated as follows:

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

First, find the partial derivatives:

  • \( \frac{\partial u}{\partial x} = 2x \)
  • \( \frac{\partial u}{\partial y} = -2y \)
  • \( \frac{\partial v}{\partial x} = 2y \)
  • \( \frac{\partial v}{\partial y} = 2x \)

Now apply the determinant formula:

\[ J = (2x \cdot 2x) - (-2y \cdot 2y) = 4x^2 + 4y^2 = 4(x^2 + y^2) \]

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

3. A critical point \((a,b)\) of \( f(x,y) \) is a saddle point if:

Correct Answer: c) \( D = f_{xx}f_{yy} - (f_{xy})^2 < 0 \)

Explanation:

To classify a critical point using the second derivative test, we calculate:

\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]

  • If \( D > 0 \) and \( f_{xx} > 0 \), the point is a local minimum.
  • If \( D > 0 \) and \( f_{xx} < 0 \), the point is a local maximum.
  • If \( D < 0 \), the point is a saddle point.
  • If \( D = 0 \), the test is inconclusive.

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

4. To maximize \( f(x,y) \) subject to \( g(x,y) = 0 \) using Lagrange multipliers, we solve:

Correct Answer: a) \( \nabla f = \lambda \nabla g \), \( g(x,y) = 0 \)

Explanation:

The method of Lagrange multipliers is used to find the extrema of a function subject to a constraint. The relationship is expressed as:

\[ \nabla f(x, y) = \lambda \nabla g(x, y) \]

Where:

  • \( \nabla f(x, y) \) is the gradient of the objective function.
  • \( \nabla g(x, y) \) is the gradient of the constraint function.
  • \( \lambda \) is the Lagrange multiplier.

Additionally, the constraint itself must be satisfied:

\[ g(x, y) = 0 \]

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

5. Which of the following is a homogeneous function of degree 3?

Correct Answer: b) \( x^2 y + y^2 z \)

Explanation:

A function \( f(x, y, z) \) is called homogeneous of degree \( n \) if:

\[ f(tx, ty, tz) = t^n f(x, y, z) \]

Let's check each option:

  • (a) \( x^3 + y^3 + z^3 \) - Applying \( t \) scaling: \( (tx)^3 + (ty)^3 + (tz)^3 = t^3(x^3 + y^3 + z^3) \). - Degree = 3 ✅
  • (b) \( x^2y + y^2z \) - Applying \( t \) scaling: \( (tx)^2(ty) + (ty)^2(tz) = t^3(x^2y + y^2z) \). - Degree = 3 ✅
  • (c) \( x + y + z \) - Applying \( t \) scaling: \( tx + ty + tz = t(x + y + z) \). - Degree = 1 ❌
  • (d) \( \frac{x^3}{y} \) - Applying \( t \) scaling: \( \frac{(tx)^3}{ty} = t^2 \frac{x^3}{y} \). - Degree = 2 ❌

Correct Answer: Option (a) and (b) are homogeneous of degree 3, but as per the intent of the question, the most straightforward homogeneous function of degree 3 is Option (a).

6. If \( f(x,y) \) has a critical point at \( (a,b) \) with \( D = f_{xx}f_{yy} - (f_{xy})^2 = 0 \), the second derivative test is:

Correct Answer: c) Inconclusive

Explanation:

The second derivative test is used to classify critical points of a function using the discriminant:

\[ D = f_{xx}(a,b)f_{yy}(a,b) - \left(f_{xy}(a,b)\right)^2 \]

  • If \( D > 0 \) and \( f_{xx}(a,b) > 0 \), the point is a local minimum.
  • If \( D > 0 \) and \( f_{xx}(a,b) < 0 \), the point is a local maximum.
  • If \( D < 0 \), the point is a saddle point.
  • If \( D = 0 \), the test is inconclusive. Further analysis is required to determine the nature of the critical point.

7. For the polar coordinate transformation \( x = r \cos \theta \), \( y = r \sin \theta \), the Jacobian determinant is:

Correct Answer: a) \( r \)

Correct Answer: a) \( r \)

Explanation:

The Jacobian matrix for the transformation \( x = r \cos \theta \) and \( y = r \sin \theta \) is given by:

\[ J = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{bmatrix} = \begin{bmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix} \]

The Jacobian determinant is calculated as:

\[ \det(J) = (\cos \theta)(r \cos \theta) - (-r \sin \theta)(\sin \theta) \]

\[ \det(J) = r \cos^2 \theta + r \sin^2 \theta \]

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ \det(J) = r \]

Hence, the correct answer is a) \( r \).

8. In Lagrange multipliers, the condition \( \nabla f = \lambda \nabla g \) implies:

Correct Answer: a) \( f \) and \( g \) are parallel

Correct Answer: a) \( f \) and \( g \) are parallel

Explanation:

In the method of Lagrange multipliers, we are given a function \( f(x, y) \) to maximize or minimize, subject to a constraint \( g(x, y) = 0 \).

The condition \( \nabla f = \lambda \nabla g \) implies that the gradients of \( f \) and \( g \) are proportional.

Since gradients point in the direction of the greatest rate of increase, and the constraint \( g(x, y) = 0 \) defines a curve or surface, the gradients being proportional means the gradients are parallel at the points of extremum.

Thus, the correct interpretation is that the gradients of the objective function and the constraint function are parallel.

9. If \( f(x,y) = x^4 + y^4 \) and \( f \) is homogeneous of degree \( n \), then \( n \) is:

Correct Answer: c) 4

Correct Answer: c) 4

Explanation:

A function \( f(x, y) \) is called homogeneous of degree \( n \) if:

\[ f(kx, ky) = k^n f(x, y) \text{ for all } k > 0 \]

Here, \( f(x, y) = x^4 + y^4 \).

If we replace \( x \) with \( kx \) and \( y \) with \( ky \), we get:

\[ f(kx, ky) = (kx)^4 + (ky)^4 = k^4x^4 + k^4y^4 = k^4 (x^4 + y^4) \]

Since we obtained \( k^4 f(x, y) \), it is evident that \( n = 4 \).

10. The global maximum of \( f(x,y) = x^2 + y^2 \) on the unit disk \( x^2 + y^2 \leq 1 \) is:

Correct Answer: c) On the boundary \( x^2 + y^2 = 1 \)

Correct Answer: c) On the boundary \( x^2 + y^2 = 1 \)

Explanation:

The given function is \( f(x, y) = x^2 + y^2 \). This is a quadratic function representing the square of the Euclidean distance from the origin.

The unit disk is defined by the inequality \( x^2 + y^2 \leq 1 \). To find the global maximum, check the boundary condition where:

\[ x^2 + y^2 = 1 \]

On this boundary, \( f(x, y) = 1 \). Inside the disk, the maximum value is always achieved when the point is on the boundary. Therefore, the global maximum is 1, occurring on the boundary of the unit disk.