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Unit5 - Subjective Questions

MTH174 • Practice Questions with Detailed Answers

Define the limit of a function of two variables, $f(x,y)$ , as $(x,y)$ approaches a point $(a,b)$ .

Definition of Limit:
A function $f(x,y)$ is said to have a limit $L$ as $(x,y)$ approaches $(a,b)$ , written as $\lim_{{(x,y) \to (a,b)}} f(x,y) = L$ , if for every number $\epsilon > 0$ , there exists a corresponding number $\delta > 0$ such that:

$|f(x,y) - L| < \epsilon$

whenever $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$ .

Key Points:

The limit must be the same regardless of the path of approach to the point $(a,b)$ .
If different paths lead to different limits, the limit does not exist.

Explain the concept of continuity for a function of two variables at a point $(a,b)$ .

Continuity of $f(x,y)$ :
A function $f(x,y)$ is continuous at a point $(a,b)$ if all of the following three conditions are satisfied:

Existence of the function value: $f(a,b)$ is defined.
Existence of the limit: $\lim_{{(x,y) \to (a,b)}} f(x,y)$ exists.
Equality: The limit of the function as it approaches the point is equal to the value of the function at that point, i.e., $\lim_{{(x,y) \to (a,b)}} f(x,y) = f(a,b)$ .

If a function is continuous at every point in a region $R$ , it is said to be continuous in that region.

Discuss how to prove that a limit does NOT exist for a function of two variables using the path method.

Proving a Limit Does Not Exist:
To show that $\lim_{{(x,y) \to (a,b)}} f(x,y)$ does not exist, we use the property that if the limit exists, it must be independent of the path taken to reach $(a,b)$ .

Method:

Choose two different paths approaching $(a,b)$ . Common paths to $(0,0)$ include straight lines $y = mx$ or parabolas $y = mx^2$ .
Evaluate the limit along the first path.
Evaluate the limit along the second path.
If the values obtained from different paths are not equal (e.g., the limit depends on the slope $m$ ), then the limit does not exist.

Example: For $f(x,y) = \frac{{xy}}{{x^2 + y^2}}$ as $(x,y) \to (0,0)$ , putting $y=mx$ gives $\frac{{m}}{{1+m^2}}$ . Since this depends on $m$ , the limit does not exist.

Define differentiability for a function of two variables $f(x,y)$ at a point $(a,b)$ .

Differentiability of $f(x,y)$ :
A function $f(x,y)$ is differentiable at $(a,b)$ if its total increment $\Delta z = f(a+\Delta x, b+\Delta y) - f(a,b)$ can be expressed in the form:

$\Delta z = A\Delta x + B\Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y$

where $A$ and $B$ are constants (specifically, $A = f_x(a,b)$ and $B = f_y(a,b)$ ), and $\epsilon_1, \epsilon_2 \to 0$ as $(\Delta x, \Delta y) \to (0,0)$ .

Alternative Form:
It can also be defined if $\lim_{{(\Delta x, \Delta y) \to (0,0)}} \frac{{\Delta z - f_x(a,b)\Delta x - f_y(a,b)\Delta y}}{{\sqrt{\Delta x^2 + \Delta y^2}}} = 0$ .

Discuss the relationship between continuity, existence of partial derivatives, and differentiability for functions of two variables.

Relationship Breakdown:

Differentiability implies Continuity: If a function $f(x,y)$ is differentiable at a point $(a,b)$ , then it is continuous at $(a,b)$ .
Continuity does NOT imply Differentiability: A function can be continuous but not differentiable (e.g., having a sharp corner in 3D space).
Partial Derivatives do NOT imply Continuity: The mere existence of partial derivatives $f_x$ and $f_y$ at a point does NOT guarantee that the function is continuous or differentiable at that point.
Sufficient Condition for Differentiability: If the partial derivatives $f_x$ and $f_y$ exist and are continuous in a neighborhood of $(a,b)$ , then $f(x,y)$ is differentiable at $(a,b)$ .

State and explain the Chain Rule for a function of two variables where the variables themselves are functions of a single independent variable.

Chain Rule (Case 1):
If $z = f(x,y)$ is a differentiable function of $x$ and $y$ , and both $x$ and $y$ are differentiable functions of a single independent variable $t$ , i.e., $x = g(t)$ and $y = h(t)$ , then $z$ is a differentiable function of $t$ .

The total derivative of $z$ with respect to $t$ is given by:

$\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}} \frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}} \frac{{dy}}{{dt}}$

Explanation:
This formula calculates the rate of change of $z$ with respect to $t$ by summing the contributions from its intermediate variables $x$ and $y$ .

Describe the Chain Rule for a function $z = f(x,y)$ where $x$ and $y$ are functions of two independent variables $u$ and $v$ .

Chain Rule (Case 2):
Suppose $z = f(x,y)$ is a differentiable function of $x$ and $y$ , and $x = g(u,v)$ and $y = h(u,v)$ are differentiable functions of the independent variables $u$ and $v$ .

Then $z$ is a differentiable function of $u$ and $v$ , and its partial derivatives are given by:

Partial with respect to $u$ :
$\frac{{\partial z}}{{\partial u}} = \frac{{\partial z}}{{\partial x}} \frac{{\partial x}}{{\partial u}} + \frac{{\partial z}}{{\partial y}} \frac{{\partial y}}{{\partial u}}$
Partial with respect to $v$ :
$\frac{{\partial z}}{{\partial v}} = \frac{{\partial z}}{{\partial x}} \frac{{\partial x}}{{\partial v}} + \frac{{\partial z}}{{\partial y}} \frac{{\partial y}}{{\partial v}}$

These represent the rate of change of $z$ in the directions of $u$ and $v$ respectively.

Explain the concept of 'Change of Variables' in multivariate calculus and its primary purpose.

Change of Variables:
Change of variables is a technique used to transform a function or a differential expression from one coordinate system to another (e.g., from Cartesian $(x,y)$ to polar $(r,\theta)$ ).

Primary Purpose:

Simplification: Certain problems, domains, or differential equations are much easier to solve in a different coordinate system. For example, circular domains are easier to handle in polar coordinates.
Procedure: It involves using transformation equations like $x = r \cos\theta$ and $y = r \sin\theta$ , and applying the chain rule to convert partial derivatives $\frac{{\partial z}}{{\partial x}}$ and $\frac{{\partial z}}{{\partial y}}$ into expressions involving $\frac{{\partial z}}{{\partial r}}$ and $\frac{{\partial z}}{{\partial \theta}}$ .

Define a homogeneous function of degree $n$ and provide an example.

Homogeneous Function:
A function $f(x,y)$ is said to be homogeneous of degree $n$ in variables $x$ and $y$ if, for any scalar $t$ , the following condition holds:

$f(tx, ty) = t^n f(x, y)$

Alternatively, it can be written in the form $x^n \phi(\frac{{y}}{{x}})$ or $y^n \psi(\frac{{x}}{{y}})$ .

Example:
Consider $f(x,y) = \frac{{x^3 + y^3}}{{x + y}}$ .
Replace $x$ with $tx$ and $y$ with $ty$ :
$f(tx, ty) = \frac{{(tx)^3 + (ty)^3}}{{tx + ty}} = \frac{{t^3(x^3 + y^3)}}{{t(x + y)}} = t^2 \frac{{x^3 + y^3}}{{x + y}} = t^2 f(x,y)$

Since $f(tx, ty) = t^2 f(x,y)$ , the function is homogeneous of degree $n = 2$ .

State and prove Euler's Theorem for a homogeneous function of two variables.

Euler's Theorem Statement:
If $z = f(x,y)$ is a homogeneous function of $x$ and $y$ of degree $n$ , having continuous partial derivatives, then:
$x \frac{{\partial z}}{{\partial x}} + y \frac{{\partial z}}{{\partial y}} = nz$

Proof:
Since $z$ is a homogeneous function of degree $n$ , it can be written as:
$z = x^n f(\frac{{y}}{{x}})$ . Let $u = \frac{{y}}{{x}}$ . Then $z = x^n f(u)$ .

Differentiating partially with respect to $x$ :
$\frac{{\partial z}}{{\partial x}} = n x^{n-1} f(u) + x^n f'(u) (-\frac{{y}}{{x^2}}) = n x^{n-1} f(u) - x^{n-2} y f'(u)$
Multiply by $x$ :
$x \frac{{\partial z}}{{\partial x}} = n x^n f(u) - x^{n-1} y f'(u)$ --- (1)

Differentiating partially with respect to $y$ :
$\frac{{\partial z}}{{\partial y}} = x^n f'(u) (\frac{{1}}{{x}}) = x^{n-1} f'(u)$
Multiply by $y$ :
$y \frac{{\partial z}}{{\partial y}} = x^{n-1} y f'(u)$ --- (2)

Adding (1) and (2):
$x \frac{{\partial z}}{{\partial x}} + y \frac{{\partial z}}{{\partial y}} = n x^n f(u) = n z$ .
Hence proved.

What is the extension (second-order) of Euler's Theorem for homogeneous functions?

Extension of Euler's Theorem:
If $z = f(x,y)$ is a homogeneous function of degree $n$ in independent variables $x$ and $y$ , and its second-order partial derivatives exist and are continuous, then:

$x^2 \frac{{\partial^2 z}}{{\partial x^2}} + 2xy \frac{{\partial^2 z}}{{\partial x \partial y}} + y^2 \frac{{\partial^2 z}}{{\partial y^2}} = n(n-1)z$

Significance:
This extended theorem provides a direct way to evaluate complex second-order partial differential expressions for homogeneous functions without explicitly computing the derivatives, simplifying many algebraic procedures.

Define the Jacobian of $u, v$ with respect to $x, y$ .

Jacobian Definition:
If $u$ and $v$ are functions of two independent variables $x$ and $y$ , then the determinant of their first-order partial derivatives is called the Jacobian of $u, v$ with respect to $x, y$ .

It is denoted by $J = \frac{{\partial(u,v)}}{{\partial(x,y)}}$ and is defined as:

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

Evaluating the determinant gives: $\frac{{\partial u}}{{\partial x}}\frac{{\partial v}}{{\partial y}} - \frac{{\partial u}}{{\partial y}}\frac{{\partial v}}{{\partial x}}$ .

State any two important properties of Jacobians.

Properties of Jacobians:

Inverse Property: If $J = \frac{{\partial(u,v)}}{{\partial(x,y)}}$ and $J' = \frac{{\partial(x,y)}}{{\partial(u,v)}}$ , then $J \cdot J' = 1$ , provided $J \neq 0$ . This implies that the Jacobian of an inverse transformation is the reciprocal of the Jacobian of the original transformation.
Chain Rule Property: If $u, v$ are functions of $r, s$ , and $r, s$ are themselves functions of $x, y$ , then:
$\frac{{\partial(u,v)}}{{\partial(x,y)}} = \frac{{\partial(u,v)}}{{\partial(r,s)}} \cdot \frac{{\partial(r,s)}}{{\partial(x,y)}}$ .
This is analogous to the chain rule for single variable derivatives.

Explain the condition for functional dependence using Jacobians.

Functional Dependence using Jacobians:
Two functions $u = f(x,y)$ and $v = g(x,y)$ are said to be functionally dependent if there exists a relation $F(u,v) = 0$ between them that does not involve $x$ and $y$ explicitly.

Condition:
The necessary and sufficient condition for $u$ and $v$ to be functionally dependent is that their Jacobian with respect to $x$ and $y$ must be identically zero.

$\frac{{\partial(u,v)}}{{\partial(x,y)}} = 0$

If the Jacobian is non-zero, the functions are functionally independent.

What are the necessary conditions for a function of two variables $f(x,y)$ to have an extreme value (maxima or minima) at a point $(a,b)$ ?

Necessary Conditions for Extrema:
For a function $f(x,y)$ to have a local maximum or a local minimum at a point $(a,b)$ , the first-order partial derivatives must vanish at that point.

Mathematically:

$\frac{{\partial f}}{{\partial x}} \bigg|_{(a,b)} = 0$
$\frac{{\partial f}}{{\partial y}} \bigg|_{(a,b)} = 0$

A point $(a,b)$ that satisfies these conditions is called a stationary point or a critical point. It is important to note that this is a necessary condition, but not a sufficient one (it could be a saddle point).

Describe the sufficient conditions to determine whether a stationary point is a maximum, a minimum, or a saddle point.

Sufficient Conditions for Extrema:
Let $(a,b)$ be a stationary point of $f(x,y)$ , meaning $f_x(a,b) = 0$ and $f_y(a,b) = 0$ .
Compute the second-order partial derivatives at $(a,b)$ :

$r = \frac{{\partial^2 f}}{{\partial x^2}}$
$s = \frac{{\partial^2 f}}{{\partial x \partial y}}$
$t = \frac{{\partial^2 f}}{{\partial y^2}}$

Evaluate the discriminant $\Delta = rt - s^2$ :

Local Minimum: If $rt - s^2 > 0$ and $r > 0$ , the function has a local minimum at $(a,b)$ .
Local Maximum: If $rt - s^2 > 0$ and $r < 0$ , the function has a local maximum at $(a,b)$ .
Saddle Point: If $rt - s^2 < 0$ , the point $(a,b)$ is a saddle point (neither max nor min).
Inconclusive: If $rt - s^2 = 0$ , the test fails, and further investigation is required.

Define a 'Saddle Point' in the context of functions of two variables.

Saddle Point Definition:
A saddle point of a function of two variables is a critical point (where first partial derivatives are zero) that is neither a local maximum nor a local minimum.

Geometric Interpretation:
At a saddle point, the surface resembles a horse's saddle. In one cross-sectional direction, the point represents a local minimum, while in another cross-sectional direction, it represents a local maximum.

Mathematical Condition:
Using the second derivative test, a critical point $(a,b)$ is a saddle point if $rt - s^2 < 0$ , where $r = f_{xx}$ , $s = f_{xy}$ , and $t = f_{yy}$ evaluated at $(a,b)$ .

Outline the step-by-step procedure to find the extreme values of a function $f(x,y)$ .

Procedure to find Extrema:

Find Partial Derivatives: Calculate $\frac{{\partial f}}{{\partial x}}$ and $\frac{{\partial f}}{{\partial y}}$ .
Locate Critical Points: Set $\frac{{\partial f}}{{\partial x}} = 0$ and $\frac{{\partial f}}{{\partial y}} = 0$ and solve the system of equations simultaneously to find all critical points $(a,b)$ .
Calculate Second Derivatives: Find $r = f_{xx}$ , $s = f_{xy}$ , and $t = f_{yy}$ .
Apply Second Derivative Test: For each critical point $(a,b)$ , evaluate $r, s, t$ and the discriminant $rt - s^2$ .
Classify:
- If $rt - s^2 > 0$ and $r < 0 \implies$ Maximum.
- If $rt - s^2 > 0$ and $r > 0 \implies$ Minimum.
- If $rt - s^2 < 0 \implies$ Saddle Point.
Find the Value: Substitute the points giving max/min back into $f(x,y)$ to find the extreme values.

Explain Lagrange's Method of Undetermined Multipliers.

Lagrange's Method of Undetermined Multipliers:
This method is used to find the maximum or minimum of a function $f(x,y,z)$ subject to a constraint equation $\phi(x,y,z) = 0$ .

Method:

Define a new auxiliary function (Lagrangian function) $F(x,y,z,\lambda)$ as:
$F(x,y,z,\lambda) = f(x,y,z) + \lambda \phi(x,y,z)$
where $\lambda$ is an unknown constant called the Lagrange multiplier.
Find the partial derivatives of $F$ with respect to $x, y, z$ and equate them to zero:
$\frac{{\partial F}}{{\partial x}} = 0, \quad \frac{{\partial F}}{{\partial y}} = 0, \quad \frac{{\partial F}}{{\partial z}} = 0$
Solve these three equations along with the constraint equation $\phi(x,y,z) = 0$ to find the values of $x, y, z$ and $\lambda$ .
The points $(x,y,z)$ obtained are the critical points where the constrained extrema occur.

What is the primary limitation or drawback of Lagrange's method of undetermined multipliers?

Limitation of Lagrange's Method:

Nature of Extrema is Unknown: The primary drawback of Lagrange's method is that it only provides the critical points (where the stationary values occur). It does not directly provide a mathematical test (like the second derivative test) to determine whether the function value at that point is a maximum, a minimum, or neither.
Resolution: To determine the nature of the extremum, one usually has to rely on the physical or geometric context of the problem, investigate the values of the function at neighboring points, or use advanced second-order conditions involving bordered Hessians.
Complexity: For systems with multiple variables and non-linear constraints, solving the resulting system of equations can be algebraically very difficult or impossible analytically.

Unit4 Unit6