Unit 4: Multivariate differentiation - Subjective Questions

Definition: The statement

means that for every , there exists a such that

implies

Path interpretation:

• If the limit exists, must approach the same value along every possible path leading to .
• Different values along two paths are sufficient to prove that the limit does not exist.
• Agreement along a finite number of paths does not prove existence because infinitely many other paths, including nonlinear paths such as or , remain unchecked.
• A proof of existence normally requires an - argument, a suitable inequality, the squeeze theorem, or conversion to polar coordinates.

Consider two different paths approaching .

Path 1:

Therefore,

Path 2:

Therefore,

Since the function approaches different values along the two paths, the limit is not unique. Hence,

A function is continuous at if:

• is defined,
• exists, and
• .

For the given function, use polar coordinates:

Then, for ,

Since ,

As , . By the squeeze theorem,

Because , the limit equals the function value. Therefore, is continuous at the origin.

A partial derivative is obtained by differentiating with respect to one independent variable while keeping the other independent variables constant.

First-order derivatives:

Second-order derivatives:

For the mixed derivatives,

and

Thus,

This equality is consistent with Clairaut's theorem because the mixed partial derivatives are continuous.

Partial derivatives:

• measures the rate of change of with respect to while is held constant.
• measures the rate of change of with respect to while is held constant.

Total derivative or total differential:

When both and vary, the combined first-order change in is

For small increments and ,

can be approximated by

Therefore,

The total differential is used for linear approximation, propagation of measurement errors, and estimating small changes in engineering quantities.

The increments are

For

the partial derivatives are

At ,

The total differential is

Therefore,

Hence, the estimated change is

Since , the estimated new value is

If , where and , the chain rule states that

For

we have

Also,

Thus,

Substituting and gives

Because depends on and through and , the chain rule gives

The required derivatives of and are

Therefore,

and

In matrix form, this can be written as

First calculate the derivatives of :

Also,

Using the chain rule,

Hence,

Substituting and ,

Similarly,

so

Therefore,

Equivalently,

A function is called homogeneous of degree if, for every admissible scalar ,

Examples include:

• , which is homogeneous of degree .
• , where defined, which is homogeneous of degree .

Euler's theorem: If is differentiable and homogeneous of degree , then

The theorem provides a differential test for homogeneity and is useful for simplifying expressions involving partial derivatives.

Since is homogeneous of degree ,

Differentiate both sides with respect to . Applying the chain rule to the left-hand side gives

Now set . Then and , so

Therefore,

This proves Euler's theorem. The proof relies on:

• the scaling property of a homogeneous function,
• differentiability of , and
• application of the chain rule with respect to the scaling parameter .

First test the scaling property:

for positive . Thus, is homogeneous of degree .

Now calculate the partial derivatives:

Therefore,

Hence,

Thus,

which verifies Euler's theorem for a homogeneous function of degree .

For an interior point at which is differentiable, a necessary condition for a local extremum is

A point satisfying these equations is called a stationary point or critical point.

Define

The second derivative test gives:

• If and , then is a local minimum.
• If and , then is a local maximum.
• If , then is a saddle point.
• If , the test is inconclusive and another method is required.

The quantity is the determinant of the Hessian matrix

Calculate the first-order partial derivatives:

At a stationary point,

Thus,

The only stationary point is .

The second-order derivatives are

Therefore,

Since and , is a local minimum. The function can also be written as

so the minimum is global. Its value is

Hence,

There is no maximum because as .

The first-order derivatives are

Setting them equal to zero gives

Substituting into gives

so

The real solutions are and . Hence, the stationary points are

The second-order derivatives are

Thus,

At ,

so is a saddle point.

At ,

and . Therefore, is a local minimum. Its value is

Hence,

The first derivatives are

Both vanish only at , so the origin is a stationary point.

The second derivatives are

At ,

and therefore

Thus, the ordinary second derivative test is inconclusive.

However,

for every , and equality holds only when and . Therefore, every nearby point other than the origin has a positive function value, while

Hence,

This example shows that does not imply the absence of an extremum; it only means that the second derivative test cannot decide the nature of the point.

For the constraint

define the Lagrangian

At a regular constrained extremum, the gradients are parallel:

Thus, solve the system

The resulting candidate points are substituted into and compared to identify constrained maxima and minima.

Geometric meaning: At a constrained extremum, the level curve of is tangent to the constraint curve. Their normal vectors, and , are consequently parallel.

Regularity condition: The standard method requires

at the candidate point. Points where must be examined separately because the multiplier equations may not capture them correctly.

Let

Define

The multiplier equations are

Thus,

Neither nor can be zero under these equations and the constraint. Multiplying the two equations gives

so

and therefore .

If , then . The constraint gives

so the points are

where .

If , then . The points are

where .

Hence,

Minimizing the distance is equivalent to minimizing its square,

subject to

The Lagrange equations are

Thus,

Therefore,

Substituting these into the constraint gives

Hence,

so

Therefore, the nearest point is

The minimum distance is

Thus,

Maximize

subject to

The Lagrange equations are

Because , divide the first two equations:

This gives

Cross-multiplying,

so

Since , . Comparing other pairs similarly gives

Apply the surface-area constraint:

so

Therefore,

Hence, the maximum-volume box is a cube with dimensions

Its maximum volume is

Thus, among rectangular boxes with a fixed surface area, the cube has the greatest volume.