Unit 4: Multivariate differentiation - Subjective Questions

Definition: A function has the limit as if, for every , there exists such that

It is written as

Path test:

• If the limit exists, must approach the same value along every path leading to .
• Common paths include , , the coordinate axes, and polar paths , .
• If two paths produce different limiting values, the multivariable limit does not exist.
• Obtaining the same value along several paths does not by itself prove that the limit exists; an inequality, polar-coordinate argument, or - proof may still be required.

Consider paths of the form . Substitution gives

This value depends on :

• Along , corresponding to , the value tends to .
• Along , corresponding to , the value tends to .

Because two paths approaching produce different limiting values,

A function is continuous at if

For the given function,

As , . By the squeeze theorem,

Since , the limit equals the function value. Therefore,

Treating as constant,

Treating as constant,

At ,

Hence,

Clairaut's theorem: If and are continuous in a neighborhood of a point, then

at that point.

For the given function,

so

Also,

and therefore

Thus,

Since these derivatives are continuous everywhere, Clairaut's theorem applies.

Partial differentiability examines change in one variable while holding the other fixed. Total differentiability requires a linear approximation

At the origin,

and similarly,

If were differentiable at the origin, then because both partial derivatives are zero, it would be necessary that

Along ,

and hence

The required limit is not zero. Therefore, the partial derivatives exist at the origin, but

At ,

The partial derivatives are

Thus,

The linear approximation is

For ,

Therefore,

Numerically,

If , where and , the chain rule states

Here,

and

Therefore,

At , and . Hence,

Thus,

The required chain-rule formulas are

Now,

and

Therefore,

while

Hence,

This agrees with the direct expression .

Let

Differentiating with respect to , while treating as a function of and , gives

Thus,

Similarly, differentiation with respect to gives

so

At ,

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

then

Proof: Define

By homogeneity,

Differentiating this expression with respect to gives

On the other hand, applying the chain rule to gives

Set . Then

Therefore Euler's theorem is proved.

First, test homogeneity:

for . Hence is homogeneous of degree .

The partial derivatives are

Therefore,

Thus,

Hence Euler's relation is verified:

Euler's theorem gives

Differentiate equation with respect to :

Therefore,

Differentiate equation with respect to :

so

Multiply equation by and equation by , then add:

Using Euler's theorem again,

Consequently,

A point is stationary if

Define the Hessian discriminant

The classification is:

• 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 higher-order terms or another argument must be used.

This test follows from the sign of the quadratic part of the Taylor expansion of near the stationary point.

The first partial derivatives are

At a stationary point,

Substitution gives , so

The real solutions are . Since , the stationary points are

The second derivatives are

Thus,

• At , , so it is a saddle point.
• At , and , so it is a local minimum.
• At , and , so it is also a local minimum.

At both minima, . Since the quartic terms dominate as , these are also global minima.

For both functions, all first derivatives vanish at , so the origin is stationary. Their second derivatives also vanish at the origin, giving

Therefore, the quadratic part of the Taylor expansion supplies no classification.

For

we have , with equality only at the origin. Hence is a strict local minimum.

For

along ,

whereas along ,

Thus, is a saddle point for .

These examples show that when , higher-order terms or direct sign analysis must be examined.

Because the disk is closed and bounded and is continuous, absolute extrema exist.

Rewrite the function as

Interior: The stationary conditions are

Thus, is an interior point, and

Boundary: On ,

Since on the circle:

• The smallest boundary value occurs at , giving at .
• The largest boundary value occurs at , giving at .

Comparing all candidates,

The constraint represents a level curve. At a constrained extremum, motion along the tangent to this curve produces no first-order change in .

The gradient is normal to the constraint curve. At the extremum, the level curve of is tangent to the constraint, so must also be normal to the constraint. Therefore, the two gradients are parallel:

In component form,

together with

Procedure:

• Form the Lagrangian .
• Solve , , and .
• Evaluate at all feasible candidates.
• Compare the values to identify maxima and minima.

The method assumes that at the candidate point; singular constraint points should be checked separately.

Let

The Lagrange equations are

which give

with

Neither nor can be zero at a solution of both multiplier equations. Multiplying the first two equations appropriately gives

so .

• If , then . The constraint gives , producing and . At these points, .
• If , then . This gives and . At these points, .

Therefore,

The volume is

subject to

The Lagrange equations are

along with

For a positive-volume interior solution, . Comparing the multiplier equations gives

and

Hence,

Using the constraint,

so

The resulting volume is

Boundary points have at least one side equal to zero and therefore have volume zero. Thus,