Unit 5: Multivariate Calculus - Subjective Questions

Definition: The statement means that for every , there exists a such that Explanation: The point can approach along infinitely many paths in the plane, not merely from the left or right as in one-variable calculus. Therefore, the limit exists only if approaches the same value along every possible path. If two paths produce different limiting values, the limit does not exist.

The function is defined for all , but it is not defined at the origin. Along the path , Along the path , Since the limiting values along these two paths are different, does not exist. Hence the function cannot be continuous at . Even if a value were assigned at the origin, continuity could not be obtained because the limit does not exist.

Partial derivatives: The partial derivatives at are defined by and Differentiability: A function is differentiable at if there exist constants and such that In that case, and , and the total differential is The existence of partial derivatives only gives information along the coordinate directions. It does not ensure that the function has a valid linear approximation from all directions. However, if the first-order partial derivatives exist in a neighborhood and are continuous at the point, then the function is differentiable there.

A function is continuous at if A commonly used sufficient condition for differentiability is that the first-order partial derivatives and exist in a neighborhood of and are continuous at . Under this condition, is differentiable at and Differentiability implies continuity, because the relation tends to zero as . Thus, continuous first partial derivatives provide a convenient sufficient condition, although they are not necessary in every case.

Let , with and . Then is a function of through both and . A small change in is given by Dividing by gives Therefore, the chain rule is More generally, if , where and ), then and

First calculate the partial derivatives of with respect to and : Also, and By the chain rule, Hence, Substituting and , Similarly, so

Suppose the variables are transformed according to A small rectangle in the -plane is transformed into a small parallelogram in the -plane. The tangent vectors of this parallelogram are and Its area is the absolute value of their determinant: Thus, The transformation should be one-to-one over the region, except possibly on boundary points, and the Jacobian should not vanish in the interior.

Use the polar transformation Then and the Jacobian is so that The circular region becomes Therefore, Evaluating,

A function is homogeneous of degree if Euler's theorem states that if is differentiable and homogeneous of degree , then Proof: Starting with differentiate both sides with respect to . By the chain rule, Put . Then Hence, Euler's theorem is proved.

The function can be written as so it is homogeneous of degree . Its partial derivatives are and Therefore, Since we obtain Thus, which verifies Euler's theorem for a homogeneous function of degree .

For two functions and , the Jacobian of with respect to is defined by It measures the local scale factor by which areas change under the transformation. If then, provided the transformation is invertible, The area element transforms according to If the Jacobian is zero, the transformation may fail to be locally invertible and may collapse area into a lower-dimensional set.

We have and Hence, Therefore, Thus, The transformation is singular when the Jacobian is zero: This occurs only at

For a function , stationary points are found by solving At each stationary point , calculate the second-order partial derivatives Form the discriminant The classification is as follows: • If and , has a local minimum. • If and , has a local maximum. • If , the point is a saddle point. • If , the test is inconclusive and another method is required. For absolute extrema on a closed bounded region, boundary points must also be examined.

First compute the first partial derivatives: Set them equal to zero: Substitution gives , so or . Corresponding points are and . The second partial derivatives are The discriminant is At , so is a saddle point. At , and , so is a local minimum. The corresponding minimum value is

The first partial derivatives are Both vanish at , so the origin is a stationary point. The second derivatives are Therefore, Since , the origin is a saddle point. This can also be seen directly: along , while along , Hence the function takes both positive and negative values near the origin.

To find constrained extrema of subject to the constraint introduce a multiplier and define the Lagrangian The necessary conditions are Equivalently, together with Thus, the equations to solve are The resulting points are candidate constrained maxima or minima. Their function values must be compared to determine the required extrema.

Let The Lagrange equations are Since we obtain Multiplying the equations gives either or . The cases or do not satisfy both equations except at points that give but are not the extreme values. From , we get or . Using the constraint: If , then , so , and If , then , , and Therefore,

Minimize the squared distance subject to The Lagrange equations are Hence Using the constraint, so Therefore, the closest point is The minimum distance is

Local maximum: A point is a local maximum if there exists a neighborhood of the point such that throughout that neighborhood. Local minimum: A point is a local minimum if in some neighborhood. Absolute maximum: A point is an absolute maximum on a domain if its function value is greater than or equal to the value at every point of the entire domain. Absolute minimum: It has a value less than or equal to the value at every point of the domain. Saddle point: A stationary point is a saddle point if the function takes both greater and smaller values than its value at that point in every neighborhood. The second derivative test identifies a saddle point when

At a stationary point , we have For small increments and , Taylor's expansion gives The quadratic form is Its determinant is If and , the quadratic form is positive definite, so the point is a local minimum. If and , it is negative definite, so the point is a local maximum. If , the quadratic form takes both positive and negative values, so the point is a saddle point. If , the quadratic form is degenerate and the test gives no conclusion.