Level Curves and Implicit Differentiation PDF

Preview

Summary

Great notes to help achieve a first class...

Description

Level Curves and Implicit Differentiation

level curves • Let f(x, y) be a function in two variables x, y, and let c be a real number. The level curve of f at c is the curve in the x-y-plane of all points (x, y) satisfying the equation: f(x, y) = c

• In economics, isocost lines and indifference curves are typical applications of level curves

Implicit differentiation • A level curve is given by a formula of the form: c = f(x,y) where c is a constant. The purpose of this section is to measure the rate of change of y depending on x as we move along this level curve.

• Although it may not be possible to determine an explicit expression y = y(x), it is still possible to treat y as a dependent variable and write y = y(x), inserting this into the previous equation yields: c = f(x, y(x))

• Taking derivatives with respect to x on both sides yields 0 on the left side and the right side is determined using the chain rule. This yields:

• Solving for dy/dx yields:

- If fy = 0 at points on the level curve, the tangent to the level curve is parallel to the y - axis

EXAMPLE: Verify that (1, —1) lies on the level curve x² — 2y³ = 3 and calculate the rate of change of y with respect to x at the point. Set f(x,y) = x² — 2y³ Evaluating f at (1, —1) yields f(1, —1) = 1 — 2(—1) = 1 + 2 = 3, which shows that (1, —1) lies on the level curve f(x,y) = 3 We have

Evaluated at (1, —1) we get that

This shows that if we start from the point (1, —1), then the rate of change y depending on x is 1/3 This can also be done by taking directly the derivative with respect to x along the level curve 3 = x² — 2y³ This yields

hence as before

Several variables Let f be a function in n + 1 variables x₁, x₂, …, xn, y. Consider y = y(x₁, x₂, …, xn, y) satisfying: f(x₁, x₂, …, xn, y) = c, where c is some constant In order to calculate the partial derivatives , we use the chain rule together with the observation

#

for 1 ≤ i,j ≤ n and i ≠ j

Provided that

we get as in the case of two variables that:

EXAMPLE: Consider the function f(x₁, x₂, y) = x₁ — 2x₂ — 3y + y² and the level curve f(x₁, x₂, y) = — 2

These are defined except if y = 3/2

Optimisation without constraint Let f(x,y) a function in two variables. A point (x0, y0) in the x-y-plane is called a stationary point of f if

or equivalently, if

These points can be classified as maxima, minima or saddle points (neither max or min) A sufficient condition for the function f to have an extremum at the stationary point (x0, y0) is the the inequality

Assuming that the second order derivatives are continuous and hence we can develop a technique to decide whether this a min or max. Thus:

concave and convex functions A function f(x, y) in two variables is called concave if:

The function f is called convex if:

If these inequalities are strict inequalities, we say that f is strictly concave or convex

• Being concave around a stationary point implies that f has a local maximum, while being convex at a stationary point means that f has a local minimum at that point.

• A sufficient criterion for a strict local extremum (either a local maximum or a local minimum) at an interior point (x0, y0) of the domain of f is the condition:

or equivalently,

• The function f has at the point (x0, y0) a local maximum if

, and a local

minimum if

• If

at (x0, y0), then (x0, y0) is a saddle point

classifying local extrema and saddle points Step 1: Determine where

and

are both zero; this yields the stationary points

Step 2: Classify the stationary points. Let (x0, y0) be a stationary point.

Step 3: Calculate the value of the function at the stationary points is required

EXAMPLE 1: Locate and classify the stationary points of the function f(x, y) = 12x³ + y³ + 12x²y — 75y We need to determine the stationary points of f; that is, the points at which and = 0. Now:

Now

=0

= 0 implies either x = 0 or 3x + 2y = 0

Therefore, x = ± 2. We have y = — 3/2x and so this gives the points (2, — 3) and (— 2, 3) Hence there are 4 stationary points, which are (0, 5), (0, 5), (2, — 3) and (—2, 3). The corresponding stationary values are —250, 250, 150, —150. We now need to determine the nature of the stationary points. To do this we need to test if or point. Also, if we also need to determine whether First we calculate

and

at each or

We can now classify the points as follows:

EXAMPLE 2: Locate and classify the stationary points of the function f(x, y) = 2x² — 2y² — x⁴ — 2x²y² — y⁴ We need to determine the stationary points of f; that is, the points at which and = 0. Now:

=0

Thus = 0 implies y = 0 = 4x(1 — x²) = 0 gives x = 0, 1, —1 Hence there are three stationary points, namely (0, 0), (1, 0) and (—1, 0). The corresponding stationary values of f at these points are 0, 1, 1 respectively

We now need to determine the nature of the stationary points. To do this we need to test if or point. Also, if we also need to determine whether

at each or

First we calculate

and

We can now classify the points as follows:

EXAMPLE 3: Consider the function f(x, y) = x⁴ — y⁴ The first partial derivatives are

Thus (0,0) is the unique stationary point The second partial derivatives are

Thus all second partial derivatives and hence also |D²f| are zero at the stationary point (0,0). Therefore the above test for local extrema does not work Along the x-axis (that is, for y = 0), the function f becomes

This function is nonnegative everywhere, and 0 at 0, hence has its absolute minimum at 0 Along the y-axis (that is, for x = 0), the function f becomes This function is nonpositive everywhere, and 0 at 0, hence has its absolute maximum at 0. This shows that (0,0) is neither a maximum nor a minimum for f but a saddle point...