Lecture Note
University
Massachusetts Institute of TechnologyCourse
Multivariable CalculusPages
3
Academic year
2022
Sporkz
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Critical point type for a quadratic function We will use the second derivative test to find a maximum or minimum value for anequation. By definition, if y'=0, then there must be a local maximum or minimum atthat point. The main problem is that we have more possible situations and haveseveral derivatives, so we need to think harder about how we will decide. However, itwill involve the second derivatives again. Let us examine an example in which we have a function that is quadratic. We willconsider the case where the function is of the form ax2+bxy+cy2. The equation has a critical point at the origin because if you take the derivative withrespect to x, and if you plug x = y = 0, you'll get 0. You can also see that all of thesenumbers are much smaller than x and y when x and y are small, so the linearapproximation, the tangent plane to the graph, is really just w = 0. So we took a look at something that starts with x squared plus 2xy plus 3y squared.We can rewrite this as x plus y squared plus 2y squared. Since each of these termsmust be non-negative, the origin will be a minimum. A generalization of this identitycan be obtained by first completing the square in order to obtain, in general, aquadratic equation. We will assume that a is non-zero so that we may proceed.Now the square of this looks like it might be the beginning of a binomial expansion.This is similar to what we did earlier with the binomial expansion a(x + y)2 = x2 +2axy + y2. If we put b over 2a times y and square this expression, then thecross-term 2xy will become b over ax by. Of course, now we also get some y2 termsout of this.
How many y squares do we get? We can multiply b squared over 4a to get thenumber of y squares, c times y squared. We want to find c minus b squared over 4a. Let's look at that again. If we expand this thing, we will get ax squared plus b over 2atimes 2xy. That's going to be our bxy. But we also get squared over 4y squared timesa. That's b squared over 4a y squared. And that cancels out with this item here. Leftwith cy squared. If we expand (x + 2)2, we get x2 + 2x(2) squared over 2a squared plus a times xsquared divided by a. Notice that the two a's cancel out, leaving us with bxy over asquared y squared plus cy over a squared minus b square over 4a y squared. Here,the a and the a simplify, and now these two terms simplify to give us just c squaredin the end. So we'll rewrite this equation in a slightly different form. ITake 4a to the secondpower and add it to x plus b over 2a times y squared. This will give us the same thingas before, but with the 4a squared term canceled out by another 4a squared term. Inother words, nothing changes numerically; we just clear the denominator on one sideof the equation by adding the same quantity to both sides. Next, subtract b squared ysquared from both sides of the equation.
The equation can be expressed as the sum of two squares, which means that theorigin is a minimum point on the graph. If you have the equation expressed in termsof the difference of two squares, then it will be a saddle point because it will haveeither positive or negative values depending on whether one of the numbers is largerthan the other.
Critical Point Type for a Quadratic Function
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