To find the extreme values of a function f(x, y) on a curve...

To find the extreme values of a function $f(x, y)$ on a curve $x=x(t), y=y(t),$ we treat $f$ as a function of the single variable $t$ and use the Chain Rule to find where $d f / d t$ is zero. As in any other single-variable case, the extreme values of $f$ are then found among the values at the a. critical points (points where $d f / d t$ is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: a. $f(x, y)=x^{2}+y^{2} \quad$ b. $g(x, y)=1 /\left(x^{2}+y^{2}\right)$ Curves: i. The line $x=t, \quad y=2-2 t$ ii. The line segment $x=t, \quad y=2-2 t, \quad 0 \leq t \leq 1$

To find the extreme values of a function $f(x, y)$ on a curve $x=x(t), y=y(t),$ we treat $f$ as a function of the single variable $t$ and use the Chain Rule to find where $d f / d t$ is zero. As in any other single-variable case, the extreme values of $f$ are then found among the values at the
a. critical points (points where $d f / d t$ is zero or fails to exist), and
b. endpoints of the parameter domain.
Find the absolute maximum and minimum values of the following functions on the given curves.
Functions:
a. $f(x, y)=x^{2}+y^{2} \quad$ b. $g(x, y)=1 /\left(x^{2}+y^{2}\right)$
Curves:
i. The line $x=t, \quad y=2-2 t$
ii. The line segment $x=t, \quad y=2-2 t, \quad 0 \leq t \leq 1$

Key Concepts

Parameterization of Curves

This concept involves representing a curve using a parameter, typically denoted by t, so that the coordinates x and y become functions of t. This allows a multivariable function to be expressed as a single-variable function by substituting x(t) and y(t) into the function, which is essential for applying single-variable calculus techniques.

Chain Rule

The chain rule is used to differentiate composite functions. When a function f(x, y) is parameterized by t through x = x(t) and y = y(t), the chain rule allows us to compute the derivative df/dt by taking into account the rates of change of x and y with respect to t. This is fundamental in finding where the derivative equals zero or does not exist, indicating potential extrema.

Critical Points

Critical points are points at which the derivative of a function is zero or fails to exist. In optimization problems, especially when the function is converted to a single-variable function via parameterization, identifying these points is crucial because they are candidates for local maxima or minima.

Endpoints and the Extreme Value Theorem

When dealing with functions over a closed interval, it is not sufficient to only consider critical points. The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval attains its absolute maximum and minimum, which could occur at the endpoints of the interval. Evaluating the function at both critical points and endpoints is necessary to determine the absolute extrema.

Transcript

00:01 This problem is kind of a pain because they give us lots of different combinations here.

00:05 They want us to look at two functions, x squared plus y squared and one over x squared plus y squared, along two different, well along a line, same line, but in this case we're going from assuming it's unbounded.

00:21 And here we're saying, okay, we have the line segment, 0 to 1, t equals 0 to 1.

00:28 Now, what's going on here is, you know, one thing we could do, and it's probably almost easier to do it this way, is just plug these into here, you know, plug them into here, and then minimize over t.

00:43 But so what we want to do is we want to find the derivative, minimize these over t, so on these curves.

00:49 So we can basically do that by saying, okay, this is the f -a -d -t is 2x -d -t plus 2y d -x -d -t, and then over here we have minus 2x all over x2 plus x2 plus x -y -squared plus y -squared, and then minus 2 -y over x -quare plus y -squared d -y -ttt.

01:15 Well, the x -d -tt, in both cases, is 1, and d -y -d -t, and d -y.

01:19 Y d t in both cases is minus 2.

01:25 So we can plug that into here and this into here.

01:29 So we get the fa d t is 2x minus 4y.

01:36 And then using x and y, that winds up being minus 8 plus 10t.

01:42 Over here, we get this.

01:45 And then using x and y, we get 8 minus 10t all over this denominator, which really don't care about as long as it isn't zero.

01:55 I'm not sure if this has real roots or not.

02:00 But when we want to make this, this and this equals zero.

02:03 So this tells that, and for this function here, we have the critical value, there's a critical value at t equals five or four -fifths.

02:18 And let's see here.

02:22 Likewise, over here we have a critical value at t equals four -fifths.

02:30 So let's see here.

02:32 Now, if t is four -fifths, that means x is four -fifths and y in the first case, and y is two -fifths.

02:46 That means that the critical value of this function is five, there's a critical value at four -fifths, and on this one there's a critical value at five -fourths.

02:58 Now we can see here that this point here, i mean, both, cases is in this domain.

03:10 So we need to consider it because it is in this domain.

03:14 It's also in this domain...

Please give Ace some feedback