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Find the points on the circle $x^{2}+y^{2}=9$ (a) closest to and (b) farthest from the point $(8,12)$.

(a) $\left(\frac{6}{\sqrt{13}}, \frac{9}{\sqrt{13}}\right),\left(\frac{-6}{\sqrt{13}}, \frac{-9}{\sqrt{13}}\right)$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 4

Applications I - Geometric Optimization Problems

Derivatives

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

06:15

The coordinates of two poi…

04:39

Solve using Lagrange multi…

03:06

01:55

Find the center and radius…

the problem was given another circle this time with a radius of three and has to find the minimum distance to the 30.8, 12. Um This is very, very similar the problem 24 23. So I'm not going to go through all the details. Um So we get lots of different solutions. So you know, x is plus or minus square root of nine minus wide square. And so we can plug that into the square which we're going to minimize and we get a plus or minus here and take the derivative and we get a plus or minus here. And that means we have we get why, why the optimal solution has plus or minus and then the why value and then X body also has plus or minus. So there's generally four solutions here and we have to just kind of either plug them back and plug them in and see which is the minimum maximums, which is probably the simplest thing to do or take a second derivative, which is kind of ugly. So if he and if you graph it, you can see pretty clearly what's going on. So here's our circle, here's our point of interest and we can see we have these four points are our solutions. Well, this one is clearly the um is clearly the minimum distance and this one is clearly the maximum distance. Now interestingly, I think you should probably prove that always um the difference between these two, the difference between this distance and this distance is going to be the diameter of this circle. So these ones are then actually saddle points, they're not mens and max. Is there somewhere in between here? The distance from here and obviously I think no, they're not. They're not. They're not equal this distance and this distance are not equal. So is the man that you can take both plus signs and the max or the minimum distance play. Take plus signs and if you take both minus signs, that gives you the maximum distance.

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