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Find the points on the curve whose equation is $x^{2}+y^{2}=16$ nearest and farthest from (a) (8,0) (b) (6,2) (c) (2,3) . Hint: The computation may be easier if you try to minimize $d^{2}$ rather than $d$.

(a) (4,0),(-4,0)(b) $\left(\frac{12}{\sqrt{10}}, \frac{4}{\sqrt{10}}\right),\left(\frac{-12}{\sqrt{10}}, \frac{-4}{\sqrt{10}}\right)$(c) $\left(\frac{8}{\sqrt{13}}, \frac{12}{\sqrt{13}}\right),\left(\frac{-8}{\sqrt{13}}, \frac{-12}{\sqrt{13}}\right)$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 4

Applications I - Geometric Optimization Problems

Derivatives

Campbell University

Harvey Mudd College

Baylor University

Boston College

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.

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03:55

this problem is a bit of a uh just there's lots of really what's a branches to it. So we want to minimize the distance between a certain point and a circle of radius. Uh for so we can solve this as our constraint for X. And so that we get a plus or minus here. So that means the squared is the quantity plus or minus where the 16 minus Y squared minus X. Not elsewhere plus, why minus? Why not squared? Right. You can take the derivative of the square with respect to why? And we get this here we get -2. Why not? Um plus or minus two. X not over two X. Not y. All over the square root of uh 16 minus y square mm. Now, if we said why you go to y one and set this equal zero, we then get why one equals plus or minus for why not All over square root effects not square because why not squared? And that means then X not X. One is plus or minus four X. Not over squared of X squared plus. Why not swear? All right. So we got one up with four different possible combinations here, plus or minus here. Plus or minus here. Now, the first point they asked us to do is to um 2.80. So right here, If we plug in 80 here we get um X is plus or minus four and why is plus or minus zero? Just zero? So we get this solution and this solution. Yeah. So obviously that looks like the minimum and that looks like the match them as we would expect Now if you plug in 6 to we get four possible solutions plus or minus six times the square root of 2/5 and plus or minus two times the square root of 2/5. Now here's our point and you can see that you know this one, the plus plus is the minimum distance and the minus minus is the maximum distance. So if we would have to take the second derivatives of this, we could have figured out that these were saddle points, not men's arm axes. So this is a minimum, this is an accident, but we can just kind of look at that from a by inspection here. Now the next point is 2, 3 which is actually inside the circle here And we have four possible solutions plus or -8 over square to 13 and plus or -12 over square to 13. And so again, we have this solution, the solution, the solution, this solution. So the minimum is the plus with plus in both both of these, the plus side plus plus. The that's the maximum distance from here is the one where we have both minus signs and then these two are seven points. Well, it's not a minimum or maximum. So again, we could have figured that out if you take second derivatives, but it's really ugly. Taking second derivatives of taking another derivative. This it doesn't get too ugly, but we could have figured that out what second test if we wanted to, but just uh kind of sketching shows us clearly what's going on. Yeah.

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