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Use the first and second derivatives to sketch the graph of the given equation. Also include the intercepts, whenever they are easily determined.$$x^{2}+4 y^{2}=16$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 3

Concavity and the Second Derivative

Derivatives

Missouri State University

Harvey Mudd College

Idaho State 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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Use the first and second d…

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Welcome today, we're gonna be talking about curve sketching. Now there are a few ways to approach this, but we're going to be using the one that focuses on calculus and using derivatives and the first derivative test is gonna be really important to us today as we go through this. Now, I'm going to skip a lot of the algebra because I want you to focus on using the derivatives instead of finding the derivatives. We already assume that you've gotten up to this point that you know how to take a derivative. So I'm going to present you with three equations, but we want you to focus on the actual use of them. So let's go through your given X squared plus four, Y squared is equal to 16. And you're asked to sketch this curve. Now we want to take derivatives of this and this is not in the best form to take derivatives. You'd have to do implicit differentiation in this case, but we want something in the form of Y equals something and that will be a lot easier to take derivatives of. So there's some algebra we're going to get Y equals plus or minus one, half the square root of 16 minus x squared. Now, all this plus or minus means is that we're going to have a if you cover up that negative, we're gonna have a positive version of this graph and a negative version of this graph. So whatever we have, we already know will be symmetric on the top and on the bottom. So it's going to either look like this on the top of the bottom or like this. You know, symmetric top and bottom are going to be symmetric. So that's useful information with that Plus or minus. Also notice that we have a square root. That square root should be screaming to you that we have a restriction on this function. We cannot take square roots of negative numbers or at least we don't want to. So we have to figure out what numbers can and can't we plug in. So let's set this equal to zero and find out those restrictions. So we're going to take this and we're gonna start setting it equal to zero to find out what the restrictions are. Because we don't want to be less than zero. We either want to be equal to zero or greater than zero. So let's start solving 16 is equal to X squared When we moved it over we added extra to both sides. We're going to take the square root of both sides and we're going to get X is equal to positive for and negative for All right. So that gave us some restrictions already, which is nice. Now, let's continue on when we do some more algebra and taking the derivative we will get Y. Prime is equal to negative X over two route 16 minus X squared. Alright, again, we see that screw. It should be screaming to us. We can't take square roots of negative numbers. We don't want to we don't like it. So we already found our restrictions there. But that's when they equal zero. If we plug in positive for, we're going to get 16 -16, that's going to equal zero. And we're in the denominator. We can't have a zero in the denominator. That is a big, big big no. So we actually have to restrict our interval to an open interval we can't equal for but we can approach it. So we're going to create a range we're gonna say, Hey, we're gonna span from -4. Less than yeah, less than for this is our refined interval Because this accounts for the fact that we cannot plug in negative for or positive for into a function. Otherwise our slopes are going to go insane and it's gonna be undefined. We already have a good idea of what this function looks like. Let's start putting down some important points on here. So we have these points right here, Open right here, this is negative for and positive for. This is where our function stretches from. We need some other information. Now, we're gonna get this from the first derivative tests and the first derivative test. We need our critical numbers to work with. Those are really important in our critical numbers. Come from setting the derivative equal to zero. So let's do that. So we're gonna take our derivative and set it equal to zero. It's a negative acts over to route 16 minus X squared. Alright. And we're going to solve by cross multiplying. So on this diagonal we're going to get zero and on the other diagonal we're gonna get negative X. And solving we get X equals zero as our critical number. We only have one this time, which is nice. So we know what X equals zero. Somewhere along here our function is going to give us a some sort of weird anomaly. We need to analyze it first. So remember at the first derivative test, we're gonna set it up like this, going to create this we're gonna put our intervals. So we're gonna do from our interval to our critical number and are critical number two our interval. So we're gonna set that up right now. We have an open bracket of negative four, two are critical number zero And from our critical number zero to our other bound, which is open. All right. So now we need to pick test numbers within each interval to test our slopes. We want to find out what it's doing. Is it increasing decreasing? We want to understand our function a little bit more. So we need to test are derivative on either side of the interval and find out whether it's increasing, decreasing what's happening. So, I'm going to pick a number in here and I'm gonna pick an easy number. We're going to pick, you know, why don't we pick negative too negative two to plug in? Because it looks like we could cancel some twos there. We're gonna plug in positive to over here. Get some nice numbers. All right. So, I'm going to start resolving this algebra quickly so you can see the results. So why Prime -2? We're gonna resolve this over here. We're gonna take this and just plug it in here. Negative negative too. Two 16 minus negative two squared. And we're really, really interested in the sign of everything, so, it doesn't matter what the actual number is for the first derivative test. We're just interested in the sign because that will tell us well for increasing or decreasing with the slope. So that's gonna give us a positive 2/2. Route 16 -4. All right. This is going to be positive. This is a positive value. So, we're going to mark that right here and we're going to do the same over here. Why Prime of positive too? Going to give us negative too? Two. Okay, two squared. All right. This right here is always going to be negative. So, we're gonna mark that down right here. We're going to say this is increasing because we had a positive slope value increasing and over here decreasing because we had a negative slope value. See ice and organized. And we're just gonna do a mock little sketch right here. That's what our silk values would look like. And this would indicate that this is a maximum. Act X equals zero because before and after a critical point we had a change. So we know that we went from positive slopes to negative slopes. We cross zero. Somewhere in our slope function, so X equals zero. We have a maximum. So somewhere here, I don't know where it could be anywhere here. We went from positive slopes to negative slopes. So rough sketch something like this. Maybe. All right now we have to go back to our function because we had to technically, and we don't wanna have to do this derivative test again. But we can justify the bottom half of this graph by saying, hey, this is symmetric. The positive and negative versions are identical is just flipped. So because we know it's symmetric, we can fill out and say, hey, our other function when we do the first negative test will give us a minimum. And this is roughly what our graph will look like. Um And yeah, so this is how we would find our sketch using derivatives. Now there's other ways that we could look at this and we could recognize our original equation is a connects section and this is the equation of the lips. Through some algebra, you can find out what your intercepts are on each access and you can find that out. But using calculus, we found out that using the first round of tests and symmetry, that we can quickly sketch these graphs. So thank you for watching.

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