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Vectors

Vector Functions

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##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

in this problem, we want to find at what point the curve y equals natural. Log of X has maximum curvature. And we want to see what happens to the curvature as X approaches infinity. So, Teoh, figure out the first part of what happened at what point? But what value X The curvature has the curve has maximum curvature. We would use this formula on the top right hand side to find the curvature of this. This curve y equals natural log of X. So we're gonna need our first and second derivatives. So I'm gonna take the first derivative of why, with respect to X, which is going to give us one over X and then my second derivative, he's going to give us negative one over X squared. So to find the curvature Kappa of X, we're going to be looking at the absolute value of Y double prime in the numerator and in the denominator, we're gonna have one plus the first derivative of a Y squared with the entire denominator raised to the three halfs power. And if we go ahead and substitute in our values, we're gonna have the absolute value of one over negative one over X squared all over one plus one over X quantity squared to the three halves power. And we're going to simplify this a bit because we need to find the the maximum for this resulting function. So we want to be able to use our first derivative test is the test that I'm going to use. So our numerator is going to become one over X squared, which is always gonna be positive and then in the denominator, going to have one plus one over X squared. If I simplify that to the three halves power, I'm gonna put the X squared from the come from the fraction in my numerator down in the denominator. And then if I add the two quantities in the denominator, this is going to become one plus x squared over X squared all to the three house power. And then, if I split the numerator and denominator of that fraction and parentheses, um, in the denominator than what we're gonna have is in my numerator, I'm gonna have X squared to the three house power. And then in the denominator, I'm gonna have X squared times one plus x squared to the three has power. And if I look at the numerator and denominator, those air going Teoh, these were considered, like terms. We're gonna combine them. So we end up with just X and the numerator, and in the denominator, we have one plus x squared 23 house power And if I want, so that way you don't have to use the quotient rule. I find the product roll a little bit easier when taking. Do If it is, I'm gonna write. This is X times one plus X squared to the negative three halves power. So that is the function that represents the curvature of X. If I want to find the maximum for this function and the point at which that maximum occurs, I'm gonna need to take the first derivative. So I'm gonna look at Kappa Prime of X and using a problem. The product rule. What we'll do is we'll have the first times the derivative of the second, which is gonna be negative. Three halves times one plus X squared to the negative five halves Power times, two exes. I take the derivative of that inside function, plus the derivative of the first which is ex, uh thing the derivative of X is just one times the second function. So one plus X squared to the negative three halfs power and we're going to simplify that just a little bit. So I see my choose here are gonna cancel out and I'm gonna be left with negative three x squared times one plus x squared to the negative five halves plus one plus x squared to the negatives three halfs power. So to find the maximum, I'm gonna set this first derivative equal to zero, and we're going to solve for X. So what we can do is we can factor out a well First will move this negative term over to the right hand side by adding it to both sides. So we end up with one plus X squared times the negative or to the negative three halfs power equals three X squared times one plus X squared to the negative five house power. If I were to divide both signs of this equation by one plus X squared to the negative five halfs power on the right hand side, I'm just gonna be left with three x squared and on the left hand side. This is going Teoh simplify to be you one plus x squared. Because if I consider the exponents, um and I move this to the numerator change the exponents toe a positive five times we're gonna be left with five house minus three halves or just one as our exponents. So now that we're here, we'll subtract Negative will subtract X squared from move science who were left with one equals two x squared when a divide by both sides by two And then if I take the square root of both sides we're gonna end up with X equals one over the square root of two I know I can take the positive square root because if I think about that original function we were given, why equals natural log of X that can Onley take positive values of X. So I know that X has to be positive one of the root you and not negative one ever route to. So we know that, sim that a critical point occurs here because the derivative is equal to zero. We do need to verify that it's a maximum so over off to the side, I'm gonna write that we are verifying that a max and not a men occurs at X equals one ever route to. So if I think about my derivative here, I think about the value of X. I like to draw a number line. We knew that X has to be positive. So the way that we can do this using the first derivative test is I think about what my, um what The value of X is where my critical point occurs. And then I want to think about a value less than that and greater than that where I can a value, evaluate the derivative and see whether it's positive or negative. So one over route to is approximately 0.7. So what I might do is evaluate the derivative of Kampe at 1/2 4.5. And once we get a point less than X and greater than X, so you can evaluate this in in a graphing calculator. So if we wanted to evaluate caba prime of 1/2 to get that first point, uh, again, you can cup Put this in your calculator, but you should get 0.286 roughly and then if you evaluate capper prime of one, you're going to get approximately negative 0.177 So what we see is that we have a positive slope or positive derivative at X equals 1/2 in a negative derivative at X equals one. So what that means is that we have a function Kappa that is increasing. It turns around at that critical point and then it's decreasing. So we know that we do have a maximum at X equals one over route to so we can say cap of X or the curvature has a max at X equals one over route to Or we can say that why equals natural Log of X has maximum curvature, which is represented by Kappa at X equals one over route to there. We've got the first part of our problem. Now. We also want Teoh discuss what happens as the value of X approaches infinity. What happens to the curvature? So weaken dio is we can look at the limit of the curvature as exit purchase infinity. And if we used the function that we evaluate here that we came up with earlier, we're looking at the limits as X approaches infinity of X over one plus X squared to the three halves power and we can see that the degree of X and our denominator is going to be greater than the degree of X on the numerator. So this limit is going to equal zero. So what we can say is that the curvature capital of X approaches zero as X approaches infinity.

Campbell University

#### Topics

Vectors

Vector Functions

##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp