Like

Report

At what point does the curve have maximum curvature? What happens to the curvature as $x \rightarrow \infty ?$

$$y=e^{x}$$

The curve $y=e^x$ has maximum curvature at $x=\ln \left(\frac{1}{\sqrt{2}}\right)$. Curvature $\kappa(x)\rightarrow 0$ as $x\rightarrow\infty$.

You must be signed in to discuss.

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

in this problem. We want Teoh evaluate what At what value of X. The curve y equals e to the power of X has maximum curvature. And we also want to take a look at what happens when X approaches infinity. Look at what happens to the curvature. So first thing, we're going to use the formula on the right hand side, Teoh, find the curvature, the formula for the curvature of this, uh, this function. Why? So we're gonna need our first and second derivative. Let's go ahead and take those. It happens to be pretty easy in this case, because the derivative of each of the power of X is eat of the power of X, and so is the second derivative. So now we can use the formula to find the formula for curvature cap. So Kappa of exc equals the absolute value of y double prime over one plus, why prime square to the power of three halves in our denominator. So if we substitute the appropriate values in, we're gonna have absolute value of each of the X power in the numerator and the denominator whereto have one plus e to the power of X squared in the entire denominator raised to the three halfs power. Let's simplify that a little bit. So in the numerator, we're just gonna have a positive, uh, E to the power of X because that quantity is always going to be positive. So we don't need to incorporate the absolute value signs anymore. And then our denominator is going to be one plus e to the power of two X all raised three house power. And because we're gonna be finding the maximum of the function. Kappa, I like to use the product rule as opposed to the quotient rule. So we're gonna go ahead and write this as e to the power of X times one plus eat of the power of two X to the negative three house power. So let's go ahead and find where that maximum occurs. We're gonna look at the first derivative of the curvature or first derivative of the function Kappa. So if we look at the first function, E g of x times, the derivative of the second or gonna look it U to the power of X times negative three halves times one plus either the power of two X to the negative five halves Power times the derivative of the inside function, which is gonna be two times e to the power of two X. Using our chain rule there and then derivative of the first function is e to the power of X. And we multiply that by the second function needed about one plus e to the power of two X to the negative three House power. And we can go ahead and write that out a little bit more simply So our twos were going to cancel out here, will have negative three times e to the power of three X Combining are like terms times one plus eat of the power of two X to the negative five House power Plus you the power of X times one plus eat of the power of two X with the entire quantity to the negative three House power. And because we're trying to find the critical point for this derivative, we're gonna refer the function Kapo, we're gonna set this equal to zero. Now, what we can do is I'm gonna add this to both sides, so we're gonna end up with each of the power of x times one plus e to the power of two X to the negative three halves power on the left hand side and three times e to the power of three x times one plus e to the power of two X to the negative five House power on the right hand side. I can divide both sides by e to the power of X and then divide both sides by one plus e to the power of two X to the negative five house power. So in the right hands are on the left hand side. Eat of the power of X. Um, that simplifies over U to the power of X that simplifies to one and then these two base terms of the same. So we're going Teoh, combine our Judum or exponents appropriately. So we're gonna end up with one plus eat of the power of two x, all of the power of one, or I can just remove those parentheses. We don't have to write the X moment of one, but on the right hand side, these air going, those terms are gonna cancel and we're gonna be left with three times. Eats the power of two X um, because of the length terms. So then we can subtract eat of the power of two X from both sides so they end up with one equals three. Ah, actually, one equals two times e to the power of two X. If I divide both sides by two, will end up with 1/2 equals e the power of two X and we still wanna solve this for X. What I'm gonna do is I'm gonna take the natural log of both sides. So on the left, I'm gonna have natural log of 1/2. Now the rate I'm gonna have natural log of e to the power of two X, which we know that the natural log, which is a log base e and the e inside it are going to cancel. Essentially. So we're gonna end up with natural log. 1/2 equals two x. So then we're gonna have if I divide both sides by two, will have X equals 1/2 times natural log of 1/2. Now remember our exponents rules allow us to take this 1/2 on the outside and write it as the exponents of the natural log argument so you could have natural log of 1/2 raised to the 1/2 power which is just the square root of 1/2. So I'm gonna rewrite this as natural log of one over squared of to I think that's the most compact way of writing it. So we know that a critical point for Capital of X occurs here at natural log one over right to, but we need to verify that it's a maximum. So verify that a max occurs at natural log of one ever squared of to. So if I think about are derivative capital prime and her values of X, our function, why equals e to the power of X that can take any value of X were not restricted on our domain here. So I'm gonna think about, um, my my number line Ask such. So I want to see what happens to my derivative around the points or at the points around natural log of one ever to which is approximately negative 10.35 And we know that at that point, the derivative of the curvature is equal to zero. So what I could do is I could look at a point a bit below that. I'm gonna look at negative one, cause it's easy toe plug in and I'm gonna look a zero. Ah, point a little bit above it to see what's happening to my derivatives. So they look at keep or Kappa prime of negative one. You can plug that in using your calculator, but you should get something that's approximately 0.195 So we see that we have a positive derivative, um, this value of X and then if I look at copper prime of zero, you should get approximately negative 0.177 So we have a negative derivative after that critical point. So we can think about that. That refers to our function Kappa as something that has a positive slope. It's increasing. It hits are critical point with the slip of zero and then starts decreasing again. So we know that we do have a maximum. So we can say that why equals eat of Power X has maximum curvature at X equals natural log of one over route to, and that's the first part of our problem. So it also asked us What happens to the curvature is exit purchase infinity. So what we can look at is the sorry about that. The limit as X approaches infinity of the curvature, which is represented by Kapo of X So we can rewrite. This is the limit as X approaches infinity of e to the power of X over one plus e to the power of two X to the three house power. And we can see that inner denominator. The degree or the power of are our value are constantly e The power is have a higher degree essentially that are numerator. So it's gonna be increasing faster and our denominator than it will in our numerator as extra purchase infinity. So what we can see is that this limit is going to equal zero. So as a result, we can say that the curvature capital of X approaches zero as x a purchase. Infinity