Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 36 Hard Difficulty

Use Theorem 10 to show that the curvature of a plane
parametric curve $x=f(t), y=g(t)$ is
$$\kappa=\frac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{\left[\dot{x}^{2}+\dot{y}^{2}\right]^{3 / 2}}$$
where the dots indicate derivatives with respect to $t$.

Answer

See proof demonstrated in video.

Discussion

You must be signed in to discuss.

Video Transcript

in this problem, we have a Parametric er that is represented by the functions X equals f of tea and why equals g of t and this is a plane curve and we want to use theorem 10 which is over on the right to show that the curvature can be written as follows The absolute value of first derivative of X times, second derivative of why minus first derivative of why time Second derivative of X over first derivative of X squared plus first derivative of why squared. And that denominator is going to be raised to the three halves power. Now, just a note about notation. Typically, we don't often right derivatives with the dot super top of the of the function variable. But we're going to hear just the number of dots correspond Teoh the number of times we're taking the derivative of that function. So the first thing that I'm gonna do to tackle this problem is right, our plane curve as a vector function. So we're gonna write this as our of tea and for a plane curve. We know that the X uh component function is going to be FFT. Why component function is going to be GFT and because it's in, it's a curve that's in a plane R z coordinate here is just gonna be zero rz a component function. So in order to be able to apply fearing 10 we know that we need to have the first and second derivatives of the SPECTRE function are so particular first derivative. But I'm going to end up with is f prime of tea G prime of tea and zero I take the derivative again are a double prime is going to be equal to F double prime of tea G double private T and zero again And we want to take the cross product of the first and second derivatives. We're not gonna take the magnitude yet and remember that you can take the cross product by looking at the determinant of a three by three matrix. Go ahead and review that if you need to, but what we're going to end up with, I'm gonna write the final form here is zero zero and then our third component function is where we're going to have f prime of tea. Times G double prime of tea, minus F double prime of tee times G Prime of tea. Like I said, make sure that you review how Teoh find the cross product between two vectors if you need to. So then we're going to take the absolute value or sorry, the magnitude of that cross product. So our prime of t cross with our double prime if tea with the magnitude of that resulting vector. So we'll take the square root, our first component function squared and our second component functions squared. Just add up to be zero, so we don't need to write that. But we do. I need to write or third component function quantity squared under our square root sign. And remember that the square root of a squared value is the absolute value of the argument there. So we're looking at the absolute value F prime of tee times G double prime of T minus F double prime of tea, G prime of tea. So there's the numerator based off a theorem tent. Now what we want to do is also take the magnitude of our prime of tea, which we're gonna need for denominator. So our prime is right here. So if we take the magnitude of that we're gonna end up with our first component functions squared, plus our second component function squared. And then third component function squares just zero. So now we have enough to use theorem 10 and see. I'm gonna write it up here and plug things in to our formula here. So in our numerator, we're going to have absolute value. F private T G double prime of tea, minus F double prime of tee times G prime of tea and then in urgent nominator, we're going to have the square root uh, f prime of T squared plus g prime of T Square. And then that entire denominator is going to be raised to the third power. So what we want to do now is right. This in terms of the first and second derivatives of X and y so in our numerator f prime of tea corresponds to the first derivative of X. So I'm gonna write. That is X with the daughter over. Top G double prime of tea corresponds to second derivative of why rece attract f double prime of T, which is the second derivative of X times. First derivative of why actually gonna write that in the opposite order. So that way it corresponds to the relationship. We're trying to get to correspondence toe this name aerator right there. And then in our denominator, what we're gonna have is f prime of T is the first derivative of X Matt Quantity squared. We're adding first derivative of why squared and that's all under the square root sign and the race of the third power which we can also a right as raised to the three house power. And what we see is that this is indeed equal. Teoh, capital of T weaken right cap of tea in this notation.

Campbell University
Top Calculus 3 Educators
Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Joseph L.

Boston College