Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 12: Problem 15
Find the curvature and the radius of curvature at the stated point. $$ x=e^{t} \cos t, y=e^{t} \sin t, z=e^{t} ; t=0 $$
Short Answer
Expert verified
Curvature is \( \frac{\sqrt{2}}{3\sqrt{3}} \); Radius of curvature is \( \frac{3\sqrt{6}}{2} \).
Step by step solution
01
Parameterize with respect to t
We are given parametric equations: \( x = e^{t} \cos t \), \( y = e^{t} \sin t \), and \( z = e^{t} \). The point of interest is where \( t = 0 \).
02
Compute the first derivatives
Calculate the first derivatives of \( x, y, z \) with respect to \( t \):\[ \frac{dx}{dt} = \frac{d}{dt}(e^{t} \cos t) = e^{t}\cos t - e^{t}\sin t \]\[ \frac{dy}{dt} = \frac{d}{dt}(e^{t} \sin t) = e^{t}\cos t + e^{t}\sin t \]\[ \frac{dz}{dt} = \frac{d}{dt}(e^{t}) = e^{t} \]
03
Compute the second derivatives
Calculate the second derivatives:\[ \frac{d^2x}{dt^2} = \frac{d}{dt}(e^{t}\cos t - e^{t}\sin t) = 2e^t\cos t - 2e^t\sin t \]\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(e^{t}\cos t + e^{t}\sin t) = 2e^t\sin t + 2e^t\cos t \]\[ \frac{d^2z}{dt^2} = \frac{d}{dt}(e^{t}) = e^{t} \]
04
Evaluate the derivatives at t = 0
Substitute \( t = 0 \) into the first and second derivatives:\[ \frac{dx}{dt} \bigg|_{t=0} = 1 \quad \frac{dy}{dt} \bigg|_{t=0} = 1 \quad \frac{dz}{dt} \bigg|_{t=0} = 1 \]\[ \frac{d^2x}{dt^2} \bigg|_{t=0} = 2 \quad \frac{d^2y}{dt^2} \bigg|_{t=0} = 2 \quad \frac{d^2z}{dt^2} \bigg|_{t=0} = 1 \]
05
Calculate curvature
The formula for curvature \( \kappa \) in 3D is:\[ \kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \]Calculate the cross-product using evaluated derivatives:\[ \mathbf{r}'(0) = (1, 1, 1) \quad \text{and} \quad \mathbf{r}''(0) = (2, 2, 1) \]\[ \mathbf{r}'(0) \times \mathbf{r}''(0) = \mathbf{i}(1 \cdot 1 - 1 \cdot 2) - \mathbf{j}(1 \cdot 1 - 1 \cdot 2) + \mathbf{k}(1 \cdot 2 - 1 \cdot 2) = -\mathbf{i} + \mathbf{j} + 0\mathbf{k} \]Magnitude of cross product:\[ \| \mathbf{r}'(0) \times \mathbf{r}''(0) \| = \sqrt{1^2 + 1^2} = \sqrt{2} \]Magnitude of \( \mathbf{r}'(0) \):\[ \| \mathbf{r}'(0) \| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]Finally, compute curvature:\[ \kappa = \frac{\sqrt{2}}{(\sqrt{3})^3} = \frac{\sqrt{2}}{3\sqrt{3}} \]
06
Calculate radius of curvature
The radius of curvature \( R \) is the reciprocal of curvature:\[ R = \frac{1}{\kappa} = \frac{3\sqrt{3}}{\sqrt{2}} \]
07
Simplify result
The radius of curvature simplifies to:\[ R = \frac{3\sqrt{6}}{2} \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a way to represent a curve or a set of points by expressing the coordinates as functions of a parameter, typically denoted as \( t \) in mathematics. When we parameterize a curve, we can move smoothly along it by varying the parameter. This is particularly useful for representing curves in a 3D space, where each of the coordinates \( (x, y, z) \) is expressed as a distinct function of \( t \).
In the given exercise, the parametric equations are:
In the given exercise, the parametric equations are:
- \( x = e^{t} \cos t \)
- \( y = e^{t} \sin t \)
- \( z = e^{t} \)
Curvature Formula
Curvature gives us a sense of how a curve bends or twists in space. In a mathematical sense, it's a measure of the rate at which the direction of the curve changes. In the world of 3D space curves, curvature helps us understand the local shape of a curve at any given point.
In the original exercise, the curvature \( \kappa \) is calculated using the formula:\[\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}\]This formula involves:
In the original exercise, the curvature \( \kappa \) is calculated using the formula:\[\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}\]This formula involves:
- The cross product of the first derivatives \( \mathbf{r}'(t) \) and second derivatives \( \mathbf{r}''(t) \)
- The magnitude of the derivative vector \( \mathbf{r}'(t) \)
Radius of Curvature
The radius of curvature is a concept that provides a physical sense to the mathematical idea of curvature. It is essentially the radius of the osculating circle, the circle that closely "kisses" or matches the curve at a certain point. This makes it useful in understanding the 'tightness' of the curve's bend at that location.
In mathematical terms, the radius of curvature \( R \) is the reciprocal of curvature:\[R = \frac{1}{\kappa}\]Conceptually, the radius of curvature tells us how large or small a circle would need to be to best approximate the curve at that point. If a curve has a large radius of curvature, it means the curve is relatively flat at that point. Conversely, a small radius indicates a sharper bend. In the exercise, once the curvature was determined, calculating the radius of curvature involved this straightforward reciprocal relationship, further simplified to obtain the final expression for \( R \).
In mathematical terms, the radius of curvature \( R \) is the reciprocal of curvature:\[R = \frac{1}{\kappa}\]Conceptually, the radius of curvature tells us how large or small a circle would need to be to best approximate the curve at that point. If a curve has a large radius of curvature, it means the curve is relatively flat at that point. Conversely, a small radius indicates a sharper bend. In the exercise, once the curvature was determined, calculating the radius of curvature involved this straightforward reciprocal relationship, further simplified to obtain the final expression for \( R \).
