Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the curvature of a straight line?

Short Answer

Expert verified
Answer: The curvature of a straight line is zero, as there is no deviation from being a straight line.

Step by step solution

01

Understand Curvature

Curvature is a measure of how much a curve deviates from being a straight line. A curve with higher curvature changes its direction more rapidly, while one with lower curvature changes its direction more slowly.
02

Determine Curvature of a Curve

The curvature of a curve can be found using the following formula, given a function y=f(x): $$k(x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{\frac{3}{2}}}$$ Where: - \(k(x)\) is the curvature at any point x - f'(x) is the first derivative of the function, which represents the slope at point x - f''(x) is the second derivative of the function, which represents the rate of change of the slope at point x
03

Determine Function of a Straight Line

A straight line can be represented by a linear function of the form \(y = mx + b\), where m is the slope of the line and b is the y-intercept.
04

Calculate the Derivatives

For a straight line with the function \(y = mx + b\), calculate the first and second derivatives: - First Derivative: \(f'(x) = m\) - Second Derivative: \(f''(x) = 0\)
05

Calculate the Curvature of a Straight Line

Plug the derivatives of the straight line into the curvature formula: \begin{align*} k(x) &= \frac{|0|}{(1 + (m)^2)^{\frac{3}{2}}} \\ &= 0 \end{align*}
06

Conclusion

The curvature of a straight line is zero, as there is no deviation from being a straight line.

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 Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=e^{3 t} \mathbf{i}+\frac{1}{1+t^{2}} \mathbf{j}-\frac{1}{\sqrt{2 t}} \mathbf{k}$$

Evaluate the following definite integrals. $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right) \mathbf{j}\right) d t$$

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. What conditions on \(\mathbf{u}\) and \(\mathbf{v}\) lead to equality in the CauchySchwarz Inequality?

Let \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\). a. Assume that \(\lim \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle,\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0 .\) Prove that \(\lim _{t \rightarrow a} f(t)=L_{1}, \quad \lim _{t \rightarrow a} g(t)=L_{2}, \quad\) and \(\quad \lim _{t \rightarrow a} h(t)=L_{3}\). b. Assume that \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2},\) and \(\lim _{t \rightarrow a} h(t)=L_{3} .\) Prove that \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0\).

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Geometric-arithmetic mean Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free