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Chapter 12: Problem 1

What is the curvature of a straight line?

Short Answer

Expert verified
Answer: The curvature of a straight line is 0, as its tangent direction remains constant along its entire length and there is no deviation from its linear slope.

Step by step solution

01

Define the Straight Line Equation

Consider a straight line equation in two-dimensional space, given by y = mx + b, where m is the slope and b is the y-intercept.
02

Calculate the Tangent Vector

The tangent vector at any point on the curve is given by the first derivative of the curve equation with respect to the parameter. Since the curve is a straight line, its tangent direction remains constant along its entire length. The derivative is given by dy/dx = m, which is a constant.
03

Calculate the Curvature

For a curve defined by y = f(x), the curvature can be calculated using the following formula: Curvature (K) = \(\frac{(1 + (\frac{dy}{dx})^2)^\frac{3}{2}}{|\frac{d^2y}{dx^2}|}\) For our straight line, the first derivative is \(\frac{dy}{dx} = m\); since it is a straight line, the second derivative (change in the tangent direction) is equal to 0, i.e., \(\frac{d^2y}{dx^2} = 0\). We can plug these values into the formula for curvature: K = \(\frac{(1 + m^2)^\frac{3}{2}}{|0|}\)
04

Determine the Curvature

Since the denominator of the curvature formula is 0, the entire curvature expression is undefined. However, it can be inferred from the intuitive meaning of curvature that the curvature of a straight line must be zero because its tangent direction never changes. Therefore, the curvature of a straight line can be considered as 0.

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Most popular questions from this chapter

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. 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 \(b \geq 0\).

Evaluate the following limits. $$\lim _{t \rightarrow \ln 2}\left(2 e^{t} \mathbf{i}+6 e^{-t} \mathbf{j}-4 e^{-2 t} \mathbf{k}\right)$$

Show that the two-dimensional trajectory $$x(t)=u_{0} t+x_{0}\( and \)y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0},\( for \)0 \leq t \leq T$$ of an object moving in a gravitational field is a segment of a parabola for some value of \(T>0 .\) Find \(T\) such that \(y(T)=0\)

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) 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 Cauchy-Schwarz Inequality?

Consider the parallelogram with adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\). a. Show that the diagonals of the parallelogram are \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\). b. Prove that the diagonals have the same length if and only if \(\mathbf{u} \cdot \mathbf{v}=0\). c. Show that the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the sides.
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