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Chapter 1: Problem 3

A parametrized curve \(\alpha(t)\) has the property that its second derivative \(\alpha^{\prime \prime}(t)\) is identically zero. What can be said about \(\alpha\) ?

Short Answer

Expert verified
The parametrized curve \( \alpha(t) \) is a straight line.

Step by step solution

01

Identify the given information

The curve \(\text{\( \alpha(t) \) } \) is given, and it is stated that its second derivative is zero: \( \alpha''(t) = 0 \).
02

Integrate the second derivative

Since \( \alpha''(t) = 0 \), integrate once with respect to \( t \) to obtain the first derivative: \( \alpha'(t) = C_1 \), where \( C_1 \) is a constant vector.
03

Integrate the first derivative

Integrate \( \alpha'(t) = C_1 \) with respect to \( t \) to get the original function: \( \alpha(t) = C_1 t + C_2 \), where \( C_2 \) is another constant vector.
04

Interpret the result

The function \( \alpha(t) = C_1 t + C_2 \) represents a straight line in the vector form, where \( C_1 \) is the direction vector and \( C_2 \) is the initial point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Geometry
Differential geometry is a field of mathematics that uses techniques of calculus and algebra to study problems in geometry. It focuses on curves, surfaces, and more complex structures. One key concept in differential geometry is the parametrized curve, which is a curve described by a function that depends on one or more parameters. For example, the curve \( \alpha(t) \) depends on the parameter \( t \). In our exercise, we are given a parametrized curve \( \alpha(t) \) and learn about its properties by examining its derivatives.
Second Derivative
The second derivative of a function, often denoted as \( \alpha''(t) \), tells us about the curvature and acceleration of a curve. If the second derivative is zero, as given in our exercise, it means the curve is not accelerating; instead, it moves at a constant velocity. Thus, the curve is a straight line. This result is derived from calculus: integrating the second derivative gives us the first derivative \( \alpha'(t) = C_1 \), a constant vector. Integrating this first derivative, in turn, gives us the original function \( \alpha(t) = C_1 t + C_2 \), representing the equation of a straight line.
Constant Vector
A constant vector remains the same regardless of changes in parameters. In the context of our exercise, the constant vectors \( C_1 \) and \( C_2 \) have specific roles. \( C_1 \) represents the direction in which the straight line moves. It indicates that no matter how \( t \) changes, the direction of \( \alpha(t) \) remains the same. The vector \( C_2 \) represents the starting point or the offset of our straight line. Together, these constants fully describe the line in vector form.
Straight Line
Understanding straight lines is crucial because they are the simplest form of curves in differential geometry. A straight line can be described in vector form as \( \alpha(t) = C_1 t + C_2 \), where \( C_1 \) is the direction vector and \( C_2 \) is the initial point. This equation shows that each point on the line can be computed by scaling the direction vector and adding it to the initial point. Because the second derivative is zero in our problem, we confirm that \( \alpha(t) \) is indeed a straight line.

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Most popular questions from this chapter

If a closed plane curve \(C\) is contained inside a disk of radius \(r\), prove that there exists a point \(p \in C\) such that the curvature \(k\) of \(C\) at \(p\) satisfies \(|k| \geq 1 / r .\)

The natural orientation of \(R^{2}\) makes it possible to associate a sign to the area \(A\) of a parallelogram generated by two linearly independent vectors \(u, v \in R^{2}\). To do this, let \(\left\\{e_{i}\right\\}, i=1,2\), be the natural ordered basis of \(R^{2}\), and write \(u=u_{1} e_{1}+u_{2} e_{2}, v=v_{1} e_{1}+v_{2} e_{2}\). Observe the matrix relation $$ \left(\begin{array}{cc} u \cdot u & u \cdot v \\ v \cdot u & v \cdot v \end{array}\right)=\left(\begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}\right)\left(\begin{array}{cc} u_{1} & v_{1} \\ u_{2} & v_{2} \end{array}\right) $$ and conclude that $$ A^{2}=\left|\begin{array}{ll} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}\right|^{2} $$ Since the last determinant has the same sign as the basis \(\\{u, v\\}\), we can say that \(A\) is positive or negative according to whether the orientation of \(\\{u, v\\}\) is positive or negative. This is called the oriented area in \(R^{2}\).

Given two planes \(a_{i} x+b_{i} y+c_{i} z+d_{i}=0, i=1,2\), prove that a necessary and sufficient condition for them to be parallel is $$ \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}, $$ where the convention is made that if a denominator is zero, the corresponding numerator is also zero (we say that two planes are parallel if they either coincide or do not intersect).

Let \(\alpha: I \rightarrow R^{3}\) be a regular parametrized curve (not necessarily by \(\operatorname{arc}\) length and let \(\beta: J \rightarrow R^{3}\) be a reparametrization of \(\alpha(I)\) by the arc length \(s=s(t)\), measured from \(t_{0} \in I\) (see Remark 2). Let \(t=t(s)\) be the inverse function of \(s\) and set \(d \alpha / d t=\alpha^{\prime}, d^{2} \alpha / d t^{2}=\alpha^{\prime \prime}\), etc. Prove that a. \(d t / d s=1 /\left|\alpha^{\prime}\right|, d^{2} t / d s^{2}=-\left(\alpha^{\prime} \cdot \alpha^{\prime \prime} /\left|\alpha^{\prime}\right|^{4}\right)\). b. The curvature of \(\alpha\) at \(t \in I\) is $$ k(t)=\frac{\left|\alpha^{\prime} \wedge \alpha^{\prime \prime}\right|}{\left|\alpha^{\prime}\right|^{3}} . $$ c. The torsion of \(\alpha\) at \(t \in I\) is $$ \tau(t)=-\frac{\left(\alpha^{\prime} \wedge \alpha^{\prime \prime}\right) \cdot \alpha^{\prime \prime \prime}}{\left|\alpha^{\prime} \wedge \alpha^{\prime \prime}\right|^{2}} $$ d. If \(\alpha: I \rightarrow R^{2}\) is a plane curve \(\alpha(t)=(x(t), y(t))\), the signed curvature (see Remark 1) of \(\alpha\) at \(t\) is $$ k(t)=\frac{x^{\prime} y^{\prime \prime}-x^{\prime \prime} y^{\prime}}{\left(\left(x^{\prime}\right)^{2}+\left(y^{\prime}\right)^{2}\right)^{3 / 2}} . $$

In general, a curve \(\alpha\) is called a helix if the tangent lines of \(\alpha\) make a constant angle with a fixed direction. Assume that \(\tau(s) \neq 0, s \in I\), and prove that: * a. \(\alpha\) is a helix if and only if \(k / \tau=\) const. *b. \(\alpha\) is a helix if and only if the lines containing \(n(s)\) and passing through \(\alpha(s)\) are parallel to a fixed plane. *c. \(\alpha\) is a helix if and only if the lines containing \(b(s)\) and passing through \(\alpha(s)\) make a constant angle with a fixed direction. d. The curve $$ \alpha(s)=\left(\frac{a}{c} \int \sin \theta(s) d s, \frac{a}{c} \int \cos \theta(s) d s, \frac{b}{c} s\right), $$ where \(c^{2}=a^{2}+b^{2}\), is a helix, and that \(k / \tau=a / b\).
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