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Chapter 12: Problem 26

Find the curvature \(K\) of the plane curve at the given value of the parameter. \(\mathbf{r}(t)=t^{2} \mathbf{j}+\mathbf{k}, \quad t=0\)

Short Answer

Expert verified
The curvature of the plane curve at \(t=0\) is undefined

Step by step solution

01

Calculate the derivative of the vector function

The vector function \(\mathbf{r}(t)=t^{2} \mathbf{j}+\mathbf{k}\). Differentiating \(\mathbf{r}(t)\) with respect to \(t\), we get \[\mathbf{v}(t)= \mathbf{r}'(t) = 2t \mathbf{j}\].
02

Calculate the velocity and acceleration at t=0

Evaluating this at \(t=0\), we find \(\mathbf{v}(0) = \mathbf{0}\). Now, calculate the second derivative of \( \mathbf{r}(t) \), i.e., take the derivative of \( \mathbf{v}(t) \). We obtain \[\mathbf{a}(t) = \mathbf{v}'(t) = 2 \mathbf{j}\].At \(t=0\), this gives \(\mathbf{a}(0) = 2\mathbf{j}\).
03

Compute the curvature

Substitute the obtained values into the formula of curvature \(K = |\mathbf{v} \times \mathbf{a}| / |\mathbf{v}|^3 \). Since \(\mathbf{v}(0) = \mathbf{0}\), the magnitude of the cross product \(|\mathbf{v} \times \mathbf{a}|\) equals 0. Therefore, the curvature at \(t=0\) is \(K = 0 / 0\), which is undefined.

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Key Concepts

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

Plane Curve
When dealing with a plane curve, imagine a smooth path on a two-dimensional surface. Just like drawing a line on paper. These curves can be represented mathematically using vector functions, providing a way to study properties like direction, length, and shape. In our given exercise, we have a plane curve defined by the vector function \( \mathbf{r}(t) = t^{2} \mathbf{j} + \mathbf{k} \). Here, the curve lies in the \( j-k \) plane. Break it down and you'll see:
  • \( t^2 \mathbf{j} \) means as \( t \) changes, the curve moves through points on the \( j \)-axis.
  • \( \mathbf{k} \) signifies movement along the \( k \)-axis.
Moving on to the next critical concept, let us talk about vector functions.
Vector Function
A vector function is a function that outputs a vector for every input value. Think of vector functions as providing a map with directions in space. In calculus, it is a powerful tool that represents a path or curve in a multi-dimensional space. For our exercise, the vector function is given by \( \mathbf{r}(t) = t^{2} \mathbf{j} + \mathbf{k} \). This means:
  • Each value of \( t \) gives a corresponding point on the plane, characterized by the combination \( (t^2, 1) \) in the \( j-k \) plane.
  • The function outputs a position vector \( (0, t^2, 1) \) in three-dimensional space.
Understanding vector functions is crucial in tackling calculus problems, which leads us to the concept of calculus problem-solving.
Calculus Problem-Solving
Solving calculus problems involves a systematic approach. It often requires understanding derivatives, rates of change, and geometric properties of functions. In the context of the original exercise:
  • We need to find how curves bend or twist, known as curvature, at a specific parameter \( t=0 \).
  • The process involves differentiating vector functions to calculate velocity and acceleration vectors.
  • For curvature \( K \), use the formula \( K = |\mathbf{v} \times \mathbf{a}| / |\mathbf{v}|^3 \).
These steps are essential for breaking down complex calculus exercises into manageable parts. Now, let's explore derivatives, a fundamental tool in this process.
Derivatives in Calculus
Derivatives are foundational elements in calculus, providing a method to understand how functions change. In the context of vector functions, derivatives help identify rates of movement and change direction. For instance, in the original exercise:
  • The first derivative \( \mathbf{v}(t) = \mathbf{r}'(t) = 2t \mathbf{j} \) represents the velocity.
  • The second derivative \( \mathbf{a}(t) = \mathbf{v}'(t) = 2\mathbf{j} \) indicates the acceleration.
These derivatives are computed at different parameter values, like \( t=0 \), to reveal critical insights into the behavior of the curve, such as its curvature. Ultimately, derivatives unlock a deeper understanding of the physical and geometric properties represented by mathematical functions.

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Most popular questions from this chapter

Find the curvature and radius of curvature of the plane curve at the given value of \(x\). $$ y=3 x-2, \quad x=a $$

Sketch the graph of the plane curve given by the vector-valued function, and, at the point on the curve determined by \(\mathbf{r}\left(t_{0}\right)\), sketch the vectors \(\mathbf{T}\) and \(\mathbf{N}\). Note that \(\mathbf{N}\) points toward the concave side of the curve. $$ \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j} \quad t_{0}=\frac{\pi}{4} $$

Consider the vector-valued function \(\mathbf{r}(t)=\left(e^{t} \sin t\right) \mathbf{i}+\left(e^{t} \cos t\right) \mathbf{j}\) Show that \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime \prime}(t)\) are always perpendicular to each other.

True or False? Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Prove that the vector \(\mathbf{T}^{\prime}(t)\) is \(\mathbf{0}\) for an object moving in a straight line.

Find the curvature \(K\) of the curve, where \(s\) is the arc length parameter. $$ \mathbf{r}(s)=\left(1+\frac{\sqrt{2}}{2} s\right) \mathbf{i}+\left(1-\frac{\sqrt{2}}{2} s\right) \mathbf{j} $$
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