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Chapter 12: Problem 28

Find the curvature \(K\) of the plane curve at the given value of the parameter. \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}, \quad t=1\)

Short Answer

Expert verified
The curvature \(K\) of the plane curve at \(t = 1\) is \(K = 2/125\).

Step by step solution

01

Compute the first derivative

To begin with, calculate the derivative of \(\mathbf{r}(t)\), which is \(\mathbf{r}'(t)\). In this case, \(\mathbf{r}'(t)= \mathbf{i} + 2t \mathbf{j}\).
02

Compute the second derivative

Once you have the first derivative, you'll need to find the second derivative of \(\mathbf{r}(t)\), which is \(\mathbf{r}''(t)\). And \(\mathbf{r}''(t)= 0 \mathbf{i} + 2 \mathbf{j}\).
03

Calculate the magnitudes of the first and second derivatives

Next, calculate the magnitudes of \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\). For \(t=1\), the magnitudes will be \(|\mathbf{r}'(t)|=\sqrt{(1)^2 +(2)^2}= \sqrt{5}\) and \(|\mathbf{r}''(t)|=\sqrt{(0)^2 +(2)^2}= 2\).
04

Find the cross product of the first and second derivatives

The cross product of \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\) gives a vector orthogonal to both vectors. Calculating it for \(t=1\), we get \(\mathbf{r}'(t) \times \mathbf{r}''(t) = 2 \mathbf{k}\). Then, find the magnitude of the cross product \(|(\mathbf{r}'(t) \times \mathbf{r}''(t))|=2\).
05

Apply the curvature formula

Finally, the curvature \(K\) is given by \(K = {{|(\mathbf{r}'(t) \times \mathbf{r}''(t))|} \over {|\mathbf{r}'(t)|^3}} = {{2} \over {(\sqrt{5})^3}} = {{2} \over {125}}\).

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Key Concepts

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

Plane Curve
A plane curve is a continuously curved line that lies in a single plane in space. It can be described mathematically by a vector-valued function, like \[ \mathbf{r}(t) = t \mathbf{i} + t^{2} \mathbf{j} \]in our given problem.
This function provides coordinates for each point on the curve in terms of a parameter, typically represented as \( t \).
  • Each \( t \) value gives a unique point \((x,y)\) on the curve.
  • The curve's path is traced out as \( t \) varies.
Plane curves are fundamental in understanding various geometric properties like curvature, which is a measure of how sharply the curve bends at a particular point. Understanding the behavior of a curve within a plane provides insights into more complex structures in vector calculus.
Derivatives
Derivatives are crucial in determining how a function behaves at any given point. For plane curves, derivatives help find the rate of change of the curve's direction and steepness.
In our case, the first derivative of the curve, \( \mathbf{r}'(t) \), represents the tangent vector at any point, reflecting the curve's direction. By differentiating \[ \mathbf{r}(t) = t \mathbf{i} + t^{2} \mathbf{j}\]yields \[\mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j}\]
  • \( \mathbf{r}'(t) \) is the rate at which the curve's positions change as \( t \) changes.
  • The length of this vector shows the speed of traversal along the curve.
To further analyze the curve's bending, we find the second derivative, \( \mathbf{r}''(t) \), which is \[0 \mathbf{i} + 2 \mathbf{j}\].
This provides information on how the tangent vector itself is changing direction.
Parameter
In mathematics, a parameter is a variable that is part of a mathematical expression or equation and can take on different values. For parametric equations of curves like \[ \mathbf{r}(t) = t \mathbf{i} + t^{2} \mathbf{j}\],the parameter \( t \) determines a specific point on the curve.
  • Changing \( t \) alters the position along the curve.
  • For \( t = 1 \), the point on the curve is given by substituting the parameter into the curve equation, providing insight into the curve's geometry at that instance.
Parameters enable precise control over how curves are traced and are crucial for calculating derived properties like the tangent and normal vectors and curvatures at any point.
They serve as tools for converting abstract mathematical concepts into tangible points or features on a graph.
Vector Calculus
Vector calculus combines calculus and vector algebra to explore complex three-dimensional problems involving vectors. In the context of this curvature problem, vector calculus helps us assess how curves behave in a plane.
Using vector calculus:
  • First, calculate the first derivative to determine the direction of the curve.
  • Next, find the second derivative to assess changes in the curve's direction.
  • Cross product calculations, as seen with \( \mathbf{r}'(t) \times \mathbf{r}''(t) \), help determine curvature by finding perpendicular vectors.
Curvature measures how a curve bends and is defined by: \[K = \frac{|(\mathbf{r}'(t) \times \mathbf{r}''(t))|}{|\mathbf{r}'(t)|^3}\]
Such formulas leverage vector operations to study shapes and motions, providing a deeper understanding of geometric properties. These concepts are essential not just in theoretical mathematics but also in physics, engineering, and computer graphics.

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Most popular questions from this chapter

Air Traffic Control Because of a storm, ground controllers instruct the pilot of a plane flying at an altitude of 4 miles to make a \(90^{\circ}\) turn and climb to an altitude of \(4.2\) miles. The model for the path of the plane during this maneuver is \(\mathbf{r}(t)=\langle 10 \cos 10 \pi t, 10 \sin 10 \pi t, 4+4 t\rangle, \quad 0 \leq t \leq \frac{1}{20}\) where \(t\) is the time in hours and \(\mathbf{r}\) is the distance in miles. (a) Determine the speed of the plane. (b) Use a computer algebra system to calculate \(a_{\mathbf{T}}\) and \(a_{\mathrm{N}}\) Why is one of these equal to 0 ?

Find all points on the graph of the function such that the curvature is zero. $$ y=1-x^{3} $$

You are asked to verify Kepler's Laws of Planetary Motion. For these exercises, assume that each planet moves in an orbit given by the vector- valued function \(\mathbf{r}\). Let \(\boldsymbol{r}=\|\mathbf{r}\|\), let \(\boldsymbol{G}\) represent the universal gravitational constant, let \(M\) represent the mass of the sun, and let \(m\) represent the mass of the planet. Prove Kepler's First Law: Each planet moves in an elliptical orbit with the sun as a focus.

Consider the helix represented by the vectorvalued function \(\mathbf{r}(t)=\langle 2 \cos t, 2 \sin t, t\rangle\) (a) Write the length of the arc \(s\) on the helix as a function of \(t\) by evaluating the integral $$ s=\int_{0}^{t} \sqrt{\left[x^{\prime}(u)\right]^{2}+\left[y^{\prime}(u)\right]^{2}+\left[z^{\prime}(u)\right]^{2}} d u $$ (b) Solve for \(t\) in the relationship derived in part (a), and substitute the result into the original set of parametric equations. This yields a parametrization of the curve in terms of the arc length parameter \(s\). (c) Find the coordinates of the point on the helix for arc lengths \(s=\sqrt{5}\) and \(s=4\) (d) Verify that \(\left\|\mathbf{r}^{\prime}(s)\right\|=1\)

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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