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Chapter 12: Problem 9

Find the point on the curve $$\mathbf{r}(t)=(5 \sin t) \mathbf{i}+(5 \cos t) \mathbf{j}+12 t \mathbf{k}$$ at a distance \(26 \pi\) units along the curve from the point (0,5,0) in the direction of increasing arc length.

Short Answer

Expert verified
The point is (0, 5, 24蟺).

Step by step solution

01

Parameterize the Curve

The curve is given as \( \mathbf{r}(t) = (5 \sin t) \mathbf{i} + (5 \cos t) \mathbf{j} + 12t \mathbf{k} \). This describes a helix with a circle in the \(xy\)-plane and linear displacement along \(z\)-axis according to time \(t\).
02

Calculate the Arc Length Integral

The arc length \( s(t) \) from \( t = 0 \) to some point \( t \) is given by \( s(t) = \int_0^t \| \frac{d\mathbf{r}}{dt} \| dt \). First compute the derivative \( \frac{d\mathbf{r}}{dt} = (5 \cos t) \mathbf{i} - (5 \sin t) \mathbf{j} + 12 \mathbf{k} \).
03

Find the Magnitude of the Derivative

The magnitude of the derivative is \( \| \frac{d\mathbf{r}}{dt} \| = \sqrt{(5\cos t)^2 + (-5\sin t)^2 + 12^2} = \sqrt{25 + 144} = 13 \).
04

Evaluate the Arc Length for General t

Substitute the magnitude into the integral: \( s(t) = \int_0^t 13 \ dt = 13t \).
05

Set Arc Length Equal to Given Distance

Solve for \( t \) using \( s(t) = 26\pi \). We have \( 13t = 26\pi \) which simplifies to \( t = 2\pi \).
06

Find the Corresponding Point on the Curve

Substitute \( t = 2\pi \) into \( \mathbf{r}(t) \): \( \mathbf{r}(2\pi) = (5 \sin 2\pi) \mathbf{i} + (5 \cos 2\pi) \mathbf{j} + 12(2\pi) \mathbf{k} = (0) \mathbf{i} + (5) \mathbf{j} + 24\pi \mathbf{k} \). Thus, the point is \( (0, 5, 24\pi) \).

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Key Concepts

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

Helix
A helix is a fascinating 3D curve resembling a spring or spiral staircase. It wraps around a cylinder or cone and moves upward with a consistent twist. In this exercise, the curve \( \mathbf{r}(t) = (5 \sin t) \mathbf{i} + (5 \cos t) \mathbf{j} + 12t \mathbf{k} \) describes a helix. Here is how we can understand this:
  • The terms \( 5 \sin t \mathbf{i} + 5 \cos t \mathbf{j} \) form a circular motion in the \( xy \)-plane. This describes a perfect circle with a radius of 5.
  • The term \( 12t \mathbf{k} \) indicates that as \( t \) increases, the helix moves upward along the \( z \)-axis. Here, 12 is the rate of rise per unit circle traveled.
Thus, the combination of circular motion and linear rise creates the helical shape. The \( z \)-component grows linearly as \( t \) increases, causing the helix to spiral upwards.
Arc Length
Arc Length is the distance along a curve from one point to another. To find it for a given curve, you typically integrate the magnitude of the curve's derivative over the interval.
  • For our helix, we start by parameterizing the position using \( \mathbf{r}(t) \).
  • The arc length, \( s(t) \), is calculated from an integral of the form \( s(t) = \int_0^t \| \frac{d\mathbf{r}}{dt} \| dt \).
This exercise requires us to measure \( 26\pi \) units along the curve. Ensuring the correct orientation is paramount; we calculate this length using the magnitude \( \| \frac{d\mathbf{r}}{dt} \| = 13 \), as found in the solution. This gives us a straightforward integration and simplifies learning how to determine distances along curves.
Parameterization
Parameterization involves expressing a curve with respect to one or more variables, typically using a parameter \( t \). It allows complex figures to be understood in simpler terms.
  • In our exercise, \( \mathbf{r}(t) = (5 \sin t) \mathbf{i} + (5 \cos t) \mathbf{j} + 12t \mathbf{k} \) parameterizes the helix.
  • Here, \( t \) acts as the parameter traverse through positions on the curve as it increases.
By parameterizing, geometrical and physical properties of the shape become more accessible. For instance, calculating velocities, positions, and angles becomes systematically simple, and one can utilize calculus to analyze these attributes efficiently.
Derivative
In calculus, the derivative of a function describes how a quantity changes as its input changes. It provides insight into the curve's direction and speed.
  • For our helix, the derivative \( \frac{d\mathbf{r}}{dt} = (5 \cos t) \mathbf{i} - (5 \sin t) \mathbf{j} + 12 \mathbf{k} \) gives the rate of change of position.
  • This derivative tells us the direction the curve moves at any point \( t \).
  • The derivative's magnitude, here computed as 13, reveals the speed of movement along the path.
Understanding the derivative is crucial when studying motion or growth patterns. It allows for determining slopes, calculating arc lengths, and even finding tangential vectors.

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Most popular questions from this chapter

In Exercises \(11-14,\) find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau$$ from Equation ( 3 ). Then find the length of the indicated portion of the curve. $$\mathbf{r}(t)=(4 \cos t) \mathbf{i}+(4 \sin t) \mathbf{j}+3 t \mathbf{k}, \quad 0 \leq t \leq \pi / 2$$

Show that the vector-valued function \(\mathbf{r}(t)=(2 \mathbf{i}+2 \mathbf{j}+\mathbf{k})\) $$+\cos t\left(\frac{1}{\sqrt{2}} \mathbf{i}-\frac{1}{\sqrt{2}} \mathbf{j}\right)+\sin t\left(\frac{1}{\sqrt{3}}\mathbf{i}+\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}\right)$$ describes the motion of a particle moving in the circle of radius 1 centered at the point (2,2,1) and lying in the plane \(x+y-2 z=2\).

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the plane curves in Exercises. $$\mathbf{r}(t)=t \mathbf{i}+(\ln \cos t) \mathbf{j}, \quad-\pi / 2

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{j}+\left(\sin ^{3} t\right) \mathbf{k}, \quad 0 \leq t \leq \pi / 2$$

In Example \(5,\) we found the curvature of the helix \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+b t \mathbf{k}\) \((a, b \geq 0)\) to be \(\kappa=a /\left(a^{2}+b^{2}\right) .\) What is the largest value \(\kappa\) can have for a given value of \(b\) ? Give reasons for your answer.
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