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Magazine Chapter 12: Problem 10
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 corresponding to decreasing \(t\) values.
Short Answer
Expert verified
The point is (0, 5, -24\pi).
Step by step solution
01
Understanding the Curve
The given curve is a parametric curve represented by \( \mathbf{r}(t) = (5 \sin t) \mathbf{i} + (5 \cos t) \mathbf{j} + 12t \mathbf{k} \). This is a helix with radius 5 in the \(xy\)-plane and height increasing linearly with \(t\) in the \(z\)-direction.
02
Calculating the Initial Point
Substitute \(t = 0\) into \(\mathbf{r}(t)\) to find the initial point: \[ \mathbf{r}(0) = 5 \sin(0) \mathbf{i} + 5 \cos(0) \mathbf{j} + 12 \cdot 0 \mathbf{k} = 0 \mathbf{i} + 5 \mathbf{j} + 0 \mathbf{k} = (0, 5, 0) \].This confirms that (0, 5, 0) is at \(t=0\).
03
Calculating the Arc Length
The arc length \(s\) from \(t=0\) to an arbitrary \(t\) is \[ s = \int_{0}^{t} \left\| \mathbf{r}'(u) \right\| du \], where \( \mathbf{r}'(t) \) is the derivative of \( \mathbf{r}(t) \).
04
Differentiating the Position Vector
Differentiate \( \mathbf{r}(t) \):\[ \mathbf{r}'(t) = \frac{d}{dt}(5 \sin t \mathbf{i} + 5 \cos t \mathbf{j} + 12t \mathbf{k}) = 5 \cos t \mathbf{i} - 5 \sin t \mathbf{j} + 12 \mathbf{k} \].
05
Finding Magnitude of \( \mathbf{r}'(t) \)
Calculate the magnitude of \(\mathbf{r}'(t)\):\[ \left\| \mathbf{r}'(t) \right\| = \sqrt{(5 \cos t)^2 + (-5 \sin t)^2 + 12^2} = \sqrt{25 \cos^2 t + 25 \sin^2 t + 144} \].Use \( \cos^2 t + \sin^2 t = 1 \):\[ \left\| \mathbf{r}'(t) \right\| = \sqrt{169} = 13 \].
06
Establishing the Arc Length Equation
The arc length formula simplifies to:\[ s = \int_{0}^{t} 13 \, du = 13t \]. We need to find the \(t\) such that \(s = 26\pi\), so set:\[ 13t = 26\pi \].
07
Solving for \(t\)
Solve for \(t\): \[ t = \frac{26\pi}{13} = 2\pi \].Since we need the point in the direction of decreasing \(t\), use \(t = -2\pi\).
08
Finding the Point on the Curve
Substitute \(t = -2\pi\) into \(\mathbf{r}(t)\) to find the point:\[ \mathbf{r}(-2\pi) = 5 \sin(-2\pi) \mathbf{i} + 5 \cos(-2\pi) \mathbf{j} + 12(-2\pi) \mathbf{k} \]Which simplifies to:\[ (0 \cdot \mathbf{i} + 5 \cdot \mathbf{j} - 24\pi \cdot \mathbf{k}) = (0, 5, -24\pi) \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Curves
A parametric curve is a way of defining a curve by using one or more parameters, typically denoted as `t`. Each point on the curve can be accessed by substituting `t` with specific values. In this scenario, the parametric curve is given as \( \mathbf{r}(t) = (5 \sin t) \mathbf{i} + (5 \cos t) \mathbf{j} + 12t \mathbf{k} \). This expression tells us:
- \( 5 \sin t \) represents the x-component of the curve.
- \( 5 \cos t \) represents the y-component.
- \( 12t \) shows how the height of the curve changes linearly in the z-direction.
Helix
A helix is a type of curve which takes the form of a spiraling shape, often resembling a spring or coil. It is characterized by its circular path in two dimensions while simultaneously maintaining a linear progression in the third dimension.
In our example, the curve \( \mathbf{r}(t) = (5 \sin t) \mathbf{i} + (5 \cos t) \mathbf{j} + 12t \mathbf{k} \) describes a helix:
In our example, the curve \( \mathbf{r}(t) = (5 \sin t) \mathbf{i} + (5 \cos t) \mathbf{j} + 12t \mathbf{k} \) describes a helix:
- The components \( 5 \sin t \) and \( 5 \cos t \) define its circular shape in the xy-plane with a radius of 5.
- The \( 12t \) component adds a vertical element to the curve, causing it to ascend as the parameter `t` increases.
Differentiation
Differentiation is the mathematical process of finding the derivative of a function. It reveals how a function changes at any point 鈥� essentially measuring its rate of change. For the parametric curve, we use differentiation to find the derivative \( \mathbf{r}'(t) \):
This derivative informs us how the curve is changing in the three-dimensional space as `t` varies. It's crucial to find arc length and understand the motion of an object along parametric curves.
- Differentiate \( 5 \sin t \rightarrow 5 \cos t \).
- Differentiate \( 5 \cos t \rightarrow -5 \sin t \).
- Differentiate \( 12t \rightarrow 12 \).
This derivative informs us how the curve is changing in the three-dimensional space as `t` varies. It's crucial to find arc length and understand the motion of an object along parametric curves.
Magnitude of a Vector
The magnitude of a vector is a measure of its length in space, akin to the "distance" the vector covers from one point to another but starting from the origin. To find the magnitude of a derivative vector, such as \( \mathbf{r}'(t) = 5 \cos t \mathbf{i} - 5 \sin t \mathbf{j} + 12 \mathbf{k} \), you calculate:
Magnitude helps us compute the arc length of a curve over an interval, which is essential in determining the actual physical distance traveled along the curve. It provides a metric to gauge the length without dissecting each component separately.
- \( 5 \cos t \) squared plus \( -5 \sin t \) squared add up to 25 using the Pythagorean identity \( \cos^2 t + \sin^2 t = 1 \).
- The component 12 is squared to 144.
Magnitude helps us compute the arc length of a curve over an interval, which is essential in determining the actual physical distance traveled along the curve. It provides a metric to gauge the length without dissecting each component separately.
