Chapter 12: Problem 10
Arc length calculations Find the length of the following two and three- dimensional curves. $$\mathbf{r}(t)=\langle 3 t-1,4 t+5, t\rangle, \text { for } 0 \leq t \leq 1$$
Short Answer
Expert verified
Answer: The arc length of the given curve is \(L = \sqrt{26}\).
Step by step solution
01
Calculate the derivative of the curve with respect to t
In order to find the arc length, we first need to find the derivative \(\mathbf{r'}(t)\). To do this, we will find the derivative of each component of the curve with respect to \(t\). So,
$$\frac{d}{dt}(3t - 1) = 3, \quad \frac{d}{dt}(4t + 5) = 4, \quad \frac{d}{dt}(t) = 1$$
Hence, the derivative of the curve with respect to \(t\) is \(\mathbf{r'}(t) = \langle 3, 4, 1\rangle\).
02
Calculate the magnitude of the derivative
Now that we have \(\mathbf{r'}(t)\), we need to find its magnitude, \(|\mathbf{r'}(t)|\). We can use the formula for the magnitude of a vector:
$$|\mathbf{r'}(t)| = \sqrt{(3^2) + (4^2) + (1^2)} = \sqrt{26}$$
03
Integrate to find the arc length
Now that we know \(|\mathbf{r'}(t)| = \sqrt{26}\), we can integrate it over the interval for \(t\) from \(0\) to \(1\) to find the arc length:
$$L = \int_{0}^{1} |\mathbf{r'}(t)| dt = \int_{0}^{1} \sqrt{26} dt = \sqrt{26}\int_{0}^{1} dt$$
Integration gives us:
$$L = \sqrt{26}(t\big|_0^1) = \sqrt{26}(1 - 0) = \sqrt{26}$$
Thus, the arc length of the given curve is \(L = \sqrt{26}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Three-Dimensional Curves
Three-dimensional curves are fascinating as they allow us to visualize and understand paths in the space around us. Unlike two-dimensional curves that lie flat on a plane, three-dimensional curves exist in space, having an x, y, and z coordinate.
In three dimensions, a curve can be represented using a vector function, \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), where \( x(t), y(t) \), and \( z(t) \) are functions of a parameter \( t \). This parameter is akin to a time variable that tells us the position of a point moving along the curve at any given moment.
Key things to remember about three-dimensional curves:
In three dimensions, a curve can be represented using a vector function, \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), where \( x(t), y(t) \), and \( z(t) \) are functions of a parameter \( t \). This parameter is akin to a time variable that tells us the position of a point moving along the curve at any given moment.
Key things to remember about three-dimensional curves:
- They provide a way to model the trajectory of objects in space.
- The vector function \( \mathbf{r}(t) \) succinctly captures the curve's path.
- Each component (x, y, z) function details how the curve progresses in each spatial direction with respect to the parameter \( t \).
Vector Calculus
Vector calculus involves the study of vector fields and differential equations that describe the motion and interaction of objects in space using vectors. It is a powerful mathematical tool used to solve problems involving curves and surfaces in three-dimensional space.
Vectors are mathematical objects having both a direction and a magnitude, and are often represented in component form, like \( \mathbf{r} = \langle x, y, z \rangle \). In the context of calculus, we use vectors to calculate things like motion along a curve, areas under surfaces, and flow within a vector field.
In our example of arc length calculation:
Vectors are mathematical objects having both a direction and a magnitude, and are often represented in component form, like \( \mathbf{r} = \langle x, y, z \rangle \). In the context of calculus, we use vectors to calculate things like motion along a curve, areas under surfaces, and flow within a vector field.
In our example of arc length calculation:
- The derivative \( \mathbf{r'}(t) \) represents the velocity of a point moving along the curve.
- The magnitude of this derivative \( |\mathbf{r'}(t)| \) tells us the speed at which this point travels along the curve.
- Integrating this speed over a given interval delivers the total arc length of the curve.
Derivative of a Curve
The derivative of a curve is a crucial concept in calculus as it represents the rate of change or slope of the curve at any point. When dealing with three-dimensional curves, derivatives help describe how each component of the curve changes with respect to a parameter like \( t \).
For a vector function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), its derivative \( \mathbf{r'}(t) = \langle x'(t), y'(t), z'(t) \rangle \) is attained by differentiating each of the component functions with respect to \( t \). This yields the tangent vector to the curve at each point, indicating both direction and speed.
Key takeaways about curve derivatives:
For a vector function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), its derivative \( \mathbf{r'}(t) = \langle x'(t), y'(t), z'(t) \rangle \) is attained by differentiating each of the component functions with respect to \( t \). This yields the tangent vector to the curve at each point, indicating both direction and speed.
Key takeaways about curve derivatives:
- The derivative \( \mathbf{r'}(t) \) is a vector that illustrates the rate of change of the curve.
- This derivative not only shows the direction of travel along the curve but also how fast that movement is.
- Finding the magnitude of this derivative and integrating it helps calculate the arc length of the curve.