Problem 3
Question
In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=t \mathbf{i}+(2 / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq 8 $$
Step-by-Step Solution
Verified Answer
The unit tangent vector is \( \frac{1}{\sqrt{1+t}} \mathbf{i} + \frac{t^{1/2}}{\sqrt{1+t}} \mathbf{k} \), and the arc length is \( \frac{52}{3} \).
1Step 1: Find the velocity vector
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Thus, we differentiate \( \mathbf{r}(t) = t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k} \) with respect to \( t \). We have: \[ \mathbf{v}(t) = \frac{d}{dt} (t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k}) = \mathbf{i} + (t^{1/2} \mathbf{k}) \]
2Step 2: Find the unit tangent vector
The unit tangent vector \( \mathbf{T}(t) \) is given by the velocity vector \( \mathbf{v}(t) \) divided by its magnitude. First, calculate the magnitude of \( \mathbf{v}(t) \): \[ \|\mathbf{v}(t)\| = \sqrt{1^2 + (t^{1/2})^2} = \sqrt{1 + t}. \] Then, find \( \mathbf{T}(t) \): \[ \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} = \frac{1}{\sqrt{1 + t}} \mathbf{i} + \frac{t^{1/2}}{\sqrt{1 + t}} \mathbf{k}. \]
3Step 3: Determine the arc length
The arc length \( s \) for \( t \) from 0 to 8 is given by the integral of the magnitude of the velocity vector: \[ s = \int_{0}^{8} \|\mathbf{v}(t)\| \ dt = \int_{0}^{8} \sqrt{1 + t} \ dt. \] Performing the integration: \[ s = \left[\frac{2}{3}(1 + t)^{3/2}\right]_{0}^{8} = \frac{2}{3} \left((1 + 8)^{3/2} - 1^{3/2}\right) = \frac{2}{3} \left(9^{3/2} - 1\right) = \frac{2}{3} (27 - 1) = \frac{2}{3} \times 26 = \frac{52}{3}. \]
Key Concepts
Velocity VectorArc Length of CurveMagnitude of Vector
Velocity Vector
A velocity vector reflects how quickly a position is changing over time, and indicates both speed and direction in which an object is moving. To find the velocity vector of a curve described by a position vector function, we need to compute its time derivative.
In mathematical terms, if we have a position vector function \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) can be found by differentiating \( \mathbf{r}(t) \) with respect to time \( t \).
Let's take \( \mathbf{r}(t) = t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k} \) from our exercise. By differentiating each component of \( \mathbf{r}(t) \), we get \( \mathbf{v}(t) = \frac{d}{dt} (t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k}) \).
In mathematical terms, if we have a position vector function \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) can be found by differentiating \( \mathbf{r}(t) \) with respect to time \( t \).
Let's take \( \mathbf{r}(t) = t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k} \) from our exercise. By differentiating each component of \( \mathbf{r}(t) \), we get \( \mathbf{v}(t) = \frac{d}{dt} (t \mathbf{i} + \frac{2}{3} t^{3/2} \mathbf{k}) \).
- Derive the \( t \mathbf{i} \) term to get \( \mathbf{i} \).
- Derive the \( \frac{2}{3} t^{3/2} \mathbf{k} \) term to yield \( t^{1/2} \mathbf{k} \).
Arc Length of Curve
The arc length of a curve can be used to measure the distance traveled along a path from one point to another. It's like finding the length of a piece of string that has been placed along the curve.
To calculate this, we integrate the magnitude of the velocity vector over the interval of interest. In this case, between \( t = 0 \) and \( t = 8 \).
More formally, the arc length \( s \) is given by the integral:
\[ s = \int_{t_0}^{t_1} \| \mathbf{v}(t) \| \, dt \]
From the exercise, we know the magnitude is \( \| \mathbf{v}(t) \| = \sqrt{1 + t} \). We now perform the integration:
\[ s = \int_{0}^{8} \sqrt{1 + t} \, dt \]
Performing the integration gives us:
\[ \frac{2}{3}(1 + t)^{3/2} \bigg|_{0}^{8} = \frac{2}{3}(27 - 1) = \frac{52}{3} \]
This result tells us the object traveled this distance along the curve between these two points in time.
To calculate this, we integrate the magnitude of the velocity vector over the interval of interest. In this case, between \( t = 0 \) and \( t = 8 \).
More formally, the arc length \( s \) is given by the integral:
\[ s = \int_{t_0}^{t_1} \| \mathbf{v}(t) \| \, dt \]
From the exercise, we know the magnitude is \( \| \mathbf{v}(t) \| = \sqrt{1 + t} \). We now perform the integration:
\[ s = \int_{0}^{8} \sqrt{1 + t} \, dt \]
Performing the integration gives us:
\[ \frac{2}{3}(1 + t)^{3/2} \bigg|_{0}^{8} = \frac{2}{3}(27 - 1) = \frac{52}{3} \]
This result tells us the object traveled this distance along the curve between these two points in time.
Magnitude of Vector
Magnitude measures the size or length of a vector, and is essential in understanding how far a vector travels rather than simply its direction. For a vector \( \mathbf{v} \), its magnitude is calculated using the Pythagorean theorem, applied to its components.
If \( \mathbf{v}(t) = a(t) \mathbf{i} + b(t) \mathbf{k} \), the magnitude \( \| \mathbf{v}(t) \| \) is given by:
\[\| \mathbf{v}(t) \| = \sqrt{a(t)^2 + b(t)^2}\]
In our solution, the vector \( \mathbf{v}(t) = \mathbf{i} + t^{1/2} \mathbf{k} \) yields a magnitude of:
\[\| \mathbf{v}(t) \| = \sqrt{1^2 + (t^{1/2})^2} = \sqrt{1 + t}\]
Here, every component squared and added forms the basis of the square root. This concept is critical as it aids in determining both the arc length of the curve and the unit tangent vector. Knowing the magnitude of a vector lets us understand its overall intensity in the space it occupies.
If \( \mathbf{v}(t) = a(t) \mathbf{i} + b(t) \mathbf{k} \), the magnitude \( \| \mathbf{v}(t) \| \) is given by:
\[\| \mathbf{v}(t) \| = \sqrt{a(t)^2 + b(t)^2}\]
In our solution, the vector \( \mathbf{v}(t) = \mathbf{i} + t^{1/2} \mathbf{k} \) yields a magnitude of:
\[\| \mathbf{v}(t) \| = \sqrt{1^2 + (t^{1/2})^2} = \sqrt{1 + t}\]
Here, every component squared and added forms the basis of the square root. This concept is critical as it aids in determining both the arc length of the curve and the unit tangent vector. Knowing the magnitude of a vector lets us understand its overall intensity in the space it occupies.
Other exercises in this chapter
Problem 3
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