Problem 6
Question
In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=6 t^{3} \mathbf{i}-2 t^{3} \mathbf{j}-3 t^{3} \mathbf{k}, \quad 1 \leq t \leq 2 $$
Step-by-Step Solution
Verified Answer
The unit tangent vector is \( \frac{6}{7} \mathbf{i} - \frac{2}{7} \mathbf{j} - \frac{3}{7} \mathbf{k} \); the arc length is 49.
1Step 1: Differentiate the Vector Function
To find the unit tangent vector, we first need the derivative of the given vector function. Let's differentiate \( \mathbf{r}(t) = 6t^3 \mathbf{i} - 2t^3 \mathbf{j} - 3t^3 \mathbf{k} \). The derivative is: \( \mathbf{r}'(t) = 18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k} \).
2Step 2: Compute the Magnitude of the Derivative
Next, compute the magnitude of \( \mathbf{r}'(t) \): \( \| \mathbf{r}'(t) \| = \sqrt{(18t^2)^2 + (-6t^2)^2 + (-9t^2)^2} = \sqrt{324t^4 + 36t^4 + 81t^4} = \sqrt{441t^4} = 21t^2 \).
3Step 3: Find the Unit Tangent Vector
The unit tangent vector \( \mathbf{T}(t) \) is given by dividing \( \mathbf{r}'(t) \) by its magnitude: \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k}}{21t^2} = \frac{6}{7} \mathbf{i} - \frac{2}{7} \mathbf{j} - \frac{3}{7} \mathbf{k} \).
4Step 4: Calculate Arc Length of Curve Portion
To find the arc length from \( t = 1 \) to \( t = 2 \), use the formula \( L = \int_{1}^{2} \| \mathbf{r}'(t) \| \, dt = \int_{1}^{2} 21t^2 \, dt \).
5Step 5: Integrate to Find Length
Compute the integral: \[ \int_{1}^{2} 21t^2 \, dt = 21 \left[ \frac{t^3}{3} \right]_1^2 = 21 \left( \frac{8}{3} - \frac{1}{3} \right) = 21 \times \frac{7}{3} = 49. \] The arc length is 49.
Key Concepts
Unit Tangent VectorArc LengthVector Function DifferentiationMagnitude of a Vector
Unit Tangent Vector
When working with curves in vector calculus, understanding the unit tangent vector is crucial. The unit tangent vector, denoted as \( \mathbf{T}(t) \), provides a way to represent the direction of a curve at any point. It essentially tells us which way the curve is heading without considering the speed or magnitude.
The process of finding the unit tangent vector begins with differentiating the given vector function \( \mathbf{r}(t) \). From our example, this results in the derivative \( \mathbf{r}'(t) = 18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k} \). This derivative represents the rate of change of the curve's position, effectively pointing in the direction the curve moves.
To obtain the unit tangent vector, we then compute the magnitude of \( \mathbf{r}'(t) \) and divide the derivative by this magnitude:
The process of finding the unit tangent vector begins with differentiating the given vector function \( \mathbf{r}(t) \). From our example, this results in the derivative \( \mathbf{r}'(t) = 18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k} \). This derivative represents the rate of change of the curve's position, effectively pointing in the direction the curve moves.
To obtain the unit tangent vector, we then compute the magnitude of \( \mathbf{r}'(t) \) and divide the derivative by this magnitude:
- Calculate \( \| \mathbf{r}'(t) \| = 21t^2 \).
- Divide \( \mathbf{r}'(t) \) by \( \| \mathbf{r}'(t) \| \) to normalize it: \( \mathbf{T}(t) = \frac{6}{7} \mathbf{i} - \frac{2}{7} \mathbf{j} - \frac{3}{7} \mathbf{k} \).
Arc Length
Calculating the arc length of a curve is about determining how long a segment of the curve is. Essentially, it measures the distance travelled along the path of the curve.
For vector functions, the arc length from one point to another is computed using the integral of the magnitude of the derivative of the vector function over the desired interval. In our case, to find the arc length from \( t = 1 \) to \( t = 2 \), we perform the following steps:
This integration process allows us to "accumulate" the distance, making it crucial for understanding shapes in space and quantifying their length.
For vector functions, the arc length from one point to another is computed using the integral of the magnitude of the derivative of the vector function over the desired interval. In our case, to find the arc length from \( t = 1 \) to \( t = 2 \), we perform the following steps:
- Determine the magnitude of the derivative, \( \| \mathbf{r}'(t) \| = 21t^2 \).
- Set up the integral: \( L = \int_{1}^{2} 21t^2 \ dt \).
- Calculate the integral: \( \int_{1}^{2} 21t^2 \ dt = 49 \).
This integration process allows us to "accumulate" the distance, making it crucial for understanding shapes in space and quantifying their length.
Vector Function Differentiation
Differentiation of vector functions is similar to differentiating scalar functions, but it is applied component-wise to vector-valued functions.
Consider a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \). Each component function \( x(t) \), \( y(t) \), and \( z(t) \) is differentiated individually:
Applying this to \( \mathbf{r}(t) = 6t^3 \mathbf{i} - 2t^3 \mathbf{j} - 3t^3 \mathbf{k} \), we find \( \mathbf{r}'(t) = 18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k} \).
This gives us a new vector describing the direction and rate of change of the original vector function, which is essential in the calculation of items like tangent vectors and curve lengths.
Consider a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \). Each component function \( x(t) \), \( y(t) \), and \( z(t) \) is differentiated individually:
- The derivative \( x'(t) \) accounts for the change in the \( \mathbf{i} \) direction.
- The derivative \( y'(t) \) accounts for the change in the \( \mathbf{j} \) direction.
- The derivative \( z'(t) \) accounts for the change in the \( \mathbf{k} \) direction.
Applying this to \( \mathbf{r}(t) = 6t^3 \mathbf{i} - 2t^3 \mathbf{j} - 3t^3 \mathbf{k} \), we find \( \mathbf{r}'(t) = 18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k} \).
This gives us a new vector describing the direction and rate of change of the original vector function, which is essential in the calculation of items like tangent vectors and curve lengths.
Magnitude of a Vector
The magnitude of a vector is a measure of its length. It tells us how "big" or "strong" the vector is, independent of its direction.
To find the magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), use the formula:
\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \]
In our specific example, the magnitude of the derivative \( \mathbf{r}'(t) = 18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k} \) is calculated as:
To find the magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), use the formula:
\[ \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \]
In our specific example, the magnitude of the derivative \( \mathbf{r}'(t) = 18t^2 \mathbf{i} - 6t^2 \mathbf{j} - 9t^2 \mathbf{k} \) is calculated as:
- \( \| \mathbf{r}'(t) \| = \sqrt{(18t^2)^2 + (-6t^2)^2 + (-9t^2)^2} \).
- Simplifies to \( \| \mathbf{r}'(t) \| = \sqrt{324t^4 + 36t^4 + 81t^4} = \sqrt{441t^4} = 21t^2 \).
Other exercises in this chapter
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