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Magazine Chapter 13: Problem 17
Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter \(t\) $$ \mathbf{r}(t)=\left\langle t^{2}-2 t, 1+3 t, \frac{1}{3} t^{3}+\frac{1}{2} t^{2}\right\rangle, \quad t=2 $$
Short Answer
Expert verified
The unit tangent vector \(\mathbf{T}(2)\) is \(\left\langle \frac{2}{7}, \frac{3}{7}, \frac{6}{7} \right\rangle.\)
Step by step solution
01
Compute the Derivative of the Vector Function
To find the unit tangent vector, we first need to calculate the derivative of the vector function \(\mathbf{r}(t)\). Differentiate each component of \(\mathbf{r}(t)\):1. The derivative of the first component: \((t^2 - 2t)' = 2t - 2\).2. The derivative of the second component: \((1 + 3t)' = 3\).3. The derivative of the third component: \((\frac{1}{3}t^3 + \frac{1}{2}t^2)' = t^2 + t\).Thus, \(\mathbf{r}'(t) = \langle 2t - 2, 3, t^2 + t \rangle\).
02
Evaluate the Derivative at the Given Value of t
Now, we evaluate \(\mathbf{r}'(t)\) at \(t = 2\):1. Substitute \(t = 2\) in the first component: \(2(2) - 2 = 2\).2. The second component is constant: \(3\).3. Substitute \(t = 2\) in the third component: \(2^2 + 2 = 6\).Thus, \(\mathbf{r}'(2) = \langle 2, 3, 6 \rangle\).
03
Calculate the Magnitude of the Derivative
Find the magnitude of \(\mathbf{r}'(2)\) using the formula for vector magnitude: \[\| \mathbf{r}'(2) \| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.\]
04
Find the Unit Tangent Vector
The unit tangent vector \(\mathbf{T}(t)\) is given by normalizing \(\mathbf{r}'(2)\). This is done by dividing each component of \(\mathbf{r}'(2)\) by its magnitude:\[\mathbf{T}(2) = \frac{1}{7} \langle 2, 3, 6 \rangle = \left\langle \frac{2}{7}, \frac{3}{7}, \frac{6}{7} \right\rangle.\]
05
Conclusion of the Unit Tangent Vector
We have determined the unit tangent vector at \(t = 2\) by normalizing the derivative of \(\mathbf{r}(t)\) at that point. The final tangent vector is:\[\mathbf{T}(2) = \left\langle \frac{2}{7}, \frac{3}{7}, \frac{6}{7} \right\rangle.\] This represents the direction of the curve at \(t = 2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Derivative
When dealing with vector functions like \( \mathbf{r}(t) \), taking the derivative involves differentiating each component of the vector with respect to the parameter \( t \). This derivative is called the **vector derivative**.
By performing the differentiation, we find the new vector \( \mathbf{r}'(t) \) which represents the rate of change of \( \mathbf{r}(t) \) at any given \( t \).
Here's a simple step-by-step on how to find the vector derivative:
By performing the differentiation, we find the new vector \( \mathbf{r}'(t) \) which represents the rate of change of \( \mathbf{r}(t) \) at any given \( t \).
Here's a simple step-by-step on how to find the vector derivative:
- **Identify components**: Break down the vector into its respective parts. For example, \( \mathbf{r}(t) = \langle t^2 - 2t, 1 + 3t, \frac{1}{3}t^3 + \frac{1}{2}t^2 \rangle \).
- **Differentiate each component**: Apply standard differentiation rules to each part. This would give you, for our example, \( \mathbf{r}'(t) = \langle 2t - 2, 3, t^2 + t \rangle \).
Vector Magnitude
Once we've got the vector derivative, the next step is to compute its magnitude. The **magnitude of a vector** gives us an idea of its length or size in the space it resides. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude is calculated using the Pythagorean theorem: \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).
To find the vector magnitude, here's what you do:
To find the vector magnitude, here's what you do:
- **Square each component**: Take each part of the vector and square it.
- **Sum the squares**: Add these squared values together.
- **Take the square root**: Finally, take the square root of this sum to find the magnitude.
Unit Vector
Understanding a **unit vector** is key because it indicates direction without reference to length. It's derived by normalizing a vector, which involves dividing the vector by its magnitude.
Finding a unit vector has several steps:
Finding a unit vector has several steps:
- **Calculate the magnitude of the original vector**. We've already covered this, but it's crucial for normalization. For \( \mathbf{r}'(2) \), the magnitude we found was 7.
- **Divide each component of the vector by this magnitude**. This ensures that the length of the resulting vector is 1.
Parametric Equations
**Parametric equations** describe a system where one or more variables are expressed as functions of a parameter. In our case, using the parameter \( t \), the position of a point in 3D space is given by the vector function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \).
Such equations allow for a versatile representation of curves and surfaces, enabling us to model dynamic systems easily.
Here鈥檚 why parametric representation is essential:
Such equations allow for a versatile representation of curves and surfaces, enabling us to model dynamic systems easily.
Here鈥檚 why parametric representation is essential:
- **Flexibility in representation**: Instead of one equation in terms of \( x, y, \) and \( z \) alone, different equations are created for each coordinate.
- **Simplifies projection**: Particularly useful in computer graphics and animation, where viewpoints frequently change.
- **Describing non-function shapes**: Allows us to represent shapes like circles or ellipses, not possible with standard single equation forms.
