Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 220
Find the curvature for the following vector functions. \(\mathbf{r}(t)=(2 \sin t) \mathbf{i}-4 t \mathbf{j}+(2 \cos t) \mathbf{k}\)
Short Answer
Expert verified
The curvature \( \kappa(t) \) is \( \frac{1}{10} \), constant and independent of \( t \).
Step by step solution
01
Find the First Derivative
Differentiating the vector function \( \mathbf{r}(t) = (2 \sin t) \mathbf{i} - 4t \mathbf{j} + (2 \cos t) \mathbf{k} \) with respect to \( t \), we get:\( \mathbf{r}'(t) = (2 \cos t) \mathbf{i} - 4 \mathbf{j} - (2 \sin t) \mathbf{k} \).
02
Compute the Second Derivative
Now find the second derivative by differentiating \( \mathbf{r}'(t) = (2 \cos t) \mathbf{i} - 4 \mathbf{j} - (2 \sin t) \mathbf{k} \):\( \mathbf{r}''(t) = (-2 \sin t) \mathbf{i} + 0 \mathbf{j} - (2 \cos t) \mathbf{k} \).
03
Find the Magnitudes
Calculate the magnitude of \( \mathbf{r}'(t) \):\[ \| \mathbf{r}'(t) \| = \sqrt{(2 \cos t)^2 + (-4)^2 + (-2 \sin t)^2} = \sqrt{4 \cos^2 t + 16 + 4 \sin^2 t} = \sqrt{4(\cos^2 t + \sin^2 t) + 16} = \sqrt{20} = 2\sqrt{5}. \]Calculate the magnitude of \( \mathbf{r}''(t) \):\[ \| \mathbf{r}''(t) \| = \sqrt{(-2 \sin t)^2 + (0)^2 + (-2 \cos t)^2} = \sqrt{4 \sin^2 t + 4 \cos^2 t} = 2. \]
04
Compute the Curvature
Curvature \( \kappa(t) \) is given by the formula:\[ \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}. \]First, find \( \mathbf{r}'(t) \times \mathbf{r}''(t) \):\[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2\cos t & -4 & -2\sin t \ -2\sin t & 0 & -2\cos t \end{vmatrix} \].Calculating the determinant:\[ = \mathbf{i}((-4)(-2\cos t) - (0)(-2\sin t)) - \mathbf{j}((2\cos t)(-2\cos t) - (-2\sin t)(-2\sin t)) + \mathbf{k}((2\cos t)(0) - (-4)(-2\sin t)) \]\[ = 8\cos t \mathbf{i} - (4\cos^2 t + 4\sin^2 t) \mathbf{j} - 8\sin t \mathbf{k}. \]This simplifies to:\[ 8\cos t \mathbf{i} - 4 \mathbf{j} - 8\sin t \mathbf{k}. \]Calculate its magnitude:\[ \| \mathbf{r}'(t) \times \mathbf{r}''(t) \| = \sqrt{(8\cos t)^2 + (-4)^2 + (-8\sin t)^2} = \sqrt{64\cos^2 t + 16 + 64\sin^2 t} = \sqrt{80} = 4\sqrt{5}. \]Finally, compute the curvature:\[ \kappa(t) = \frac{4\sqrt{5}}{(2\sqrt{5})^3} = \frac{4\sqrt{5}}{40\sqrt{5}} = \frac{1}{10}. \]
05
Conclusion: Interpret the Result
The curvature, \( \kappa(t) = \frac{1}{10} \), is constant and independent of \( t \), indicating that the curve has a uniform shape along its path.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
In vector calculus, the first derivative of a vector function gives us the velocity vector. This indicates the rate of change of the vector with respect to a parameter, typically time \( t \). For a vector function \( \mathbf{r}(t) = (2 \sin t) \mathbf{i} - 4t \mathbf{j} + (2 \cos t) \mathbf{k} \), the first derivative \( \mathbf{r}'(t) \) is calculated by differentiating each component with respect to \( t \). Here, we differentiate:
- \( 2 \sin t \) with respect to \( t \), resulting in \( 2 \cos t \).
- \( -4t \) with respect to \( t \), yielding \( -4 \).
- \( 2 \cos t \) with respect to \( t \), resulting in \( -2 \sin t \).
Second Derivative
The second derivative of a vector function, denoted as \( \mathbf{r}''(t) \), offers insight into the acceleration of the vector. It reflects how the velocity is changing over time. Differentiating the velocity vector \( \mathbf{r}'(t) = (2 \cos t) \mathbf{i} - 4 \mathbf{j} - (2 \sin t) \mathbf{k} \), we find:
- The derivative of \( 2 \cos t \) is \( -2 \sin t \).
- The derivative of \( -4 \) is 0 since it is constant.
- The derivative of \( -2 \sin t \) is \( -2 \cos t \).
Cross Product
The cross product is a crucial tool in vector calculus, as it helps find a vector perpendicular to two given vectors. In the context of curvature, the cross product of the first and second derivatives \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \) gives a vector related to how the original curve twists in space.To compute \( \mathbf{r}'(t) \times \mathbf{r}''(t) \):
- We set up a determinant with the unit vectors and the components of \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \).
- Compute the determinant using cofactor expansion, which involves the interactions between corresponding terms of the vectors to find each coordinate of the result.
Magnitude Calculation
Calculating the magnitude of a vector is vital as it provides the length of the vector, showing the size irrespective of direction. For any vector \( \mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k} \), its magnitude is given by \( \| \mathbf{a} \| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).Consider the magnitudes for both derivatives:
- The magnitude of the first derivative \( \mathbf{r}'(t) \) ensures it is normalized before use in curvature formulae.
- The magnitude of the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) relates directly to the twisting potential of the curve.
