91Ó°ÊÓ

In this exercise we verify the curvature formula $$ \kappa=\frac{\left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|^{3}} $$ a. Explain why $$ \left|\mathbf{r}^{\prime}(t)\right|=\frac{d s}{d t} $$ b. Use the fact that \(\mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}\) and \(\left|\mathbf{r}^{\prime}(t)\right|=\frac{d s}{d t}\) to explain why $$ \mathbf{r}^{\prime}(t)=\frac{d s}{d t} \mathbf{T}(t) $$ c. The Product Rule shows that $$ \mathbf{r}^{\prime \prime}(t)=\frac{d^{2} s}{d t^{2}} \mathbf{T}(t)+\frac{d s}{d t} \mathbf{T}^{\prime}(t) $$ Explain why $$ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)=\left(\frac{d s}{d t}\right)^{2}\left(\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right) $$ d. In Exercise 9.8 .5 .14 we showed that \(|\mathbf{T}(t)|=1\) implies that \(\mathbf{T}(t)\) is orthogonal to \(\mathbf{T}^{\prime}(t)\) for every value of \(t\). Explain what this tells us about \(\left|\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right|\) and conclude that $$ \left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|=\left(\frac{d s}{d t}\right)^{2}\left|\mathbf{T}^{\prime}(t)\right| $$ e. Finally, use the fact that \(\kappa=\frac{\left|\mathbf{T}^{\prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|}\) to verify that $$ \kappa=\frac{\left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|^{3}} $$

Step by step solution

01

Derive the arc length formula

The arc length, s, along a curve is given by: \(s(t) = \int_a^t \left|\mathbf{r}^{\prime}(u)\right| du\) Differentiating both sides with respect to t, we get: \(\frac{d s}{d t} = \frac{d}{d t}\left[\int_a^t \left|\mathbf{r}^{\prime}(u)\right| du\right]\) By the fundamental theorem of calculus, we get the desired result: \(\frac{d s}{d t} = \left|\mathbf{r}^{\prime}(t)\right|\) #b. Explain expression using given facts#

02

Substituting given information

Given that \(\mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}\) and \(\left|\mathbf{r}^{\prime}(t)\right|=\frac{d s}{d t}\), we can write \(\mathbf{r}^{\prime}(t)\) as: \(\mathbf{r}^{\prime}(t) = \left|\mathbf{r}^{\prime}(t)\right|\mathbf{T}(t)\) Now substitute the expression found in part a: \(\mathbf{r}^{\prime}(t) = \frac{d s}{d t} \mathbf{T}(t)\) #c. Show the result using the Product Rule#

03

Apply the product rule

Using the Product Rule, we have: \(\mathbf{r}^{\prime\prime}(t) = \frac{d^2 s}{d t^2} \mathbf{T}(t) + \frac{d s}{d t} \mathbf{T}^{\prime}(t)\) Now compute the cross product of \(\mathbf{r}^{\prime}(t)\) and \(\mathbf{r}^{\prime\prime}(t)\): \(\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\) \(= \left(\frac{d s}{d t} \mathbf{T}(t)\right) \times \left(\frac{d^2 s}{d t^2} \mathbf{T}(t) + \frac{d s}{d t} \mathbf{T}^{\prime}(t)\right)\) As \(\mathbf{T}(t) \times \mathbf{T}(t) = 0\), the result is: \(\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) = \left(\frac{d s}{d t}\right)^2\left(\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right)\) #d. Determine the expression for cross product of derivatives#

04

Use the orthogonality condition

Since \(|\mathbf{T}(t)|=1\) implies that \(\mathbf{T}(t)\) is orthogonal to \(\mathbf{T}^{\prime}(t)\) for every value of \(t\), we get: \(\mathbf{T}(t) \cdot \mathbf{T}^{\prime}(t) = 0\) The magnitude of the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin{\theta}\), where \(\theta\) is the angle between the vectors. Here, \(\theta = 90^{\circ}\) because the two vectors are orthogonal. Therefore, \(\sin{\theta} = 1\), and we obtain: \(\left|\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right| = |\mathbf{T}(t)||\mathbf{T}^{\prime}(t)|\) Since \(|\mathbf{T}(t)| = 1\), we have: \(\left|\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right| = |\mathbf{T}^{\prime}(t)|\) From part c, substitute the obtained value: \(\left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right| = \left(\frac{d s}{d t}\right)^2 |\mathbf{T}^{\prime}(t)|\) #e. Verify the curvature formula#

05

Verify the curvature formula

Now we need to show that \(\kappa = \frac{\left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|^{3}}\). Using the results from part d and the given expression \(\kappa = \frac{|\mathbf{T}^{\prime}(t)|}{|\mathbf{r}^{\prime}(t)|}\), we can write: \(\kappa = \frac{\left(\frac{d s}{d t}\right)^2 |\mathbf{T}^{\prime}(t)|}{\left(\frac{d s}{d t}\right)^3}\) Simplify the expression: \(\kappa = \frac{|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)|}{|\mathbf{r}^{\prime}(t)|^3}\) Thus, we have verified the curvature formula.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arc Length

The concept of arc length is essential in understanding curves in calculus. Imagine a piece of string laid along a curved path. The length of this string from one point to another on the curve is the arc length. In mathematical terms, arc length, often denoted as \( s \), can be calculated using the integral:\[ s(t) = \int_a^t \left|\mathbf{r}^{\prime}(u)\right| du \]Where \( a \) is the starting point, and \( \mathbf{r}^{\prime}(t) \) is the derivative of the position vector along the curve.

• To find the length of the curve, integrate the magnitude of the velocity vector.
• Differentiating \( s(t) \) with respect to \( t \) gives the rate of change of arc length concerning time:

\[ \frac{d s}{d t} = \left|\mathbf{r}^{\prime}(t)\right| \]This equation ties into the curvature formula by showing how the velocity relates to the arc length through differentiation.

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and differential operations. In the context of curvature, vectors such as \( \mathbf{r}^{\prime}(t) \) and \( \mathbf{r}^{\prime\prime}(t) \) represent the tangent and change of the tangent to the curve, respectively.

• The vector \( \mathbf{r}^{\prime}(t) \) is essential for finding the direction and speed of a particle moving along a curve.
• The unit tangent vector \( \mathbf{T}(t) \) normalizes \( \mathbf{r}^{\prime}(t) \) to give just the direction:

\[ \mathbf{T}(t) = \frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|} \]Understanding how these vectors interact through operations like differentiation and cross products is crucial for vector calculus, especially in computing curvature.

Orthogonality

In vector calculus, orthogonality describes the relationship between vectors when they intersect at right angles. Ensuring the orthogonality of vectors such as \( \mathbf{T}(t) \) and \( \mathbf{T}^{\prime}(t) \) is important for understanding the characteristics of the curve.

• Vectors \( \mathbf{T}(t) \) and \( \mathbf{T}^{\prime}(t) \) are orthogonal if their dot product is zero:

\[ \mathbf{T}(t) \cdot \mathbf{T}^{\prime}(t) = 0 \]

• This relationship implies a right angle between \( \mathbf{T}(t) \) and \( \mathbf{T}^{\prime}(t) \), often resulting in maximum separation.
• The orthogonality ensures that the cross product \( \mathbf{T}(t) \times \mathbf{T}^{\prime}(t) \) is purely related to the magnitude of \( \mathbf{T}^{\prime}(t) \):

\[ \left|\mathbf{T}(t) \times \mathbf{T}^{\prime}(t)\right| = \left|\mathbf{T}^{\prime}(t)\right| \]

Cross Product

The cross product is a fundamental vector operation in vector calculus, allowing for the calculation of a new vector perpendicular to two given vectors. For curvature, the cross product is used to determine the orientation of a curve's acceleration.

• Given two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a vector orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).
• Magnitude of the cross product gives the area of the parallelogram formed by \( \mathbf{a} \) and \( \mathbf{b} \).

In the case of:\( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \), the result is significant in understanding how sharply a path is curving, as it relates directly to the curvature formula:\[ \left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t)\right| = \left(\frac{d s}{d t}\right)^{2}\left|\mathbf{T}^{\prime}(t)\right| \]

Product Rule

The product rule is a crucial differentiation technique used in calculus to differentiate products of two functions. When applied to vector functions, it helps understand how vectors like \( \mathbf{r}^{\prime}(t) \) change over time.For two differentiable functions \( f(t) \) and \( g(t) \), the derivative of their product is:\[ \frac{d}{d t}[f(t)g(t)] = f^{\prime}(t)g(t) + f(t)g^{\prime}(t) \]

• When applied to vector functions, the product rule helps determine \( \mathbf{r}^{\prime\prime}(t) \) :

\[ \mathbf{r}^{\prime\prime}(t) = \frac{d^{2} s}{d t^{2}} \mathbf{T}(t) + \frac{d s}{d t} \mathbf{T}^{\prime}(t) \]By applying the product rule, we can further interpret how derivatives of position vectors relate through curvature and how the arc length factor \( \frac{d s}{d t} \) plays into differentials in the curvature formula.