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Chapter 10: Problem 22

Find the curvature \(K\) of the curve, where \(s\) is the arc length parameter. $$ \mathbf{r}(s)=(3+s) \mathbf{i}+\mathbf{j} $$

Short Answer

Expert verified
The curvature of the curve is 1.

Step by step solution

01

Find the Derivative of \(\mathbf{r}(s)\)

The derivative of \(\mathbf{r}(s)\) with respect to \(s\) is denoted by \(\mathbf{r}'(s)\). So, the derivative of \(\mathbf{r}(s)\) is \(\mathbf{r}'(s) = \mathbf{i}\).
02

Find the Magnitude of the Derivative

The magnitude of a vector \(\mathbf{y} = y_1 \mathbf{i} + y_2 \mathbf{j}\) is given by \(\sqrt{y_1^2 + y_2^2}\). In our case, the magnitude of the derivative vector \(\mathbf{r}'(s)\) is \(\sqrt{1^2+0^2}=1\).
03

Apply the Definition of Curvature

In general, the curvature of a curve parameterized by arc length \(s\) is given by \(K = \|\mathbf{r}'(s)\|\). So, in this case, the curvature, \(K\), of the curve is \(1\).

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Most popular questions from this chapter

In Exercises 41 and \(42,\) use the definition of the derivative to find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=(3 t+2) \mathbf{i}+\left(1-t^{2}\right) \mathbf{j} $$

Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(t)=\frac{2 t}{8+t^{3}} \mathbf{i}+\frac{2 t^{2}}{8+t^{3}} \mathbf{j} $$

Find the tangential and normal components of acceleration for a projectile fired at an angle \(\theta\) with the horizontal at an initial speed of \(v_{0}\). What are the components when the projectile is at its maximum height?

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True or False? Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \text { The acceleration of an object is the derivative of the speed. } $$
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