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Chapter 11: Problem 3

Given a tangent vector on an oriented curve, how do you find the unit tangent vector?

Short Answer

Expert verified
Answer: To find the unit tangent vector for a curve given by parametric equations x(t), y(t), and z(t), follow these main steps: 1. Find the tangent vector T(t) by taking the derivatives of x(t), y(t), and z(t) with respect to the parameter t and combining them into a single vector. 2. Calculate the magnitude of the tangent vector |T(t)|. 3. Normalize the tangent vector by dividing its components by its magnitude to find the unit tangent vector U(t).

Step by step solution

01

Find the tangent vector

To find the tangent vector, we need to find the derivative of the position vector function, which describes the curve. If the curve is given by the parametric equations x(t), y(t), and z(t), then the tangent vector T(t) can be found by taking the derivatives of these functions with respect to the parameter t and combining the resulting components in a single vector. T(t) = (dx/dt, dy/dt, dz/dt)
02

Calculate the magnitude of the tangent vector

To normalize the tangent vector and find the unit tangent vector, we need to find the magnitude of the tangent vector. The magnitude of a vector is the square root of the sum of the squares of its components. For the tangent vector T(t) = (dx/dt, dy/dt, dz/dt), the magnitude |T(t)| is given by: |T(t)| = sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)
03

Normalize the tangent vector

Finally, to find the unit tangent vector, we need to normalize the tangent vector by dividing its components by its magnitude. The unit tangent vector U(t) can be found by: U(t) = (1/|T(t)|) * T(t) U(t) = (dx/dt/|T(t)|, dy/dt/|T(t)|, dz/dt/|T(t)|) The unit tangent vector U(t) is now a vector with a magnitude of 1 that is parallel to the curve's tangent at any point specified by the parameter t.

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Most popular questions from this chapter

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\frac{t}{t^{2}+1} \mathbf{i}+t e^{-t^{2}} \mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}} \mathbf{k} ; \mathbf{r}(0)=\mathbf{i}+\frac{3}{2} \mathbf{j}-3 \mathbf{k}$$

Proof of Cross Product Rule Prove that $$\frac{d}{d t}(\mathbf{u}(t) \times \mathbf{v}(t))=\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)+\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)$$ There are two ways to proceed: Either express \(\mathbf{u}\) and \(\mathbf{v}\) in terms of their three components or use the definition of the derivative.

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. What conditions on \(\mathbf{u}\) and \(\mathbf{v}\) lead to equality in the CauchySchwarz Inequality?

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Geometric-arithmetic mean Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

Show that the (least) distance \(d\) between a point \(Q\) and a line \(\mathbf{r}=\mathbf{r}_{0}+t \mathbf{v}\) (both in \(\mathbb{R}^{3}\) ) is \(d=\frac{|\overrightarrow{P Q} \times \mathbf{v}|}{|\mathbf{v}|},\) where \(P\) is a point on the line.
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