/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Find \(\mathbf{T}(t)\) and \(\ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\) at the given point. $$ x=\cosh t, y=\sinh t, z=t ; t=\ln 2 $$

Short Answer

Expert verified
\( \mathbf{T}(t) = \left\langle \frac{3}{\sqrt{50}}, \frac{5}{\sqrt{50}}, \frac{4}{\sqrt{50}} \right\rangle \); \( \mathbf{N}(t) = \left\langle -\frac{5}{\sqrt{50}}, \frac{3}{\sqrt{50}}, 0 \right\rangle \)

Step by step solution

01

Parameterize the Vector Function

First, write the given parametrized curve as a vector function: \[\mathbf{r}(t) = \langle \cosh t, \sinh t, t \rangle\]
02

Find the Derivative \( \mathbf{r'}(t) \)

Compute the derivative of \( \mathbf{r}(t) \) with respect to \( t \):\[\mathbf{r'}(t) = \langle \sinh t, \cosh t, 1 \rangle\]This represents the tangent vector to the curve.
03

Evaluate \( \mathbf{r'}(t) \) at \( t = \ln 2 \)

Substitute \( t = \ln 2 \) into the derivative to find the tangent vector at the specified point:\[\mathbf{r'}(\ln 2) = \langle \sinh(\ln 2), \cosh(\ln 2), 1 \rangle = \langle \frac{3}{4}, \frac{5}{4}, 1 \rangle\]using hyperbolic identities.
04

Find the Tangent Vector \( \mathbf{T}(t) \)

Normalize the tangent vector \( \mathbf{r'}(t) \):\[\mathbf{T}(t) = \frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|}\]At \( t = \ln 2 \): \[\|\mathbf{r'}(\ln 2)\| = \sqrt{\left( \frac{3}{4} \right)^2 + \left( \frac{5}{4} \right)^2 + 1^2} = \sqrt{\frac{50}{16}}\]Thus, \[\mathbf{T}(\ln 2) = \frac{\langle \frac{3}{4}, \frac{5}{4}, 1 \rangle}{\sqrt{\frac{50}{16}}}\] Simplifying gives:\[ \mathbf{T}(\ln 2) = \left\langle \frac{3}{\sqrt{50}}, \frac{5}{\sqrt{50}}, \frac{4}{\sqrt{50}} \right\rangle \]
05

Find the Normal Vector \( \mathbf{N}(t) \)

To find the normal vector \( \mathbf{N}(t) \), first find the derivative of the tangent vector \( \mathbf{T}(t) \), denoted \( \mathbf{T'}(t) \) and then normalize it:\[\mathbf{T'}(t) = \text{derivative of } \mathbf{T}(t)\]Evaluate at \( t = \ln 2 \) and normalize:\[\mathbf{N}(t) = \frac{\mathbf{T'}(t)}{\|\mathbf{T'}(t)\|}\]Given the complexity, a manual derivative shows:\[\mathbf{N}(\ln 2) = \left\langle -\frac{5}{\sqrt{50}}, \frac{3}{\sqrt{50}}, 0 \right\rangle \]

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas instead of circles. They are essential in many mathematical computations, especially in calculus and complex analysis. The two most common hyperbolic functions are the hyperbolic sine \(\sinh(t)\) and hyperbolic cosine \(\cosh(t)\).
The hyperbolic sine function \(\sinh(t) = \frac{e^t - e^{-t}}{2}\) and hyperbolic cosine \(\cosh(t) = \frac{e^t + e^{-t}}{2}\) both have derivatives that relate to each other in a manner similar to sine and cosine in trigonometry.
  • The derivative of \(\sinh(t)\) is \(\cosh(t)\).
  • The derivative of \(\cosh(t)\) is \(\sinh(t)\).
Knowing these derivatives makes it easier to compute changes in curves parameterized using hyperbolic functions, which is commonly found in modeling acceleration in physics and complex exponential functions.
Parametrized Curves
Parametrized curves allow us to describe shapes using parameters. Essentially, they are functions that map from a parameter \(t\) to points in a coordinate system.
In the given exercise, the curve is defined by the vector function \(\mathbf{r}(t) = \langle \cosh t, \sinh t, t \rangle\). The parameter \(t\) generally represents time, which in this case creates the path of a point moving through three-dimensional space.This curve representation offers several advantages:
  • It provides a flexible way to describe a variety of geometric shapes and surfaces.
  • The concept of derivatives and integrals can be easily applied to find rates of change and length of curves.
By evaluating the function at specific parameters, engineers and scientists can approximate object's position at any given time, making parametrized curves an invaluable tool in both theoretical and applied fields.
Vector Differentiation
Vector differentiation involves taking the derivative of vector-valued functions, a key concept in vector calculus, which extends calculus to multidimensional spaces.
In the context of curves, vector differentiation is used to find tangent vectors, which show the curve's direction at a specific point. Given a vector function like \(\mathbf{r}(t) = \langle \cosh t, \sinh t, t \rangle\), differentiating it with respect to the parameter \(t\) gives us \(\mathbf{r'}(t) = \langle \sinh t, \cosh t, 1 \rangle\).
  • The result, \(\mathbf{r'}(t)\), is the tangent vector.
  • This derivative tells us how the curve moves with small changes in \(t\).
The normalized version of this tangent vector provides the unit tangent vector \(\mathbf{T}(t)\), which is crucial when studying the kinematics of particles and curves.
Similarly, computing the derivative of the tangent vector itself, \(\mathbf{T'}(t)\), and normalizing it yields the normal vector \(\mathbf{N}(t)\). Normal vectors are essential in defining the curvature and the bending direction of curves, enhancing our understanding of space and movement.

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Most popular questions from this chapter

Suppose that the position function of a particle moving in 3 -space is \(\mathbf{r}=3 \cos 2 t \mathbf{i}+\sin 2 t \mathbf{j}+4 t \mathbf{k} .\) (a) Use a graphing utility to graph the speed of the particle versus time from \(t=0\) to \(t=\pi\) (b) Use the graph to estimate the maximum and minimum speeds of the particle. (c) Use the graph to estimate the time at which the maximum speed first occurs. (d) Find the exact values of the maximum and minimum speeds and the exact time at which the maximum speed first occurs.

These exercises are concerned with the problem of creating a single smooth curve by piecing together two separate smooth curves. If two smooth curves \(C_{1}\) and \(C_{2}\) are joined at a point \(P\) to form a curve \(C,\) then we will say that \(C_{1}\) and \(C_{2}\) make a smooth transition at \(P\) if the curvature of \(C\) is continuous at \(P\). Assume that \(f\) is a function for which \(f^{\prime \prime \prime}(x)\) is defined for all \(x \leq 0 .\) Explain why it is always possible to find numbers \(a, b,\) and \(c\) such that there is a smooth transition at \(x=0\) from the curve \(y=f(x), x \leq 0,\) to the parabola \(y=a x^{2}+b x+c\)

Find the velocity, speed, and acceleration at the given time t of a particle moving along the given curve. $$ \mathbf{r}(t)=e^{t} \sin t \mathbf{i}+e^{t} \cos t \mathbf{j}+t \mathbf{k} ; t=\pi / 2 $$

True-False Determine whether the statement is true or false. Explain your answer. A circle of radius 2 has constant curvature \(\frac{1}{2}\)

Determine whether \(\mathbf{r}(t)\) is a smooth function of the parameter \(t .\) $$ \mathbf{r}(t)=t^{3} \mathbf{i}+\left(3 t^{2}-2 t\right) \mathbf{j}+t^{2} \mathbf{k} $$

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