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Chapter 1: Problem 47

Consider the vector \(\mathbf{a}(t)=\langle\cos t, \sin t\rangle\) with components that depend on a real number \(t\). As the number \(t\) varies, the components of \(\mathbf{a}(t)\) change as well, depending on the functions that define them. a. Write the vectors \(\mathbf{a}(0)\) and \(\mathbf{a}(\pi)\) in component form. b. Show that the magnitude \(\|\mathbf{a}(t)\|\) of vector \(\mathbf{a}(t)\) remains constant for any real number \(t\). c. As \(t\) varies, show that the terminal point of vector \(\mathbf{a}(t)\) describes a circle centered at the origin of radius 1 .

Short Answer

Expert verified
a. \(\mathbf{a}(0) = \langle 1, 0 \rangle\), \(\mathbf{a}(\pi) = \langle -1, 0 \rangle\). b. The magnitude is constant, \(\|\mathbf{a}(t)\| = 1\). c. The terminal point describes a unit circle centered at the origin.

Step by step solution

01

Evaluate the vector at t=0

To find the vector \(\mathbf{a}(0)\), substitute \(t = 0\) into the vector components: \(\mathbf{a}(t) = \langle \cos t, \sin t \rangle\). Thus, \(\mathbf{a}(0) = \langle \cos 0, \sin 0 \rangle = \langle 1, 0 \rangle\).
02

Evaluate the vector at t=蟺

To find the vector \(\mathbf{a}(\pi)\), substitute \(t = \pi\) into the vector components. We have \(\cos \pi = -1\) and \(\sin \pi = 0\). Therefore, \(\mathbf{a}(\pi) = \langle \cos \pi, \sin \pi \rangle = \langle -1, 0 \rangle\).
03

Show that the magnitude is constant

The magnitude of vector \(\mathbf{a}(t)\) is given by \(\|\mathbf{a}(t)\| = \sqrt{(\cos t)^2 + (\sin t)^2}\). Using the Pythagorean identity, \((\cos t)^2 + (\sin t)^2 = 1\). Therefore, the magnitude \(\|\mathbf{a}(t)\| = \sqrt{1} = 1\), which is constant for any real number \(t\).
04

Describe the path of the vector

As \(t\) varies, \(\cos t\) and \(\sin t\) trace out the unit circle \((x^2 + y^2 = 1)\) in the coordinate plane. Hence, the terminal point of \(\mathbf{a}(t)\) describes a circle centered at the origin with radius 1, since the components \(\cos t\) and \(\sin t\) correspond to the x and y coordinates of a point on the unit circle.

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Key Concepts

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

Vector Functions
Vector functions are a central topic in calculus and are used to describe vectors that depend on one or more variables, typically time or another parameter like \(t\). In the exercise, we have a vector function \(\mathbf{a}(t) = \langle \cos t, \sin t \rangle\).
  • The parameter \(t\) acts as a variable that influences the x and y values of the vector.
  • Such a vector function describes a path or a curve within a plane or space.
  • In this case, as \(t\) changes, the vector's terminal point describes a path on the unit circle.
This makes it easier to visualize how vector functions can represent diverse shapes and motions, providing a dynamic way to describe mathematical and physical phenomena.
Unit Circle
The unit circle is a fundamental concept in trigonometry and calculus, providing a basis for understanding circular motion and trigonometric functions. It's a circle with a radius of 1, centered at the origin of a coordinate system. In this exercise, the vector \(\mathbf{a}(t) = \langle \cos t, \sin t \rangle\) is crucial.
  • The components \(\cos t\) and \(\sin t\) are direct references to x and y coordinates of any point on the unit circle.
  • Because the radius is 1, the equation \(x^2 + y^2 = 1\) perfectly defines the unit circle.
  • As \(t\) changes, it essentially acts as an angle measure in radians, causing the point on the unit circle to rotate smoothly through all possible positions.
This is why the terminal point of the vector functions moves along a perfect circle, making the concept of the unit circle vital for understanding this exercise.
Magnitude of a Vector
The magnitude of a vector is a measure of its 'length' or 'size', independent of its direction. It is given by the formula \(\|\mathbf{v}\| = \sqrt{x^2 + y^2}\) for a 2-dimensional vector \(\mathbf{v} = \langle x, y \rangle\).
  • In our exercise, the magnitude of \(\mathbf{a}(t)\) is computed as \(\|\mathbf{a}(t)\| = \sqrt{(\cos t)^2 + (\sin t)^2}\).
  • Thanks to the Pythagorean identity, \((\cos t)^2 + (\sin t)^2 = 1\), the calculation simplifies to \(\|\mathbf{a}(t)\| = \sqrt{1} = 1\).
  • This constancy means that regardless of the change in \(t\), the vector's length does not change, reinforcing its representation as motion along the unit circle.
Understanding the magnitude helps clarify why the motion remains on the unit circle, maintaining a constant distance from the origin.

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Most popular questions from this chapter

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