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Chapter 13: Problem 5

(a) Sketch the plane curve with the given vector equation. (b) Find \(\mathbf{r}^{\prime}(t) .\) (c) Sketch the position vector \(\mathbf{r}(t)\) and the tangent vector \(\mathbf{r}^{\prime}(t)\) for the given value of \(t\) . $$ \mathbf{r}(t)=\sin t \mathbf{i}+2 \cos t \mathbf{j}, \quad t=\pi / 4 $$

Short Answer

Expert verified
The plane curve is an ellipse. \( \mathbf{r}'(t) = \cos t \mathbf{i} - 2\sin t \mathbf{j} \). At \( t=\pi/4 \), plot \( \mathbf{r}(t)=\frac{\sqrt{2}}{2} \mathbf{i}+\sqrt{2} \mathbf{j} \) and \( \mathbf{r}'(t)=\frac{\sqrt{2}}{2} \mathbf{i}-\sqrt{2} \mathbf{j} \).

Step by step solution

01

Sketch the Plane Curve

The given vector equation is \( \mathbf{r}(t) = \sin t \mathbf{i} + 2\cos t \mathbf{j} \). This is a parametric equation in terms of \( t \) for a curve in the plane. To sketch this, we notice that \( x = \sin t \) and \( y = 2\cos t \).The parameter \( t \) can take any real value, but \( \sin t \) and \( \cos t \) are periodic with a period of \( 2\pi \). By squaring both terms and using the Pythagorean identity, we align it as: \((x)^2 + \left(\frac{y}{2}\right)^2 = 1\), which describes an ellipse.The ellipse is centered at the origin with a semi-major axis (along the y-axis) of 2 and a semi-minor axis (along the x-axis) of 1. Draw an ellipse symmetric about the origin with its major axis along the y-axis.
02

Compute the Derivative \( \mathbf{r}^{\prime}(t) \)

To find \( \mathbf{r}^{\prime}(t) \), differentiate each component of \( \mathbf{r}(t) = \sin t \mathbf{i} + 2\cos t \mathbf{j} \) with respect to \( t \).- The derivative of \( \sin t \) is \( \cos t \). Thus, the x-component of \( \mathbf{r}^{\prime}(t) \) is \( \cos t \mathbf{i} \).- The derivative of \( 2\cos t \) is \( -2\sin t \). Thus, the y-component of \( \mathbf{r}^{\prime}(t) \) is \( -2\sin t \mathbf{j} \).Therefore, \( \mathbf{r}^{\prime}(t) = \cos t \mathbf{i} - 2\sin t \mathbf{j} \).
03

Evaluate and Sketch at \( t=\pi/4 \)

First, substitute \( t=\pi/4 \) into \( \mathbf{r}(t) \) and \( \mathbf{r}^{\prime}(t) \).- For \( \mathbf{r}(t) \): \( \sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2} \). So, \( \mathbf{r}(\pi/4) = \frac{\sqrt{2}}{2} \mathbf{i} + 2 \left( \frac{\sqrt{2}}{2} \right) \mathbf{j} = \frac{\sqrt{2}}{2} \mathbf{i} + \sqrt{2} \mathbf{j} \).- For \( \mathbf{r}^{\prime}(t) \): \( \cos(\pi/4) = \frac{\sqrt{2}}{2} \) and \( \sin(\pi/4) = \frac{\sqrt{2}}{2} \). Thus, \( \mathbf{r}^{\prime}(\pi/4) = \frac{\sqrt{2}}{2} \mathbf{i} - 2\left( \frac{\sqrt{2}}{2} \right) \mathbf{j} = \frac{\sqrt{2}}{2} \mathbf{i} - \sqrt{2} \mathbf{j} \).Finally, on your ellipse sketch from Step 1, mark the point where \( \mathbf{r}(t) \) is located at \( \left( \frac{\sqrt{2}}{2}, \sqrt{2} \right) \). Draw a position vector from the origin to this point. Then draw the tangent vector \( \mathbf{r}^{\prime}(t) \) starting from that point in the direction of \( \left( \frac{\sqrt{2}}{2}, -\sqrt{2} \right) \).

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Key Concepts

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

Parametric Equations
Parametric equations are a powerful tool in vector calculus enabling you to describe complex curves in the plane or space using a parameter, usually denoted as \(t\). Each coordinate of the curve is expressed as a function of \(t\). For example, in the given exercise, the vector function \(\mathbf{r}(t) = \sin t \mathbf{i} + 2\cos t \mathbf{j}\) describes a plane curve, where:
  • \(x = \sin t\)
  • \(y = 2\cos t\)
The parameter \(t\) typically represents time, allowing you to follow the path in the direction of increasing \(t\). Understanding parametric equations is crucial because they allow more flexibility compared to Cartesian equations; they can describe paths that may loop back on themselves or have other complex behavior. In the exercise, you discover that the square relation \((\sin t)^2 + (\cos t)^2 = 1\) can transform into the familiar ellipse equation through clever manipulation, showcasing how parametric descriptions can unveil hidden geometric properties.
Derivative of Vector Functions
Derivatives are fundamental in calculus, and the derivative of a vector function shows how the vector changes as its parameter changes. For the vector function \(\mathbf{r}(t) = \sin t \mathbf{i} + 2\cos t \mathbf{j}\), taking the derivative involves differentiating each component separately:
  • The derivative of \(\sin t\) with respect to \(t\) is \(\cos t\), yielding the x-component \(\cos t \mathbf{i}\).
  • The derivative of \(2\cos t\) is \(-2\sin t\), giving the y-component \(-2\sin t \mathbf{j}\).
Combining these results, the derivative of the vector function is \(\mathbf{r}^{\prime}(t) = \cos t \mathbf{i} - 2\sin t \mathbf{j}\). This derivative, \(\mathbf{r}^{\prime}(t)\), is important because it represents the tangent vector to the curve at each point \(t\). By understanding this change, you gain insights into the curve's velocity and direction at any given point.
Ellipse
An ellipse is a curve on a plane that surrounds two focal points, and for any point on the ellipse, the sum of the distances to the two foci is constant. In the exercise, \(\mathbf{r}(t) = \sin t \mathbf{i} + 2\cos t \mathbf{j}\) defines an ellipse. By squaring the components and using the identity \((\sin t)^2 + (\cos t)^2 = 1\), we obtain:\[(x)^2 + \left(\frac{y}{2}\right)^2 = 1\]This equation reveals the geometric shape of an ellipse centered at the origin. The ellipse has:
  • A semi-major axis of length 2 along the y-axis.
  • A semi-minor axis of length 1 along the x-axis.
Visualizing an ellipse helps you understand the curve's properties, such as where trajectories are closest or furthest from the center. This understanding becomes particularly useful in both practical applications and further mathematical analysis.
Tangent Vector
A tangent vector at a given point on a curve is a vector that is tangent to the curve at that point, reflecting the curve's immediate direction at that location. For our vector function \(\mathbf{r}(t)\), the tangent vector is determined by its derivative, \(\mathbf{r}^{\prime}(t) = \cos t \mathbf{i} - 2\sin t \mathbf{j}\). This vector points in the direction of the curve's path as \(t\) increases.At \(t = \pi/4\), both \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) can be evaluated:
  • \(\mathbf{r}(\pi/4) = \frac{\sqrt{2}}{2} \mathbf{i} + \sqrt{2} \mathbf{j}\)
  • \(\mathbf{r}^{\prime}(\pi/4) = \frac{\sqrt{2}}{2} \mathbf{i} - \sqrt{2} \mathbf{j}\)
The tangent vector tells you the instantaneous velocity and direction of movement along the ellipse at the point corresponding to \(t = \pi/4\). Drawing both the position vector and the tangent vector on the graph helps visualize the orientation and movement dynamics of objects traveling along the path described by the vector function.

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Most popular questions from this chapter

(a) Find the unit tangent and unit normal vectors \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\) (b) Use Formula 9 to find the curvature. \(\mathbf{r}(t)=\left\langle\sqrt{2} t, e^{t}, e^{-t}\right\rangle\)

The position function of a particle is given by \(\mathbf{r}(t)=\left\langle t^{2}, 5 t, t^{2}-16 t\right\rangle .\) When is the speed a minimum?

(a) Find the unit tangent and unit normal vectors \(\mathbf{T}(t)\) and \(\mathbf{N}(t)\) (b) Use Formula 9 to find the curvature. \(\mathbf{r}(t)=\left\langle t, \frac{1}{2} t^{2}, t^{2}\right\rangle\)

A projectile is fired with an initial speed of 500 \(\mathrm{m} / \mathrm{s}\) and angle of elevation \(30^{\circ} .\) Find (a) the range of the projectile, (b) the maximum height reached, and (c) the speed at impact.

If a particle with mass \(m\) moves with position vector \(\mathbf{r}(t)\) , then its angular momentum is defined as \(\mathbf{L}(t)=m \mathbf{r}(t) \times \mathbf{v}(t)\) and its torque as \(\tau(t)=m \mathbf{r}(t) \times \mathbf{a}(t) .\) Show that \(\mathbf{L}^{\prime}(t)=\tau(t)\) .Deduce that if \(\tau(t)=0\) for all \(t\) then \(\mathbf{L}(t)\) is constant. This is the Iaw of conservation of angular momentum.\()\)
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