Question

Consider the curve described by the vector function \(\mathbf{r}(t)=\left(50 e^{-t} \cos t\right) \mathbf{i}+\left(50 e^{-t} \sin t\right) \mathbf{j}+\left(5-5 e^{-t}\right) \mathbf{k},\) for \(t \geq 0\). a. What is the initial point of the path corresponding to \(\mathbf{r}(0) ?\) b. What is \(\lim _{t \rightarrow \infty} \mathbf{r}(t) ?\) c. Sketch the curve. d. Eliminate the parameter \(t\) to show that \(z=5-r / 10\), where \(r^{2}=x^{2}+y^{2}\).

Short answer

Question: Find the initial point, the limit as t goes to infinity, sketch the curve and eliminate the parameter t for the following vector function: \( \mathbf{r}(t) = (50 e^{-t} \cos t)\mathbf{i} + (50 e^{-t} \sin t)\mathbf{j} + (5 - 5 e^{-t})\mathbf{k} \) Answer: 1. Initial point: \(\boxed{(50,0,0)}\) 2. Limit as t goes to infinity: \(\boxed{(\text{DNE}, \text{DNE}, 5)}\) 3. This step requires graphing software or calculator. 4. Eliminate t: \(z = 5 - \frac{r^2}{10}\)

Step-by-step solution

Find \(\mathbf{r}(0)\)

To compute the initial point of the path, we must evaluate the vector function \(\mathbf{r}(t)\) at \(t=0\). \[ \mathbf{r}(0) = (50 e^{-0} \cos 0) \mathbf{i} + (50 e^{-0} \sin 0)\mathbf{j} + (5 - 5 e^{-0})\mathbf{k}.\]

Simplify the expression

Upon simplifying the expression, we get \[ \mathbf{r}(0) = (50(1)(1))\mathbf{i} + (50(1)(0))\mathbf{j} + (5 - 5(1))\mathbf{k}.\] Therefore, the initial point is \(\mathbf{r}(0) = 50\mathbf{i} + 0\mathbf{j} - 0\mathbf{k} = \boxed{(50,0,0)}\). #b: Limit as t goes to infinity#

Take the limit of each component of \(\mathbf{r}(t)\) as \(t \rightarrow \infty\)

We will find the limit of each component of \(\mathbf{r}(t)\) as \(t\) goes to infinity: \[\lim_{t \rightarrow \infty} 50 e^{-t} \cos t, \lim_{t \rightarrow \infty} 50 e^{-t} \sin t, \text{and} \lim_{t \rightarrow \infty} 5-5e^{-t}.\]

Simplify the expression

Taking the limit of both the \(x\) and \(y\) components results in infinity multiplied by an oscillatory term, which does not have a limit. Thus, the limit does not exist for x and y. For the \(z\) component, we get \[\lim_{t \rightarrow \infty} 5 - 5e^{-t} = 5 - 5(0) = 5,\] so the limit is \(\boxed{(\text{DNE}, \text{DNE}, 5)}\). #c: Sketch the curve# This step requires graphing software or calculator. #d: Eliminate the parameter t#

Solve for \(e^t\) in terms of x and y

First, we can rewrite the x and y components of \(\mathbf{r}(t)\) as \[x = 50e^{-t}\cos t \quad \text{and} \quad y = 50e^{-t}\sin t.\] We then divide the equation for y by the equation for x to cancel out the \(e^{-t}\) terms: \[\frac{y}{x} = \frac{50e^{-t}\sin t}{50e^{-t}\cos t} = \frac{\sin t}{\cos t} = \tan t.\] Now, we can use arctangent to find \(t\) in terms of x and y by taking the arctangent of both sides: \[t = \arctan\left(\frac{y}{x}\right).\] To eliminate t, we must express \(e^{-t}\) in terms of x and y. We can find \(e^{-t}\) from the x component equation: \[x = 50e^{-t}\cos t \Rightarrow e^{-t} = \frac{x}{50\cos t}.\]

Substitute \(t\) in terms of x and y

Now, we substitute \(t = \arctan\left(\frac{y}{x}\right)\) into the expression for \(e^{-t}\): \[e^{-t} = \frac{x}{50\cos(\arctan\left(\frac{y}{x}\right))}.\] Note that \(x = 50 e^{-t} \cos t\), it follows that \(50 \cos t = x e^t.\)

Substitute into z equation

Now, let's substitute our expression for \(e^{-t}\) into the equation for the z component: \[z = 5 - 5e^{-t} = 5 - 5 \cdot \frac{x}{50\cos(\arctan\left(\frac{y}{x}\right))}.\] Recall that \(50 \cos t = x e^t.\) Thus, \(5 e^{-t} =\frac{x}{10}.\) After substitution, we get: \[z = 5 - \frac{x}{10},\] which gives us \[z = 5 - \frac{r^2}{10}.\] Finally, we get the desired relation: \[z = 5 - \frac{r^2}{10}.\]

Key concepts

Parametric Equations

Parametric equations are a fascinating way to represent curves using a set of equations that express the coordinates as functions of a single parameter, often denoted as \( t \). Instead of describing a curve with a single equation relating \( x \) and \( y \) (or more dimensions), parametric equations provide a step-by-step pathway tracing out the curve using each increment of \( t \). This can be especially useful in representing complex paths or motion in 3D space.
For example, in the vector function given, \( \mathbf{r}(t) = (50e^{-t} \cos t) \mathbf{i} + (50e^{-t} \sin t) \mathbf{j} + (5 - 5e^{-t}) \mathbf{k} \), each component \( (x, y, z) \) is defined in terms of \( t \). This allows us to understand how the coordinates vary over time or space, such as in physics when tracking the motion of an object.

• \( x(t) = 50e^{-t} \cos t \)
• \( y(t) = 50e^{-t} \sin t \)
• \( z(t) = 5 - 5e^{-t} \)

By understanding parametric equations, one can describe and analyze paths that would be difficult or impossible to explain using only Cartesian equations.

Limit of a Function

Limits play a crucial role in understanding the behavior of functions as the input approaches a particular value, often infinity. They help us determine the "end behavior" of parametric equations to see where the path they describe is headed.
In this problem, we examine the limit of each component of the vector function \( \mathbf{r}(t) \) as \( t \to \infty \). The expressions \( 50e^{-t} \cos t \) and \( 50e^{-t} \sin t \) include oscillating trigonometric functions multiplied by exponential decay terms, leading to highly oscillatory behavior that does not settle at a finite value for x and y.
However, for the z-component \( 5 - 5e^{-t} \), it simply approaches 5 as \( e^{-t} \to 0 \) since the exponential term vanishes over time. This gives us valuable insight showing that the curve ultimately flattens out in the z-plane at \( z = 5 \).
Understanding limits in the context of vector functions can highlight the approach, direction, and final state of dynamic systems.

Eliminating Parameters

Eliminating parameters involves expressing the relationship between coordinates directly, removing the parameter \( t \) from parametric equations. This conversion often allows us to better understand the geometric properties of the curve by simplifying it into a single equation.
In this exercise, we connect the parametric description to a Cartesian equation \( z = 5 - \frac{r^2}{10} \), where \( r^2 = x^2 + y^2 \). This is accomplished by using trigonometric identities and algebraic manipulation:

• By expressing \( t \) in terms of \( x \) and \( y \) using the relationship \( \tan t = \frac{y}{x} \), we can substitute back to eliminate \( t \).
• The norms \( r = \sqrt{x^2 + y^2} \) give this familiar and simplified relation of a cylinder.

This process transforms a complex, dynamic description into a static one, making it easier to analyze without advanced tools or trigonometry.

3D Curve Sketching

Sketching 3D curves can be daunting as it requires visualizing multiple dimensions simultaneously. However, using parametric equations, this task becomes manageable by incrementally plotting points according to the path \( \mathbf{r}(t) \) defines.
For this particular vector function, the path begins at \( (50, 0, 0) \) when \( t = 0 \) and progresses endlessly. With tools like graphing calculators or software, you can visualize how this curve moves in space:

• The \( xy \) components \( (50e^{-t} \cos t, 50e^{-t} \sin t) \) suggest a spiral that contracts as \( t \to \infty \), due to the exponents \( e^{-t} \).
• Simultaneously, the \( z \) component moves from 0 to approach 5, pulling the spiral upwards.

Software or graphing utilities are useful here since visualization in three dimensions can be complex by hand or imagination. Understanding the structure through technology helps you grasp deeper the spatial representation of mathematical functions.