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Determine the interval(s) on which the vector-valued function is continuous. \(\mathbf{r}(t)=\left\langle e^{-t}, t^{2}, \tan t\right\rangle\)

Short Answer

Expert verified
The vector-valued function is continuous on the intervals \((n\pi-\frac{\pi}{2}, n\pi+\frac{\pi}{2})\) for all integers \(n\).

Step by step solution

01

Analyze the first component of the vector

The first component of the vector-valued function, \(e^{-t}\), is an exponential function. Exponential functions are always continuous for all real numbers since they do not have any points of discontinuity. This means that for the first component of the vector the interval of continuity is \((-\infty, \infty)\) or all real numbers.
02

Analyze the second component of the vector

The second component, \(t^2\), is a polynomial function. Polynomial functions are continuous at every number in their domains, and the domain of \(t^2\) is all real numbers. This means that for the second component of the vector the interval of continuity is \((-\infty, \infty)\), or all real numbers.
03

Analyze the third component of the vector

The third component is \(\tan t\). The function \(\tan t\) is not defined for all real numbers. There exist vertical asymptotes (discontinuities) at \(\frac{(2n+1)\pi}{2}\), where n is an integer. So, \(\tan t\) is continuous on the interval \((n\pi-\frac{\pi}{2}, n\pi+\frac{\pi}{2})\) where \(n\) is an integer.

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Key Concepts

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

Exponential Function Continuity
When it comes to the continuity of exponential functions, students can take solace in the simplicity of the concept. The exponential function, particularly in the form of the first component of the given vector-valued function, \(e^{-t}\), displays a defining property of being continuous across its entire domain—that is, for all real numbers. What does this mean in simpler terms? No matter what real value you substitute for \(t\), the function will yield a result without any 'holes' or 'jumps'.

This remarkable property of exponential functions stems from their never touching the x-axis, yet forever approaching it as they extend towards negative infinity on a graph. Not having points where the function suddenly 'breaks' or becomes undefined means that \(e^{-t}\) is continuous over the interval \( (-\infty, \infty) \), and there are no real numbers where the function doesn't have a well-defined value. This can be quite reassuring for students tackling calculus problems.
Polynomial Function Continuity
Similarly, polynomial functions offer a breath of fresh air with their predictability in terms of continuity. Take for example the second component of our vector-valued function, \(t^2\). It is a quadratic polynomial function, which, like all polynomials, is continuous everywhere within its domain. Again, this means no surprises for students on the lookout for sudden changes in the graph of \(t^2\).

The continuity of polynomial functions like \(t^2\) is due to their smooth and unbroken nature. Whether the function plots a straight line, a parabola, or some more complex curve, there are no holes, jumps, or vertical asymptotes to worry about. As such, students can confidently say that the interval of continuity for \(t^2\) is \( (-\infty, \infty) \), making polynomial functions reliable and straightforward to work with in calculus.
Tangent Function Discontinuities
Unlike exponential and polynomial functions, the continuity of trigonometric functions like the tangent function, \(\tan t\), can present students with a bit more of a challenge. The reason behind this is the existence of discontinuities or points where the function isn't defined, commonly illustrated by vertical asymptotes on a graph.

The third component of our vector-valued function, \(\tan t\), is continuous between its vertical asymptotes, which occur at \(\frac{(2n+1)\pi}{2}\), where \(n\) is an integer. Thus, for each integral value of \(n\), \(\tan t\) is continuous on the open interval \( (n\pi-\frac{\pi}{2}, n\pi+\frac{\pi}{2}) \). What these intervals indicate is a smooth, connected behavior of the function until a point is reached where it theoretically 'shoots off' to infinity. Students should note these intervals when considering the overall continuity of the tangent function and keep an eye out for those critical asymptotic values to avoid miscalculations.

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Most popular questions from this chapter

The position vector \(r\) describes the path of an object moving in space. Find the velocity, speed, and acceleration of the object. $$ \mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+\sqrt{9-t^{2}} \mathbf{k} $$

Consider a particle moving on a circular path of radius \(b\) described by $$ \begin{aligned} &\mathbf{r}(t)=b \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j}\\\ &\text { where } \omega=d \theta / d t \text { is the constant angular velocity. } \end{aligned} $$ Find the acceleration vector and show that its direction is always toward the center of the circle.

Use the model for projectile motion, assuming there is no air resistance. A bale ejector consists of two variable-speed belts at the end of a baler. Its purpose is to toss bales into a trailing wagon. In loading the back of a wagon, a bale must be thrown to a position 8 feet above and 16 feet behind the ejector. (a) Find the minimum initial speed of the bale and the corresponding angle at which it must be ejected from the baler. (b) The ejector has a fixed angle of \(45^{\circ} .\) Find the initial speed required for a bale to reach its target.

The graph of the vector-valued function \(\mathbf{r}(t)\) and a tangent vector to the graph at \(t=t_{0}\) are given. (a) Find a set of parametric equations for the tangent line to the graph at \(t=t_{0}\) (b) Use the equations for the tangent line to approximate \(\mathbf{r}\left(t_{0}+\mathbf{0 . 1}\right)\) $$ \mathbf{r}(t)=\left\langle t, \sqrt{25-t^{2}}, \sqrt{25-t^{2}}\right\rangle, \quad t_{0}=3 $$

Find \(\mathbf{r}(t)\) for the given conditions. $$ \mathbf{r}^{\prime}(t)=t e^{-t^{2}} \mathbf{i}-e^{-t} \mathbf{j}+\mathbf{k}, \quad \mathbf{r}(0)=\frac{1}{2} \mathbf{i}-\mathbf{j}+\mathbf{k} $$

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