Chapter 3: Problem 158
Given the following position functions, find the velocity, acceleration, and speed in terms of the parameter \(t .\) $$ \mathbf{r}(t)=e^{-t} \mathbf{i}+t^{2} \mathbf{j}+\tan t \mathbf{k} $$
Short Answer
Expert verified
Velocity: \(-e^{-t} \mathbf{i} + 2t \mathbf{j} + \sec^2 t \mathbf{k}\); Acceleration: \(e^{-t} \mathbf{i} + 2 \mathbf{j} + 2\sec^2 t \tan t \mathbf{k}\); Speed: \(\sqrt{e^{-2t} + 4t^2 + \sec^4 t}\).
Step by step solution
01
Find the velocity vector
The velocity vector \( \mathbf{v}(t) \) is obtained by taking the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Differentiate each component: \[ \mathbf{v}(t) = \frac{d}{dt}(e^{-t}) \mathbf{i} + \frac{d}{dt}(t^2) \mathbf{j} + \frac{d}{dt}(\tan t) \mathbf{k} \] This results in: \[ \mathbf{v}(t) = -e^{-t} \mathbf{i} + 2t \mathbf{j} + \sec^2 t \mathbf{k} \]
02
Find the acceleration vector
The acceleration vector \( \mathbf{a}(t) \) is obtained by taking the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Differentiate each component:\[ \mathbf{a}(t) = \frac{d}{dt}(-e^{-t}) \mathbf{i} + \frac{d}{dt}(2t) \mathbf{j} + \frac{d}{dt}(\sec^2 t) \mathbf{k} \]This results in:\[ \mathbf{a}(t) = e^{-t} \mathbf{i} + 2 \mathbf{j} + 2\sec^2 t \tan t \mathbf{k} \]
03
Find the speed
Speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Calculate the magnitude:\[ \text{Speed} = \left| \mathbf{v}(t) \right| = \sqrt{(-e^{-t})^2 + (2t)^2 + (\sec^2 t)^2} \]Simplify:\[ \text{Speed} = \sqrt{e^{-2t} + 4t^2 + \sec^4 t} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity Vector
In vector calculus, the velocity vector is a crucial concept when studying motion. It describes the direction and rate of change of an object's position over time. For a position function like \[ \mathbf{r}(t) = e^{-t} \mathbf{i} + t^{2} \mathbf{j} + \tan t \mathbf{k} \] the velocity vector, \(\mathbf{v}(t)\), is found by differentiating this position vector with respect to the parameter \(t\). This is because the derivative measures how a function changes as its variable changes, providing a snapshot of motion at any given instant. To find the velocity vector, we differentiate each component of the position vector:
- \( \frac{d}{dt}(e^{-t}) \mathbf{i} = -e^{-t} \mathbf{i}\)
- \( \frac{d}{dt}(t^2) \mathbf{j} = 2t \mathbf{j}\)
- \( \frac{d}{dt}(\tan t) \mathbf{k} = \sec^2 t \mathbf{k}\)
Acceleration Vector
Acceleration is defined as the rate at which an object’s velocity changes over time. In other words, it's a measure of how quickly the velocity vector itself is changing. To find the acceleration vector, \(\mathbf{a}(t)\), we differentiate the velocity vector with respect to \(t\). Continuing with our example:
- The derivative of \(-e^{-t} \mathbf{i}\) with respect to \(t\) is \(e^{-t} \mathbf{i}\)
- The derivative of \(2t \mathbf{j}\) is \(2 \mathbf{j}\)
- The derivative of \(\sec^2 t \mathbf{k}\) requires applying the chain rule, yielding \(2\sec^2 t \tan t \mathbf{k}\)
Speed Calculation
Speed is a scalar quantity, which means it has magnitude but no direction. It can be extracted from the velocity vector by calculating its magnitude. Using the velocity vector from our example:\[ \mathbf{v}(t) = -e^{-t} \mathbf{i} + 2t \mathbf{j} + \sec^2 t \mathbf{k} \]The formula for calculating speed is the square root of the sum of the squares of its components:\[ \text{Speed} = \sqrt{(-e^{-t})^2 + (2t)^2 + (\sec^2 t)^2} \]Upon simplifying:\[ \text{Speed} = \sqrt{e^{-2t} + 4t^2 + \sec^4 t} \]This value provides the magnitude of velocity, effectively describing how fast the object is moving regardless of its direction.
Differentiation
Differentiation is a fundamental tool in calculus used to determine how a function changes as its input changes. It is particularly useful in vector calculus when analyzing position functions to find velocity and acceleration vectors.
- Derivative of a constant: The derivative of a constant function is zero, as constants do not change.
- Power Rule: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
- Exponential Functions: The derivative of \(e^{-t}\) is \(-e^{-t}\).
- Trigonometric Functions: The derivative of \(\tan t\) is \(\sec^2 t\), derived from the quotient rule for trigonometric identities.
Position Function
The position function \(\mathbf{r}(t)\) describes the location of an object at any given time \(t\). It's a vector comprised of components along the \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) directions, often corresponding to three-dimensional coordinates.
- Exponential Component: \(e^{-t} \mathbf{i}\) might imply a decaying movement along the x-axis, as the exponential term decreases over time.
- Polynomial Component: \(t^2 \mathbf{j}\) indicates motion along the y-axis, with the speed increasing as time progresses due to the power of the polynomial.
- Trigonometric Component: \(\tan t \mathbf{k}\) represents the movement along the z-axis, influencing by the behavior of the tangent function.