Question

The position vector \(r\) describes the path of an object moving in the \(x y\) -plane. Sketch a graph of the path and sketch the velocity and acceleration vectors at the given point. $$ \mathbf{r}(t)=\left\langle e^{-t}, e^{t}\right\rangle,(1,1) $$

Short answer

The velocity vector at the point (1,1) is \(-\mathbf{i} + \mathbf{j}\) and the acceleration vector at the point (1,1) is \(\mathbf{i} + \mathbf{j}\). The graph of the position vector is an exponential curve that shrinks in the positive x-direction and expands in the positive y-direction.

Step-by-step solution

Determine the Velocity and Acceleration Vectors

We can find the velocity vector \(v(t)\), which is the derivative of the position vector \(r(t)\), and the acceleration vector \(a(t)\), which is the derivative of the velocity vector \(v(t)\). This can be computed as follows: The velocity vector \(v(t) = \frac{d}{dt}r(t) = \langle \frac{d}{dt}(e^{-t}), \frac{d}{dt}(e^{t}) \rangle = \langle -e^{-t}, e^{t} \rangle \). The acceleration vector \(a(t) = \frac{d}{dt}v(t) = \frac{d}{dt} \langle -e^{-t}, e^{t} \rangle = \langle e^{-t}, e^{t} \rangle \).

Sketch the Graph for the Path of the Object

Now we'll sketch a graph of the path expressed by \(r(t) = \langle e^{-t}, e^{t} \rangle \). This path represents an exponential growth and decay curve in the two-dimensional plane. As \( t \to \infty \), the curve will grow ever larger in the y-direction, while dwindling in the x-direction. And as \( t \to -\infty \), the curve will grow larger in the x-direction, while diminishing in the y-direction.

Add Velocity and Acceleration Vectors to the Graph

Velocity and acceleration vectors are depicted as arrows pointing in their respective directions. It's important to remember that these vectors are tangent to the curve at the point where they are positioned. At the given point (1,1), we plot the velocity and acceleration vectors. The vector \(v(t) = \langle -1, 1 \rangle \) will be pointing in the second quadrant. The vector \(a(t) = \langle 1, 1 \rangle \) will be pointing in the first quadrant.