Answer:
(a) v(t) = <2t, -2/t³>
(b) a(t) = <2, 6/t⁴>
(c) ||v(t)|| = √(4t² + 4/t⁶)
(d) Refer to the attached photo for the sketch.
Explanation:
To find the velocity, acceleration, and speed of a particle with the given position function, we'll need to apply some basic principles of calculus and vector operations. Let's break it down step by step:
(a) Velocity (v(t)):
Velocity represents the rate of change of position with respect to time and is defined as the derivative of the position vector r(t) with respect to time, t.

Given the position function r(t) = <t², 1/t²>, we can find the velocity vector by taking the derivative of each component with respect to time:
![\Longrightarrow v(t) = \Big < (d)/(dt)[t^2], (d)/(dt)[(1)/(t^2) ] \Big > \\\\\\\\\therefore \boxed{v(t) = \Big < 2t, -(2)/(t^3) \Big > }](https://img.qammunity.org/2024/formulas/mathematics/college/tzzolrfawhecjhz4ba2lrmuyd71cwrrk7l.png)
(b) Acceleration (a(t)):
Acceleration represents the rate of change of velocity with respect to time and is defined as the derivative of the velocity vector v(t) with respect to time, t.

Taking the derivative of each component of the velocity vector v(t) = <2t, -2/t³> with respect to time gives us the acceleration vector:
![\Longrightarrow a(t) = \Big < (d)/(dt) [2t], (d)/(dt) [-(2)/(t^3)] \Big > \\\\\\\\\therefore \boxed{a(t) = \Big < 2, (6)/(t^4) \Big > }](https://img.qammunity.org/2024/formulas/mathematics/college/z149w7t0xuspdbwso48vq0i91kma2rpvd3.png)
(c) Speed (||v(t)||):
Speed is the magnitude of velocity, which measures the rate at which the particle is moving without considering its direction.
The magnitude of the velocity vector v(t) is given by:

(d) Sketching the graph:
Now, let's address the sketch of the path of the particle and draw the velocity and acceleration vectors for t = 1.
To sketch the path of the particle, we need to look at the position vector r(t) = <t^2, 1/t^2> and see how it changes as t varies. Since t can take both positive and negative values, the path will be in both quadrants I and IV of the Cartesian plane.
For t = 1:

So, at t = 1, the particle is located at the point (1, 1) in the Cartesian plane.
Now, let's find the velocity and acceleration vectors at t = 1:


So, at t = 1:
Velocity vector (v) is <2, -2>.
Acceleration vector (a) is <2, 6>.
To draw these vectors on the sketch, plot the point (1, 1) and draw the vectors with their tails at this point. The vector <2, -2> represents the velocity, and the vector <2, 6> represents the acceleration. Make sure to label the vectors for clarity. Refer to the attached photo for the sketch.