Answers
The y-coordinate of point P is approximately 11.609.
The speed at time t = 2 is approximately 12.163.
How to find the coordinates(a) To find the y-coordinate of point P, we need to solve for y(2) using the given parametric equation:
y(2) = -ln(1 + (2 - 4)^2) = -ln(1 + 4) = -ln(5) ≈ -1.609
Now, using the initial position of the object (-13, 5), we can find the displacement of the object in the y-direction from time 0 to time 2:
Δy = y(2) - 5 ≈ -1.609 - 5 ≈ -6.609
Therefore, the y-coordinate of point P is:
y-coordinate of P = 5 + Δy ≈ 5 - 6.609 ≈ -1.609
y-coordinate of P = 5 + |Δy| ≈ 5 + 6.609 ≈ 11.609
Therefore, the y-coordinate of point P is approximately 11.609.
(b) To find the acceleration vector at time t = 2, we need to take the second derivative of the position vector with respect to time:
r''(t) = (-6t, 1/(1 + (t-4)^2) - 2/(1 + (t-4)^2)^2)
r''(2) = (-12, 0.0789)
So the acceleration vector at time t = 2 is approximately (-12, 0.0789).
To find the speed at time t = 2, we need to find the magnitude of the velocity vector at that time:
r'(t) = (12 - 6t, -1/(1 + (t-4)^2))
|r'(2)| = ✓((12 - 6(2))^2 + (-1/(1 + (2-4)^2))^2) ≈ 12.163
Therefore, the speed at time t = 2 is approximately 12.163.
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Option b. The database should not store any sensitive or personal information that Sarah does not actually need in order to schedule a job (i.e. parents' political affiliations, previous addresses, etc.).
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