At noon, Al is 6km north of point O traveling south at 10km/h. Also at noon, Barb is 2km east of point ) traveling east at 5km/h.

A) express the distance "d" between Al and Barb as a function of time "t" hours after noon.
B) Find the time at which the distance is a minimum

as usual, draw a diagram.

Al's distance a is 6+10t
Barb's distance b is 2+5t

d^2 = (6+10t)^2 + (2+5t)^2
= 5(25t^2 + 28t + 8)

Now just set dd/dt=0 to find the minimum.

A) To express the distance "d" between Al and Barb as a function of time "t" hours after noon, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.

In this case, the positions of Al and Barb can be represented by their respective coordinates on a Cartesian plane. Al is traveling south, so his position can be represented by the point (0, -6t) after t hours. Barb is traveling east, so her position can be represented by the point (2t, 0) after t hours.

The distance "d" between the two points (Al and Barb) can be calculated using the formula:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Plugging in the coordinates, we have:

d = sqrt((2t - 0)^2 + (0 - (-6t))^2)
d = sqrt(4t^2 + 36t^2)
d = sqrt(40t^2)
d = 2sqrt(10)t

Therefore, the distance "d" between Al and Barb as a function of time "t" hours after noon is given by d = 2sqrt(10)t.

B) To find the time at which the distance between Al and Barb is at a minimum, we need to find the minimum point of the function d = 2sqrt(10)t.

Since this function is a linear function, the distance "d" is always increasing as time "t" increases. Therefore, there is no time at which the distance is a minimum. The distance between Al and Barb will continue to increase indefinitely as time passes.