SOLUTION: At 9 P.M. an oil tanker traveling west in the ocean at 18 kilometers per hour passes the same spot as a luxury liner that arrived at the same spot at 8 P.M. while traveling north a

Question 999183: At 9 P.M. an oil tanker traveling west in the ocean at 18 kilometers per hour passes the same spot as a luxury liner that arrived at the same spot at 8 P.M. while traveling north at 23 kilometers per hour. If the "spot" is represented by the origin, find the location of the oil tanker and the location of the luxury liner t hours after 8 P.M. Then find the distance D between the oil tanker and the luxury liner at that time.
D(t) = ?

At what time were the ships closest together? (Hint: Minimize the distance (or the square of the distance!) between them.)
The time is _?_:_?_
P.M. (Round down to the nearest minute.)
Please explain how this is solved
Thank you
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the x-coordinate of the tanker is equal to -18t + 162.
the y-coordinate of the tanker is 0.

the x-coordinate of the liner is 0.
the y-coordinate of the liner is 23t - 184.

t is the time in hours.

let x1 = -18t + 162
let y1 = 0

let x2 = 0
let y2 = 23t - 184

distance between the tanker and the liner at any given time is equal to:

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

this equation becomes:

d(t) = sqrt((-18t+162)^2 + (-23t+184)^2)

to graph this equation, we set y = d(t) and we set x = t.

the equation becomes:

y = sqrt((-18x+162)^2 + (-23x+184)^2)

if you simplify this, you get:

y = sqrt(853x^2 - 14296x + 60100)

that's a quadratic equation.

you can find the x-coordinate of the minimum point on that graph by using x = sqrt(-b/2a) formula.

the y-value of the minimum point on the graph is sqrt(f(-b/2a))

that point is shown on the graph as (8.38,14.8)

the points used to determine the distance equation were taken from a table created for t from t = 0 to t = 11 or 12.

the position of the tanker was (x1,y1)
the position of the liner was (x2,y2)

the distance equation became d(t) = sqrt((x1-x2)^2 + (y1-y2)^2)

for the tanker, y1 = 0
for the liner, x1 = 0

for the tanker, x1 = -18t + 162
for the liner, y2 = 23t + 184

these equations were derived by applying some logic.
for example:

when t = 9, the tanker was at (0,0)
the tanker went -18 units on the x-axis every hour.
therefore, when t = 0, the tanker has to be at x = 162 because:
f(t) = -18(0) + 162 = 162
when t = 9, the tanker has to be at x = 0.
f(t) = -18(9) + 162 = -162 + 162 = 0

simkilar logic was applied to the liner to make sure it was at y = 0 when t = 8.
g(t) = 23t - 184 was the equation.
when t = 8, 23t - 184 = 184 - 184 = 0

the equations were tested to make sure they were reliable.
once determined to be reliable, they were used to find the distance between the position of the tanker and the position of the liner at any given point in time.

that graph is shown below:

$$$

any questions, just send me an email.

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