SOLUTION: I put quotes to ask question for the parts I didn't understand.
Problem: At 9 P.M. an oil tanker traveling west in the ocean at 18 kilometers per hour passes the same spot as a
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Problem: At 9 P.M. an oil tanker traveling west in the ocean at 18 kilometers per hour passes the same spot as a
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Question 999205: I put quotes to ask question for the parts I didn't understand.
Problem: 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) =?
Soltn:
T=18(t-1) "Why is it set up in this fashion, sure it's basic but I don't understand this set up"
L=23t
T=18t-18 km west of the origin
L=23t km north of the origin
d=((18t-18)^2+(23t)^2)^(1/2) "Why exactly are we using the distance formula here?"
d=(324t^2-648t+324+529t^2)^(1/2)
d=(853t^2-648t+324)^(1/2)
dd/dt=(853t-324)/(853t^2-648t+324)^(1/2... "Why was the derivative taken here?
dd/dt=0 only when 853t=324
t=0.3798h
d(.3798)=14.175km "Why was this plugged back into the original function? what does that cause?"
t = number of hours after 8PM
we'll make t = 0 represent 8 PM since it says "t hours after 8 P.M". This is the starting point in time.
D(t) = distance between the oil tanker and the luxury liner (as shown in the image)
T = distance oil tanker travels west
L = distance luxury liner travels north
"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"
The time for the tanker is simply not "t". It's "t-1" because we have to subtract off an hour (since it's an hour late).
So the distance the oil tanker sails west is T = 18(t-1) kilometers
The luxury liner travels at 23 km/hr. Its distance is measured starting at t = 0 and onward, so it travels north L = 23t kilometers.
Drawing the picture gives something like this
"Then find the distance D between the oil tanker and the luxury liner at that time."
so at any given time t, let's find the ditance between the oil tanker and the luxury liner. Once you have the triangle drawn out (image above), you can see that it's simply the same as finding the hypotenuse of the right triangle. This is exactly how the distance formula works.
So we use the pythagorean theorem to find the hypotenuse D(t)
Plug in the given triangle sides
Plug in T = 18(t-1) and L = 23t
Apply the square root to both sides to solve for D
The square root is the same as the exponent of 1/2.
Now as for why they take the derivative, I do not know. My experience with calculus says that when it comes to problems like these, they usually ask "at what time are the two vessels the closest?". In other words, "what is the smallest value of D? when does this occur?". You didn't post these instructions, but IF your homework is asking, then you would apply the derivative, set it equal to 0 and solve for t.
Why? because the min distance occurs when the derivative of the distance function is zero. This is where the tangent line is horizontal.
Doing all that leads to t = 0.3798
Now you plug t = 0.3798 back into the D(t) function to get D(t) = 14.175 km
So in summary:
The two ships are closest when the time is 0.3798 hours after 8PM. This closest distance is 14.175km. They do not get any closer than this (only focus on time values t > 0)
Note: the time value and distance value given in the summary are approximations
You can put this solution on YOUR website!
distance = speed * time
Tanker distance after tme t = 23 t
Tanker distance after time (t-1) = 18 (t-1)
Since they are travelling at right angles
apply Pythagoras theorem
Velocity is the change in position, so it's the slope of the position.
the slope of the position function, it is velocity.
slope is dy/dx
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