Related Rates Ship Distance Problem

In summary, the distance between the ships changes by 4000t when the ships travel at 80 and 100 knots.
  • #1
bobjefri
2
0
Please answer this, it's so hard.

Problem:
At 12 noon, ship A travels due west at 100 knotsshiop B. After 4hours, ship B leaves the same port on the direction N30'W at 80 knots. How fast is the distance between two ships changing at 8PM in the same day?

Reply please.I can't solve it really.
Thanks.
 
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  • #2
Ships traveling at 80 and 100 knots? I can see hydroplane boats going that fast but ships?

Any way: ship A travels at 100 knots starting at noon. In t hours it will have gone a distance 100t nautical miles. Ship B travels at 80 knots, starting 4 hours later. In t hours after noon, it will have gone 80(t- 4) nautical miles. Since A goes due west and B goes "N30W" (you have ' after the 30 and normally that means "minutes" (1/60 of a degree) but that would be a very unusual way of giving a direction- I am going to assume it is "North 30 degrees West". That makes the angle between the two paths 90- 30= 60 degrees. The third side of the triangle, the distance between the two ships is given by the "cosine rule"- if triangle ABC sides AC of length a and BC of length b then the third side, c, is given by [itex]c^2= a^2+ b^2- 2abcos(C)[/itex].

Here, that is [itex]c^2= 10000t^2+ 6400(t- 4)^2- 2(100t)(80(t- 4))cos(60)[/itex]. Of course, cos(60)= 1/2 so that is just [itex]c^2= 10000t^2+ 6400(t- 4)^2- 4000t(t-4)[/itex]. Differentiate with respect to t to find the speed with which that is changing.
 
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  • #3
HallsofIvy, thank you so much.I think it's not complete or was cutted,idk..
About the 'knots' covered by the ships,hehe,well,that's what the reference book gives..
thank you so much.
oh, sorry for the ' that i used in N30'W, that's actually degrees.i did not found the degree sign.hehe.im so sorry.
 
  • #4
No, there was nothing cut off. What happened was that I had suggested a solution using coordinates, then said "not using coordinates" and started showing how to use the cosine law, but then went back and changed the original to "cosine law" solution. I have now taken off that last paragraph.
 

1. What is the "Related Rates Ship Distance Problem"?

The "Related Rates Ship Distance Problem" is a common problem in calculus where the distance between a ship and a lighthouse is changing and the rates at which the ship's distance and the angle of elevation are changing must be determined.

2. What is the main concept behind solving the "Related Rates Ship Distance Problem"?

The main concept behind solving the "Related Rates Ship Distance Problem" is using the chain rule from calculus to find the rate of change of one variable with respect to another variable.

3. What are the necessary steps to solve the "Related Rates Ship Distance Problem"?

The necessary steps to solve the "Related Rates Ship Distance Problem" are:1. Draw a diagram to visualize the problem.2. Identify the given and unknown variables.3. Write an equation that relates the given and unknown variables.4. Use the chain rule to find the derivative of the equation.5. Substitute the given values and solve for the unknown rate of change.

4. What are some common mistakes when solving the "Related Rates Ship Distance Problem"?

Some common mistakes when solving the "Related Rates Ship Distance Problem" include:- Forgetting to label the given and unknown variables in the diagram.- Using incorrect units for the rates or variables.- Forgetting to take the derivative of the equation.- Not properly substituting the given values in the equation.- Not simplifying the equation before solving for the unknown rate of change.

5. How can the "Related Rates Ship Distance Problem" be applied in real life?

The "Related Rates Ship Distance Problem" can be applied in real life scenarios such as tracking the movement of a boat or plane using radar, calculating the rate of change of a rocket's distance from the launchpad, or determining the speed and direction of a car based on its position in a GPS system.

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