Related Rates Ship Distance Problem

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Homework Help Overview

The problem involves two ships traveling from the same port at different speeds and angles, with the goal of determining how fast the distance between them is changing at a specific time. The subject area relates to related rates in calculus.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the speeds of the ships and their respective paths. One participant outlines a method using the cosine rule to relate the distances traveled by each ship and the angle between their paths. There is also a clarification regarding the notation used for angles.

Discussion Status

The discussion includes attempts to clarify the problem setup and the mathematical approach. Some participants express confusion about the notation and the feasibility of the speeds mentioned. There is no explicit consensus on the method being used, but various interpretations and approaches are being explored.

Contextual Notes

Participants note potential issues with the problem's parameters, such as the speeds of the ships and the notation for angles. There is a mention of a reference book providing the speeds, which may influence the discussion.

bobjefri
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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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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 c^2= a^2+ b^2- 2abcos(C).

Here, that is c^2= 10000t^2+ 6400(t- 4)^2- 2(100t)(80(t- 4))cos(60). Of course, cos(60)= 1/2 so that is just c^2= 10000t^2+ 6400(t- 4)^2- 4000t(t-4). Differentiate with respect to t to find the speed with which that is changing.
 
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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.
 
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.
 

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