Distance formula maximization problem

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SUMMARY

The discussion focuses on solving a distance maximization problem involving an oil tanker and a luxury liner. The oil tanker travels west at 18 kilometers per hour, while the luxury liner travels north at 23 kilometers per hour. The distance function derived is D(t) = sqrt((18t - 18)^2 + (23t)^2), which simplifies to D(t) = sqrt(853t^2 - 648t + 324). The critical point for minimum distance occurs at t = 0.3798 hours after 8 P.M., yielding a distance of approximately 11.311 kilometers. Participants confirm the calculations and address minor arithmetic errors in the final evaluation.

PREREQUISITES
  • Understanding of basic calculus, specifically derivatives and critical points.
  • Familiarity with distance formula in a Cartesian coordinate system.
  • Knowledge of algebraic manipulation and simplification of expressions.
  • Experience with real-world applications of motion and relative speed.
NEXT STEPS
  • Review calculus concepts related to optimization problems.
  • Practice deriving distance functions from motion equations.
  • Explore applications of the distance formula in physics and engineering.
  • Investigate numerical methods for solving distance-related problems in real-time scenarios.
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Students studying calculus, particularly those focusing on optimization problems, as well as educators and tutors assisting with physics and mathematics homework related to motion and distance calculations.

Niaboc67
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Homework Statement


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) =

The Attempt at a Solution


T=18(t-1)

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)

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...

dd/dt=0 only when 853t=324

t=0.3798h

d(.3798)=
So, here is where I am unsure about the answer
Plugging back into the original sqrt( (18t-18)^2 + (23t)^2 )
d(0.3798) = sqrt( (18(0.3798) - 18)^2 + (23(0.3798))^2 ) = 11.311

Is this what you guys got? I am on my last try on my homework and am very unsure about the work here.

Thank you
 
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Niaboc67 said:

Homework Statement


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) =

The Attempt at a Solution


T=18(t-1)

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)

d=(324t^2-648t+324+529t^2)^(1/2)

d=(853t^2-648t+324)^(1/2)
You can stop here. All that is asked for is the distance between the two ships as a function of t. I don't know why you're taking the derivative in the following steps.
Niaboc67 said:
dd/dt=(853t-324)/(853t^2-648t+324)^(1/2...

dd/dt=0 only when 853t=324

t=0.3798h

d(.3798)=
So, here is where I am unsure about the answer
Plugging back into the original sqrt( (18t-18)^2 + (23t)^2 )
d(0.3798) = sqrt( (18(0.3798) - 18)^2 + (23(0.3798))^2 ) = 11.311

Is this what you guys got? I am on my last try on my homework and am very unsure about the work here.

Thank you
 
Niaboc67 said:

Homework Statement


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) =

The Attempt at a Solution


T=18(t-1)

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)

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...

dd/dt=0 only when 853t=324

t=0.3798h

d(.3798)=
So, here is where I am unsure about the answer
Plugging back into the original sqrt( (18t-18)^2 + (23t)^2 )
d(0.3798) = sqrt( (18(0.3798) - 18)^2 + (23(0.3798))^2 ) = 11.311

Is this what you guys got? I am on my last try on my homework and am very unsure about the work here.

Thank you
Looks like you left part of the problem off. Are you trying to find t where the ships were a minimum distance apart? You have the correct value of t for the global minimum of the distance function., but you did something wrong when plugging in. Simple arithmetic error http://www.wolframalpha.com/input/?i=sqrt(+(18(0.3798)+-+18)^2+++(23(0.3798))^2+)