Distance formula maximization problem

[Niaboc67]

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

 

[Mark44]

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.

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

 

[krebs]

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+)