Word problem two ships departing -- One north the other west....

[Niaboc67]

Question:
Two ships leave port at the same time. One travels north at 80 knots (that is, 80 nautical miles per hour), and the other west at 80 knots. The distance between the ships increases at a constant rate.

At what rate is the distance between the two ships increasing?

The Attempt at a Solution

The distance traveled by the north bound ship = 80t [because distance traveled = speed x time taken]

The distance traveled by the west bound ship = 80t

Both leave port at the same time ... so they are making an ever increasing right triangle

Let the distance between the boats at time t be d

so d = √[(80t)² + (80t)²]

d = √12800t²

so dd/dt = 80sqrt(2)

Why is this incorrect?[/B]

 

[Mark44]

Question:
Two ships leave port at the same time. One travels north at 80 knots (that is, 80 nautical miles per hour), and the other west at 80 knots. The distance between the ships increases at a constant rate.

At what rate is the distance between the two ships increasing?

The Attempt at a Solution

The distance traveled by the north bound ship = 80t [because distance traveled = speed x time taken]

The distance traveled by the west bound ship = 80t

Both leave port at the same time ... so they are making an ever increasing right triangle

Let the distance between the boats at time t be d

so d = √[(80t)² + (80t)²]

d = √12800t²

so dd/dt = 80sqrt(2)

Why is this incorrect?[/B]

To minimize confusion, call the distance D, not d.
Edit: Incorrect response edited out...
BTW, questions about derivatives should be posted in the Calculus section. I have moved your post to that section.

 

[Ray Vickson]

Question:
Two ships leave port at the same time. One travels north at 80 knots (that is, 80 nautical miles per hour), and the other west at 80 knots. The distance between the ships increases at a constant rate.

At what rate is the distance between the two ships increasing?

The Attempt at a Solution

The distance traveled by the north bound ship = 80t [because distance traveled = speed x time taken]

The distance traveled by the west bound ship = 80t

Both leave port at the same time ... so they are making an ever increasing right triangle

Let the distance between the boats at time t be d

so d = √[(80t)² + (80t)²]

d = √12800t²

so dd/dt = 80sqrt(2)

Why is this incorrect?[/B]

Why do you think it is incorrect?

 

[Niaboc67]

The program I keep using says it is incorrect

 

[HallsofIvy]

Exactly what form does the program expect? (That is the problem with such formats- it is not enough to get the correct answer, you must enter it in the right way.)

However, here, if I were a teacher and a student were to enter "$80\sqrt{2}$" as the answer, I would not mark it as completely correct. The correct answer is "$80\sqrt{2}$ knots" since it is a speed.

(And I disagree that it was necessary to post this in "Calculus". Since everything is linear here, it can be done as an algebra problem without taking the derivative.)

 

[Ray Vickson]

The program I keep using says it is incorrect

Well $80 \sqrt{2}$ is numerically correct, but perhaps the program you are using wants a decimal approximation $80 \sqrt{2} \doteq 125.865007 \doteq 125.86 \doteq 125.9 \doteq 126$, depending on accuracy requested. And, of course, as HallsofIvy has indicated, you (may) need to supply units as well.

 

[cheemaftw]

Wait a sec...
D = sqrt(12800)t^2
= sqrt(6400) * sqrt(2) * t^2
= 80 * sqrt(2) * t^2

dD/dt = 160 * sqrt(2) * t

Isn't it just an error with your arithmetic?

 

[Ray Vickson]

Wait a sec...
D = sqrt(12800)t^2
= sqrt(6400) * sqrt(2) * t^2
= 80 * sqrt(2) * t^2

dD/dt = 160 * sqrt(2) * t

Isn't it just an error with your arithmetic?

No, he got it right: $D = \sqrt{12800 \, t^2} = \sqrt{12800} \: t$.

 

[cheemaftw]

oops didnt see the brackets my bad :S

 

[Niaboc67]

I will see if my teacher can give me an extension on this assignment. Because this does not make any sense. The program I am using is called WebAssign and it is nortious for it's bad programming.