# Math Help - What does this derivative represent?

1. ## What does this derivative represent?

Two boats 100 km horizontally apart travel in opposite directions. Boat 1 travels north at 25 km/h, Boat 2 travels south at 35 km/h. At 4pm, how fast does the distance change between them?

Ok.. I know how to actually solve this, but I'm having trouble understanding what the derivative actually represents. The problem itself says it is "how fast the distance changes between them." The boats are travelling at a constant velocity, so isn't the distance between them changing at the rate of the sum of their velocities?

2. Originally Posted by Phire
Two boats 100 km horizontally apart travel in opposite directions. Boat 1 travels north at 25 km/h, Boat 2 travels south at 35 km/h. At 4pm, how fast does the distance change between them?

Ok.. I know how to actually solve this, but I'm having trouble understanding what the derivative actually represents. The problem itself says it is "how fast the distance changes between them." The boats are travelling at a constant velocity, so isn't the distance between them changing at the rate of the sum of their velocities?
Not sum of velocity because velocity is a vector. If you added the velocities you would have 25 + (-35) = -10 km/hr (taking North as positive).

It's sum of speed: 25 + 35 = 60 km/hr.

3. Originally Posted by mr fantastic
Not sum of velocity because velocity is a vector. If you added the velocities you would have 25 + (-35) = -10 km/hr (taking North as positive).

It's sum of speed: 25 + 35 = 60 km/hr.
Right, I should've said speed. I believe the problem presents a distance as a function of time, which I think results in a derivative of velocity, however I'm not sure if that actually applies to this problem. The problem asks how fast the distance changes between them 4 hours later, but both of the boats are traveling at a constant speed, therefore I would guess that the rate is always 60 km/hr, but what the book did is this:

$d = \sqrt{100^2 + (25t + 35t)^2}$
(d as in distance)

It uses the distance formula.

The derivative comes out to:

$\frac{dd}{dt} = 3600t(10000+3600t^2)^{-1/2}$

4 hours later, the distance changes at 55 km/s, but like I said above, if the boats are traveling at a constant speed, why would there be a difference at any moment of time?

4. Originally Posted by Phire
Right, I should've said speed. I believe the problem presents a distance as a function of time, which I think results in a derivative of velocity, however I'm not sure if that actually applies to this problem. The problem asks how fast the distance changes between them 4 hours later, but both of the boats are traveling at a constant speed, therefore I would guess that the rate is always 60 km/hr, but what the book did is this:

$d = \sqrt{100^2 + (25t + 35t)^2}$
(d as in distance)

It uses the distance formula.

The derivative comes out to:

$\frac{dd}{dt} = 3600t(10000+3600t^2)^{-1/2}$

4 hours later, the distance changes at 55 km/s, but like I said above, if the boats are traveling at a constant speed, why would there be a difference at any moment of time?
It would seem that the boats are not collinear:

<---------------------- Boat 1
................................|
................................| 100 km apart at t = 0.
................................|
-----------------Boat 2 ---------------------->

5. Originally Posted by mr fantastic
It would seem that the boats are not collinear:

<---------------------- Boat 1
................................|
................................| 100 km apart at t = 0.
................................|
-----------------Boat 2 ---------------------->
Oh, I forgot about the horizontal distance. Well, that explains that . Thanks.