Hypothetical Situation + Math Query

[studentxlol]

1. Suppose a cruise ship docked at a port. Passengers enter and leave the ship through a door located halfway up the ship. At the bottom of the doorframe, there is a square piece of of metal pertruding out from the ship that bridges the gap between the doorframe and the external stairwell located on the dock.

Suppose a freak wave caused the ships height to increase in such a way that the square piece of metal rises as the ship rises. Such a rise cause a 10 cm gap between the top of the square piece of metal and the external stairwell.

If I placed a metal object in bewteen this gap, would it be crushed under the immense pressure of the ship when the ship returned to it's original height? Or would it actually support the entire ships weight?

What if I put my hand under that gap?




2. If I integrate a simple function such as 2x^2.

∫2x^2 dx. The 'dx' imples integrate 2x^2 with respect to x.

BUT. Suppose I wanted to integrate an integer such as 5. How would this be written?

∫5. Would there be any dx involved here, or simply ∫5=5x?

 

[tommw]

1. Your hand will be crashed.

2. The 'dx' is not just some formal residual. Its in fact the difference between two adjusting nodes in the formal Rieman Integration. You definitely need it and you can even use it to calulate with it (e.g. when x itself is a function of some other variable).

Writing ∫5 makes no sense at all. What would be the answer... 5x, 5z, 5t, or even 10ε ?
You really need to write ∫5 dx in order to obtain 5x as a result.
Also, you might want to think about what ∫5x dt would result in.

 

[Clever-Name]

1) This depends on whether or not the door is locked in place when it's in the 'open' position. The amount of weight the metal, or your hand, would be able to support depends on how the door is connected to the ship. If it's a freely rotating hinge of some sort then the metal will only have to support the weight of the door and not the weight of the ship. If the door is locked in place by some sort of mechanism then it will have to support a weight up to the breaking point of that mechanism. In either case if your hand were there it will most likely be crushed. The weight of the door would be heavy enough.

2) tommw answers this sufficiently.