How much work does the fish do?

Rappaccini

Let's say there's a creek.

Down in the creek, there is a current.

The vector field that describes this current is

Cur[x y z] = [(ðy)^x]i + [(y^4)+(xyz)]j + (2z + e^z)k

Nota bene:
ð = pi

The force is in Newtons.

X, Y, and Z are the spatial dimensions in meters, whose origin is a piece of bait in this case.
You see, there's also this guy who's fishin' in the creek. His bait's down there, situated on the origin.

A fish sees it, and circles around it one complete time. The fish is unsure during this period, and maintains a distance of one meter.

This motion takes exactly seconds for the fish.
So, the motion can be described as a vector-valued function of t, time (sec.)

Fis(t) = [cos(t)]i + [sin(t)]j + k

How much work is done by the fish?

Ok... so I made this problem up... that's why it's so weird. :)

I need help setting it up. I know I need to use a line integral.

The upper limit, t, in seconds, will be , while the lower will obviously be 0.

So, first off, I need to find the integrand, which is the dot product of Cur[Fis(t)] and Fis'(t).

To begin

Cur[Fis(t)] = [(ð*sin(t))^cos(t)]i + [sin(t)^4]j + 1k

But here's some trouble for me... I'm not certain on how to differentiate Fis(t).

Tell me, O somebody-who-is-doubtlessly-wiser-than-I, would

Fis'(t) = [-sin(t)]i + [cos(t)]j + 0k ?

If that is so, I'll continue to find the dot product, and then begin the actual integration.

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Rappaccini

Since no one seems willing to help right now, I'll continue the calculation, whether or not it's right so far.

Cur[Fis(t)] = [(ðsin(t))^(cos(t))]i + [sin(t)^4]j + 1k

Fis'(t) = [-sin(t)]i + [cos(t)]j + 0k

Therefore,

Cur[Fis(t)] * Fis'(t) = -sin(t)(ðsin(t))^(cos(t)) + (cos(t))(sin(t)^4)

So...

&int; -sin(t)(ðsin(t))^(cos(t)) + cos(t)sin(t)^4 dt

from 0 to 2ð

As Shmoe, of my native forum, informed me, the force vector function along the curve Fis has an imaginary i component part of the time.

The 'force * velocity' function(t) isn't continuous, due to that.

Well... it's continuous on the domain (0, pi).

With the help of my computer, I can integrate on that domain and determine half of the work the fish does, I guess.
It says approximately -2.1927, which is correct as far as I can tell.

Half-Ass Conclusion:
During the first half of his journey, the fish is helped by the current. He is helped 2.1927 Newton-meters.

Now... is there any way to get the rest of the work?

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