- #1
Rappaccini
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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 2ð 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 + [0]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 2ð, 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.
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 2ð 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 + [0]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 2ð, 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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