In driving a motor boat, the petrol burnt per hour varies directly as the cube of its velocity. Find the most economical trip of the boat when going against a current of 2 m/hr
The velocity v relative to the current must be > 2 meters/hr if one is to get anywhere.
Write v = 2+x. Assumption: x is constant.
The time T to reach the destination is inversely proportional to x, while the amount y
of fuel burnt is proportional to v^3 T = (x+2)^3 / x.
Note that y --> 0 as x --> 0 or oo. y will have a minimum for some positive x.
dy/dx = 3(x+2)^2 /x - (x+2)^3 / x^2 = (x+2)^2 /x [ 3 - (x+2)/x]
dy/dx = 0 when x = 1 ; therefore most economical trip is at 3 meters/hr relative to
the current.
Long trip, unless you only want to travel a meter or two.
On the other hand, if there is no current, petrol burnt is proportional to x^2.
Therefore you should creep along as slowly as possible.
Conclusion: This model, with constant velocity and petrol burnt per hour proportional
to cube of velocity, is absurd, at least for low velocities.
One must allow x to be a function of time t. Then petrol burnt during time interval
from t to t + Dt is proportional to [(x(t) +2)^3 / x(t)] Dt
and one has a calculus of variations problem: minimize an appropriate integral
of the form Int [(x(t) + 2)^3 /x(t)] dt. But the endpoints are 0 and T, which is proportional
to the integral of 1/x(t).
At this point I have exceeded my level of competence
Just as I thought, I started with 1920x1080 as a safe resolution, and since almost all modern wide-screen TVs are using the aspect-ratio of 16:9, there wouldn't be any problem of lowering the resolution.
Last edited by MarisAnder; 5th November 2014 at 01:58 PM.
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