Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at
the banks. consider a long straight stretch of river flowing north, with parallel banks 40 m apart. if the maximum water speed is 3 m/s, we can use a quadratic function as a basic model for the rate of water flow x units from the west bank: f(x) = 3 400 x(40 − x). (a) a boat proceeds at a constant speed of 5 m/s from a point a on the west bank while maintaining a heading perpendicular to the bank. how far down the river on the opposite bank will the boat touch shore?
1 answer:
We can rewrite the function f(x) to be a function of time, where x =5t.
f(x) = (3/400)(x)(40 - x)
f(5t) = (3/400)(5t)(40 - 5t) . . . . substituting 5t for x
Simplifying, we have the current speed as a function of the time spent in the river.
f(t) = (3/16)·t·(8 -t)
The area under this curve is the product of speed and time, thus is distance. To find the distance, we integrate over the time it takes to get from one bank to the other, that is from t=0 to t=8.
∫[0, 8] (3/16)·(8t -t²)dt = 3/16·(8·8²/2 -8³/3) = 8³/32 = 16 . . . . meters
The boat ends up 16 meters downriver.
f(x) = (3/400)(x)(40 - x)
f(5t) = (3/400)(5t)(40 - 5t) . . . . substituting 5t for x
Simplifying, we have the current speed as a function of the time spent in the river.
f(t) = (3/16)·t·(8 -t)
The area under this curve is the product of speed and time, thus is distance. To find the distance, we integrate over the time it takes to get from one bank to the other, that is from t=0 to t=8.
∫[0, 8] (3/16)·(8t -t²)dt = 3/16·(8·8²/2 -8³/3) = 8³/32 = 16 . . . . meters
The boat ends up 16 meters downriver.
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