Water traveling along a straight portion of a river normally flows fastest in th

Question

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 140 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 units from the west bank:

(b) Suppose we would like to pilot the boat to land at the point B on the east bank directly opposite A. If we maintain a constant speed of 5 m/s and a constant heading, find the angle at which the boat should head. (Round the answer to one decimal place.)

Explanation / Answer

he first thing you need to do is find the equation for the river flow rate as a function of x (in meters from the West bank). They said this is a quadratic function - the flow will be 0 when x = 0 or 140 meters, and a max of 3m/s when x=70 meters.

So v = a (x-70)^2 +3, and a(70^2) = -3. So a = -3/4900

v = -3/4900 (x-70)^2 + 3

Now, you can integrate to find out how far downstream the boat will drift. The drift downstream can be integrated as:

(a)
d = v dt = v (dx / 5m/s) = Integral (1/5 v dx)
d = 1/5 (-3/4900 (1/3) (x-70)^3 + 3x) from [x=0 to 140]
d = 1/5 (-1/4900 70^3 + 3(140) - (-1/4900 (-70^3) + 3(0)))
d = 1/5 (420 - 2/4900 (70^3)) = 1/5 (420 - 140) = 1/5 (280) = 56 meters

(b) This is a touch harder - it involves sin and cos functions. You do it the same way though - just take the integral. I'll let you take this one.

Ed: Ok ... the boat takes a direction of k degrees upriver. In that case, its horizontal velocity is 5 cos k and its upriver velocity is 5 sin k. So the downstream drift will be:

d = v dt - 5 sin k dt = (v - 5 sin k) (dx / (5 cos k) m/s)
= Integral (1/5 (v - 5 sin k) / cos k dx)

You want to find the value of k such that d=0.

d = 1/5 Integral (v sec k - 5 tan k) dx
d = 1/5 Integral ((-3/4900) ((x-70)^2 +3) sec k - 5 tan k) dx

You finish 'er up.

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