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 stretch of river flowing north, with parallel banks 40 m apart. If the maximum water speed is 3 m/s, a reasonable model for the water speed of a river x units from the west bank is a sine function:f(x) = 3 sin (πx/40).If a boater would like to cross theriver from a point on the west bank and land at a point exactly opposite with constant headin and a constant speed of 5 m/s, determine the angle at which the boat should head.

Answer 1

Answer:

The angle at which the boat must head is $- 22.47^{\circ}$

Solution:

As per the solution:

Distance between the parallel banks, d = 40 m

The maximum speed of water, v' = 3 m/s

constant speed, u' = 5 m/s

Also,

The speed of water of the river at a distance of 'x' units from the  west bank is given as a sine function:

$f(x) = 3sin(\frac{\pi x}{40})$          (2)

Now, to determine the angel at which the boat must head:

The velocity of the engine of the boat:

v = $u'cos\theta\hat{i} + u'sin\theta\hat{j}$

v = $5cos\theta \hat{i} + 5sin\theta\hat{j}$

The abscissa of the boat at time t:

v = $5cos\theta t\hat{i}$

Now, from above and eqn(1) , we can write:

$f(5cos\theta t) = 3sin(\frac{\pi \times 5cos\theta t}{40})$

Now, boat's velocity at time t:

v = $5cos\theta \hat{i} + (5sin\theta + 3sin(\frac{\pi \times 5cos\theta t}{40})\hat{j}$

In order to obtain the position of the boat, we integrate both the sides, we get:

r = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}$ + C             (3)

Now, at r = 0:

0 = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}$ + C

C = $\frac{24}{\pi cos\theta}\hat{j}$

Now, from eqn (3)

r = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40} + \frac{24}{\pi cos\theta})\hat{j}$                      (4)

the baot will reach the point at y = 0 and x = 40

Now,

40 = $5cos\theta t$

$t = \frac{8}{cos\theta}$

Substituting the above value of 't' in eqn (4):

r = $5sin\theta \frac{8}{cos\theta}\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta \frac{8}{cos\theta}}{40} + \frac{24}{\pi cos\theta})\hat{j}$

We get:

$48 + 40\pi sin\theta = 0$

$\theta = sin^{- 1}(\frac{- 48}{40\pi}) = - 22.47^{\circ}$