Vector Calculus Boat Traversing a river problem

[Qbit42]

Homework Statement

Consider a long straight river flowing north with parallel banks 40m apart. Let us use
the function u(x) = 3 sin(\pix/40) to model the rate of water flow x units from the west bank.A boat proceeds at a constant speed of 5m/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?

The Attempt at a Solution

The velocity vector is V(x) = <5, u(x)>.

As far as my understanding of the problem goes I should use V(x) to develop a vector function for dispacement D(x), parameterize it to D(t), and then find the arc length by integrating
\int ||D&#039;(t)||dt.
My initial attempt was to multiply V(x) by a variable for time (T) and parameterize with x = t, however the bounds I have are for x (0 and 40) not time so that didn't get me anywhere. My next thought was that I remembered that displacement was the integral of velocity and I should just integrate V(x) with x = t, lower bound t =0, and upper bound t = 40.
\int V(t)dt
This gives me <200,240/pi> which means the answer is 214 (applying the typical norm) which seems pretty high

 

[Dick]

For one thing you are given the perpendicular velocity of the boat as 5m/s. That means it will cross the 40m wide river in less that 40s. How long will it take? For another thing it's not asking you for arc length. It's asking you for the displacement parallel to the flow of the river. You only have to integrate the component of velocity parallel to the river flow.

 

[Qbit42]

For one thing you are given the perpendicular velocity of the boat as 5m/s. That means it will cross the 40m wide river in less that 40s How long will it take

I'm afraid I don't see how that is true, if the river is flowing perpendicular to the boat then the boat will be moving along a hypotenuse, so if the river is flowing fast enough then couldn't it take longer than 40s for the boat to reach the other side? And without knowing the displacement I don't see how I can arrive at the time it took to cross the river. I'm sorry if I'm missing something.

It's asking you for the displacement parallel to the flow of the river. You only have to integrate the component of velocity parallel to the river flow.

Thats what I did in the second part, well sorta. I integrated the velocity vector of the boat based on the fact that it would be traveling 5m/s i + u(x) j. So if that's the case then my answer is just the second portion of my vector expression (240/pi)

 

[Dick]

I'm afraid I don't see how that is true, if the river is flowing perpendicular to the boat then the boat will be moving along a hypotenuse, so if the river is flowing fast enough then couldn't it take longer than 40s for the boat to reach the other side? And without knowing the displacement I don't see how I can arrive at the time it took to cross the river. I'm sorry if I'm missing something.



Thats what I did in the second part, well sorta. I integrated the velocity vector of the boat based on the fact that it would be traveling 5m/s i + u(x) j. So if that's the case then my answer is just the second portion of my vector expression (240/pi)

You, in fact, can treat the perpendicular and parallel motions of the boat independently. You do this all the time in problems like falling objects. Since the perpendicular velocity is a constant 5m/s it will take you the same time to cross the river regardless of the parallel flow. The 240/pi answer looks like it comes from integrating u(x) dx with an upper limit of x=40. That's not right. You want to integrate dt, and the upper limit isn't t=40.

 

[Qbit42]

You, in fact, can treat the perpendicular and parallel motions of the boat independently. You do this all the time in problems like falling objects. Since the perpendicular velocity is a constant 5m/s it will take you the same time to cross the river regardless of the parallel flow. The 240/pi answer looks like it comes from integrating u(x) dx with an upper limit of x=40. That's not right. You want to integrate dt, and the upper limit isn't t=40.

Ok here's my attempt:

0 <= t <= 8 where t is time
x is the horizontal distance across the river so it my be expressed as x = 5t
the vertical velocity is therefore u(5t) = 3sin((pi)t/8).

Integrating over dt I get a result of 48/pi, is that what you were getting at?

 

[Dick]

Ok here's my attempt:

0 <= t <= 8 where t is time
x is the horizontal distance across the river so it my be expressed as x = 5t
the vertical velocity is therefore u(5t) = 3sin((pi)t/8).

Integrating over dt I get a result of 48/pi, is that what you were getting at?

Exactly.

 

[Qbit42]

Thanks.

There is a second part to this problem that asks at what angle the boat should be launched in order to arrive directly across from the starting position. Here's what I tried.

V(t) = <5cos(theta), u(5t) - 5sin(theta)>

Does it still take him 8 sec to reach the other side or will it be 8/cos(theta) because the horizontal velocity isn't 5 anymore?

I proceeded with the assumption that it was the same since I thought it would be too complicated to get trig of trig (since t is an argument of u(5t).

I first tried to enforce

\int 5cos(theta)dt = 40 (I)

but all that got me was theta = 0 which is obviously wrong. So I tried enforcing

\int [u(5t) - 5sin(theta)]dt = 0

since the displacement in the y direction is zero. That got me theta = 22.45 degrees. This seems to conflict with the fact that (I) should be true

 

[Dick]

Thanks.

There is a second part to this problem that asks at what angle the boat should be launched in order to arrive directly across from the starting position. Here's what I tried.

V(t) = <5cos(theta), u(5t) - 5sin(theta)>

Does it still take him 8 sec to reach the other side or will it be 8/cos(theta) because the horizontal velocity isn't 5 anymore?

I proceeded with the assumption that it was the same since I thought it would be too complicated to get trig of trig (since t is an argument of u(5t).

I first tried to enforce

\int 5cos(theta)dt = 40 (I)

but all that got me was theta = 0 which is obviously wrong. So I tried enforcing

\int [u(5t) - 5sin(theta)]dt = 0

since the displacement in the y direction is zero. That got me theta = 22.45 degrees. This seems to conflict with the fact that (I) should be true

Yes, time to reach the other side is now 8/cos(theta). You have the right general approach. But you still put x=5*t into u(x). That's not right, is it?

 

[Qbit42]

But you still put x=5*t into u(x). That's not right, is it?

Oh yeah it should be x = 5cos(theta)t, although it didn't change my answer in the long run. I guess that first answer was just luck then. Thanks for all the help!