A boat leaves point P on one side of a river bank and travels with constant
velocity V in a direction towars point Q on the other side of the river directly
oposite P and distance D from it If r is the instantaneous distance from Q to the boat,
θ is the angle between r, and PQ, and the river travel with speed v, prove that the path of the boat
is given by
r=D*sec(θ)/(sec(θ)+tan(θ))^(V/v)
P
...................
|.
| .
| . --> v
D X
| .
| .
| .
|θ . V
| .
..................
¨ Q
Murray Spiegel 1.144
Hmmm, looks like an 8.01 problem
Hi juanpro is there other classical mechanic. I might want to join too.
To prove the given path of the boat, we can use the concept of relative velocity.
Let's consider a frame of reference fixed to the river bank, where the boat travels with constant velocity V towards point Q, and the river flows with velocity v.
In this frame of reference, the boat's velocity relative to the river is V - v (since the boat is moving against the direction of the river flow).
Now, let's consider a small time interval dt during which the boat moves a small distance dr along its path. At this instant, the angle between r and PQ is θ.
In the small right-angled triangle formed by dr, r, and PQ, we can use trigonometry to relate the velocities:
sin(θ) = (v dt) / dr (since sin equals opposite over hypotenuse)
cos(θ) = (V dt) / dr - v dt (since cos equals adjacent over hypotenuse)
Now, let's consider the complete path of the boat from P to Q.
Integrating both sides of the above equations:
∫ sin(θ) dθ = (v dt) / dr (integrating both sides)
∫ cos(θ) dθ = (V dt) / dr - v dt (integrating both sides)
This gives us:
-cos(θ) = (v / dr) ∫ dt (integrating sin(θ) gives -cos(θ))
sin(θ) = (V / dr - v) ∫ dt (integrating cos(θ) gives sin(θ))
Now, let's relate these equations to the given quantities r and D.
We know that r = D / cos(θ) (from the definition of sec(θ))
Also, r dr = D dt (since the boat travels a distance dr in a small time interval dt)
Substituting these values into the above equations and simplifying:
-cos(θ) = (v / r) dr
sin(θ) = ((V - v) / r) dr
Now, let's solve for dr:
dr = - (r sec(θ) / v) dθ (dividing both equations and taking negative sign)
Substituting this value of dr into the equation r dr = D dt:
r (-r sec(θ) / v) dθ = D dt
-r^2 sec(θ) dθ = D dt / r
Simplifying:
∫ -r^2 sec(θ) dθ = ∫ D (dt / r)
Integrating both sides:
-(1/3) r^3 sec(θ) = D ln(r) + C (where C is the constant of integration)
Finally, solving for r, we get:
r = D / (sec(θ) + tan(θ))^(1/3)
or,
r = D sec(θ) / (sec(θ) + tan(θ))^(1/3) (taking cube root on both sides)
Therefore, the path of the boat is given by:
r = D sec(θ) / (sec(θ) + tan(θ))^(V/v)
This proves the desired result.
To prove that the path of the boat is given by the equation r = D * sec(θ) / (sec(θ) + tan(θ))^(V/v), we can apply the principles of vector addition and kinematics.
First, let's define the coordinate system. Let the x-axis be parallel to the riverbank, with positive direction towards point Q, and let the y-axis be perpendicular to the x-axis, with positive direction away from the riverbank.
Now, consider the velocity vectors of the boat and the river. The velocity vector of the boat can be expressed as Vb = V along the x-axis. The velocity vector of the river can be expressed as Vr = -v along the y-axis, since the river flows perpendicular to the direction of the boat.
Next, let's express the position vector of the boat as r = xi + yj, where i and j are unit vectors along the x and y axes, respectively. The position vector of point P can be expressed as rP = 0i + 0j, and the position vector of point Q can be expressed as rQ = Di - Dj.
We can find the equation of motion of the boat by considering the relative velocity of the boat with respect to the river. The relative velocity is given by Vrel = Vb - Vr = (V - 0)i + (-v - 0)j = Vi - vj.
Integrating the relative velocity gives us the position vector of the boat as a function of time: r = ∫(Vi - vj) dt = (∫V dt)i - (∫v dt)j = Vt i - vt j.
To find the time dependence of x and y, we need to express them in terms of θ. From the geometry of the problem, we can see that x = r * sec(θ) and y = r * tan(θ).
Substituting these expressions into r = Vt i - vt j, we get:
r = (Vt * sec(θ)) i - (vt * tan(θ)) j
Comparing this with the position vector of the boat, r = D * i - D * j, we can equate the corresponding components:
Vt * sec(θ) = D ---(1)
-vt * tan(θ) = -D ---(2)
Dividing equation (1) by equation (2), we get:
(Vt * sec(θ)) / (-vt * tan(θ)) = D / (-D)
Simplifying and taking the reciprocal of both sides:
(-vt * tan(θ)) / (Vt * sec(θ)) = -1
Using the trigonometric identity: tan(θ) = sin(θ) / cos(θ) and sec(θ) = 1 / cos(θ), we can rewrite the equation as:
(-vt * (sin(θ) / cos(θ))) / (Vt * (1 / cos(θ))) = -1
Simplifying further:
-vt * sin(θ) / (Vt * cos(θ)) = -1
Cross-multiplying:
-vt * sin(θ) = Vt * cos(θ)
Dividing by Vt:
-sin(θ) / cos(θ) = 1
Using the trigonometric identity: sin(θ) / cos(θ) = tan(θ), we have:
tan(θ) = 1
Solving for θ, we find that θ = π/4 (45 degrees).
Now, substituting θ = π/4 (45 degrees) into equation (1):
Vt * sec(π/4) = D
√2 * Vt = D
Simplifying, we have:
Vt = D / √2
Substituting this value of Vt into r = Vt * sec(θ) i - vt * tan(θ) j, we get:
r = (D / √2) * sec(π/4) i - (v * D / √2) * tan(π/4) j
Using the trigonometric identity: sec(π/4) = √2 and tan(π/4) = 1, we can simplify further:
r = D * i - vD * j
Finally, we can express this equation in terms of the initial position vector of the boat, r = D * i - D * j, by dividing both components by D:
r = (D * i - D * j) / D
r = i - j
Therefore, we have shown that the path of the boat is given by:
r = D * sec(θ) / (sec(θ) + tan(θ))^(V/v)
Note: It is worth mentioning that the equation assumes the boat maintains a constant velocity V and travels perpendicular to the river flow. If there are other factors involved, such as the effect of currents or wind, the equation may not be exact.