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Curious3141 said:Hint: think of a circle centred at (100,0) with time-varying radius intersecting the y-axis at some point. The circle represents the locus of all the points that Margo might reach at time t. What conditions can you impose so that Margo reaches Ursula?
Pranav-Arora said:You are talking about Sven, right?
But I still don't have any idea about what to do.
The only thing I can write down is the equation of circle. ##(x-100)^2+y^2=r^2##. Should this circle intersect the position of Ursula?
Pranav-Arora said:You are talking about Sven, right?
But I still don't have any idea about what to do.
The only thing I can write down is the equation of circle. ##(x-100)^2+y^2=r^2##. Should this circle intersect the position of Ursula?
Pranav-Arora said:Homework Statement
Homework Equations
The Attempt at a Solution
Calculating ##t_m## is easy but I have no idea how would I calculate ##t_s##. I understand what kind of path would Sven take but I don't know how to form equations in her case.
Sorry about that.Curious3141 said:Sorry, Pranav, maybe you should ignore what I said. I thought you had difficulty with Margo, not Sven. (I was confused because you used "her case"...
##v_x^2+v_y^2=4 \Rightarrow v_xdv_x=-v_ydv_y##Curious3141 said:First equation is the velocity relationship. Sven has an x component and a y component to his velocity at any point (x,y). The magnitude of the resultant (vector sum) is a constant. Hint: Pythagoras.
[tex]\frac{dy}{dx}=\frac{y-t}{x}[/tex]Second equation relates the tangent of Sven's path at (x,y) to the gradient of the line joining that point to Ursula's position (0,t).
Pranav-Arora said:Sorry about that. ##v_x^2+v_y^2=4 \Rightarrow v_xdv_x=-v_ydv_y##
[tex]\frac{dy}{dx}=\frac{y-t}{x}[/tex]
Sorry, but I don't know how to solve second order D.Es. :(Curious3141 said:But maybe you'll have better luck if you work it out by hand, because I haven't done this yet.
At beginning, Sven faces the origin?EDIT: Figured out how to work both constants out. That equation ##x\frac{d^2y}{dx^2} = -\frac{dt}{dx}## comes in handy here when you realize what happens right at the beginning of Sven's trajectory.
Pranav-Arora said:Sorry, but I don't know how to solve second order D.Es. :(
At beginning, Sven faces the origin?
What I get is the following:voko said:Before you say you can't, you should at least write it down.
Why?
Pranav-Arora said:Sorry, but I don't know how to solve second order D.Es. :(At beginning, Sven faces the origin?
Curious3141 said:Yes, but from that you can deduce that ##\frac{dx}{dt}## = -2 so ##\frac{dt}{dx}## is the reciprocal of that.
Pranav-Arora said:What I get is the following:
[tex]1+\left(\frac{dy}{dx}\right)^2=4x^2\left(\frac{d^2y}{dx^2}\right)^2[/tex]
It is given in the question that Sven always faces Ursula until he reaches Ursula, so at t=0, Sven will face origin, right?
Pranav-Arora said:How do you even get that?
Curious3141 said:OK, had more time today. Solved the d.e. by hand, and it's actually very easy. Pranav - use Voko's hint to let ##z = \frac{dy}{dx}## and proceed from there.
Pranav-Arora said:Let ##z=y'##, the D.E becomes
[tex]2x\frac{dz}{dx}=\sqrt{1+z^2}[/tex]
Solving this gives:
[tex]x=c_1(\sqrt{1+z^2}+z)[/tex]
Re-arranging and squaring both the sides
[tex]\left(\frac{x}{c_1}-z\right)^2=1+z^2[/tex]
[tex]\Rightarrow \frac{x^2}{c_1^2}-1=\frac{2xz}{c_1}[/tex]
[tex]\Rightarrow \frac{x}{c_1^2}-\frac{1}{x}=\frac{2z}{c_1}[/tex]
Substituting z=dy/dx and solving the D.E
[tex]\frac{x^2}{2c_1^2}-\ln x=\frac{2y}{c_1}+c_2[/tex]
Looks good?
How should I find the constants? Should I differentiate with respect to ##t##?
Curious3141 said:There's a much simpler route to take here.
First thing - I always have a habit of imposing the boundary conditions (initial, final or whatever's known) as a bound when I do the integration. This way, you don't have to worry about constants later. You should consider doing this.
So what I had is ##\int_0^z 2 d(\arcsinh z) = \int_100^x \frac{dx}{x}##
from which you can immediately get ##2\arcsinh{z} = \ln{\frac{x}{100}}## without worrying about constants.
Bring the 2 over to the RHS, and bring it up to the exponent: ##\arcsinh{z} = \ln{(\frac{x}{100})}^{\frac{1}{2}}##
Now take sinh of both sides. Remember the exponential formula for sinh, allowing you to simplify the RHS very easily into a purely algebraic expression.
Pranav-Arora said:I tried your suggestion, this time I get:
[tex]z=5\left(\frac{\sqrt{x}}{100}-\frac{1}{\sqrt{x}}\right)[/tex]
Substituting z=y' and integrating x from 100 to zero and y from 0 to t, I get ##t_s=200/3##. Is this what you get?
Also,do you get, ##t_m=100/\sqrt{3}##?
Thank you for your patience Curious!
Pranav-Arora said:Sorry but I should have asked this before.
I did not understand when you chose the limits in the following step:
[tex]\int_0^z 2d(\text{arcsinh} z)=\int_{100}^x \frac{dx}{x}[/tex]
Can you please explain how did you choose those limits?
I got this already but I couldn't understand the way you chose the limits.Curious3141 said:Sure. The d(arcsinh z) is shorthand for the substitution I made (which was z = sinh θ). I didn't want to bring another variable into it, is all.
Thank you once again Curious!The lower bound is the initial condition. When x = 100 (start of Sven's path), the gradient of the tangent to the curve (which is z) is zero because he "points" at the origin. The upper bound is the time-variable one, this should be self-evident.
Curious3141 said:For interest's sake, here's the plot of Sven's (blue) and Margo's (red) trajectories (attached).
Pranav-Arora said:Thanks, I too thought of the same trajectories. :)
Is it possible to find the equation of path taken by Sven? We had:
[tex]\frac{dy}{dx}=5\left(\frac{\sqrt{x}}{100}-\frac{1}{\sqrt{x}}\right)[/tex]
Can I use this to find the equation of path taken by Sven?
Curious3141 said:I'm curious (like my name suggests), but if you didn't solve that d.e., how exactly did you arrive at the time for Sven? I solved that d.e., and that's how I found that time. That curve I plotted is exact, and uses that equation.
haruspex said:##s \dot \theta = - v \sin \theta ##, ## \dot s = v \cos \theta - u ##
In the line joining the two, target is moving away from pursuer with speed v cos θ, while pursuer advances with speed u: ##\dot s=vcosθ−u##. In the perpendicular direction, the target moves with speed ##v \sin \theta##. From the pursuer's perspective that's a tangential speed, tending to reduce theta. The 'radius' is s: ##v \sin \theta = -s \dot \theta## (as in the familiar v = rω).Pranav-Arora said:Sorry but I really can't follow this, can you please explain how you got those two equations?
##\dot{s}## represents the velocity but how do you get the RHS of the second equation? And I don't have any idea about the first equation. ##s\dot{\theta}## has dimensions of angular velocity but I am confused.
haruspex said:In the line joining the two, target is moving away from pursuer with speed v cos θ, while pursuer advances with speed u: ##\dot s=vcosθ−u##. In the perpendicular direction, the target moves with speed ##v \sin \theta##. From the pursuer's perspective that's a tangential speed, tending to reduce theta. The 'radius' is s: ##v \sin \theta = -s \dot \theta## (as in the familiar v = rω).
To calculate the time difference in pursuit situations, you need to know the distance between the pursuer and the target, as well as the speed of both the pursuer and the target. The time difference can be calculated by dividing the distance by the difference in speed between the two objects.
The formula for calculating time difference in pursuit situations is: Time difference = Distance / (Pursuer's speed - Target's speed).
Yes, the time difference in pursuit situations can be negative. This would occur when the pursuer is slower than the target, resulting in a negative time difference.
The angle of pursuit does not directly affect the time difference. However, it can impact the distance between the pursuer and the target, which in turn can affect the time difference calculation.
Yes, there are other factors that can affect the accuracy of the time difference calculation. These include external forces such as wind or obstacles, as well as human error in measuring distance or speed.