Micromass' big October challenge

[Pepper Mint]

That looks nice actually.
It is $$ \int \sqrt {25 - (x - 5)^2}\ dx$$
not
$$ \int \sqrt {25 - (x^2 - 5)^2}\ dx$$

So if you let $$u=x-5$$ then $$du=dx$$
You set it again by letting $$u=5sint$$ and go on with with it.

 

[ibkev]

It is $$ \int \sqrt {25 - (x - 5)^2}\ dx$$
not
$$ \int \sqrt {25 - (x^2 - 5)^2}\ dx$$

Tru dat - whoops! On the bright side, the $x^2-5$ term is a Latex typo but it seems I can't fix it anymore.

So if you let $u=x-5$ then $du=dx$. You set it again by letting $u=5sint$ and go on with with it.

Thanks for the tip - I'm in the process of brushing up on my integration techniques.

 

[Chestermiller]

I almost have Problem 4 solved, but I need some help evaluating an integral. This is basically the last step:

$$\int{\frac{\sec^2{\theta}}{(\sec{\theta}+\tan{\theta})^{V/v}}d\theta}$$

 

[Charles Link]

I almost have Problem 4 solved, but I need some help evaluating an integral. This is basically the last step:

$$\int{\frac{\sec^2{\theta}}{(\sec{\theta}+\tan{\theta})^{V/v}}d\theta}$$

I looked at problem #4 just now, and I wrote a vector equation for $d \vec{R}/dt$ which can be split into two components: $dx/dt$ and $dy/dt$ both of which had the same, but rather complex denominator. I then got a little clever and wrote $dy/dx=(dy/dt)/(dx/dt)$ which gave something rather simple: $dx/x=dy/(y-v_o t )$. I thought I should be able to integrate this, (natural logs), but it isn't giving sensible results. editing... And I just figured out what the problem is with this method=The t needs to be replaced by some function of x and y in order for this to be valid...t is not a constant.

 

[Chestermiller]

I looked at problem #4 just now, and I wrote a vector equation for $d \vec{R}/dt$ which can be split into two components: $dx/dt$ and $dy/dt$ both of which had the same, but rather complex denominator. I then got a little clever and wrote $dy/dx=(dy/dt)/(dx/dt)$ which gave something rather simple: $dx/x=dy/(y-v_o t )$. I thought I should be able to integrate this, (natural logs), but it isn't giving sensible results. editing... And I just figured out what the problem is with this method=The t needs to be replaced by some function of x and y in order for this to be valid...t is not a constant.

Right. If I have some time today, I'll write out what I have done so far to get to the point that I'm at.

 

[person_random_normal]

6) Thanks to Math_QED Consider the integrals I and J.

I=π2∫0sinxcosxx+1dxI=∫0π2sin⁡xcos⁡xx+1dxI = \int\limits_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{x+1}dx
J=π∫0cosx(x+2)2dxJ=∫0πcos⁡x(x+2)2dxJ = \int\limits_0^{\pi}\frac{\cos x}{(x+2)^{2}}dx

What is I in function of J?

https://mail.google.com/mail/u/0/?u...68608010&rm=15794b3f1927dbe0&zw&sz=w1366-h662

https://mail.google.com/mail/u/0/?ui=2&ik=8afd5aa52a&view=fimg&th=15794b3f1927dbe0&attid=0.1&disp=inline&realattid=1547350341991792640-local0&safe=1&attbid=ANGjdJ9AL_gRE5sXmUYaevAhKSEVWeCDGMrM2iIlWdWoI_HQXuUtcdRj5gRNTsNVhx1fD7uUlFXVMOekFVnhgzPXCNVO8-rWTr1U5G2r1pmTcTPjii6g4adAZZBM6Ic&ats=1475668608010&rm=15794b3f1927dbe0&zw&sz=w1366-h662
Is it right ??

 

[mfb]

You cannot link to your private mails. Even if you can see the image (I don't know), no one else can. You can use the "Upload" button to upload images here.

 

[Charles Link]

Continuing problem #4 with my post 39 above, I think I now have most of the solution: Writing it as $x(dy/dx)=y-v_o t$ and letting $dy/dx=tan(\phi)$ and taking $d/ds$ on both sides gives $(dx/ds)tan(\phi)+x \, d \, tan(\phi)/ds=dy/ds-v_o/v_1$. Now $dx/ds=cos(\phi)$ and $dy/ds=sin(\phi)$. This gives
$x sec^2(\phi)(d \phi/ds)=-v_o/v_1$ Now $ds=dx/cos(\phi)$ so that $d \phi/cos(\phi)=-(v_0/v_1) dx/x$. The solution of this is $x=x_o e^{(-v_o/v_1)ln|sec(\phi)+tan(\phi)|}$. It will need a little more work to finish up, but I think I'm almost there. editing... $\\$ $x=x_o |sec(\phi)+tan(\phi)|^{-v_o/v_1}$. ( $v_o$ is speed of good ship; $v_1$ is speed of pirate ship.) $\\$ Using $x (dy/dx)=y-v_o t$ we have $\\$ $y=v_o t +x \, tan(\phi)$. $\\$ If $v_o=v_1$ we have $\\$ $y=v_o t -x_o (|tan(\phi)/(sec(\phi)+tan(\phi))|)$ $\\$ where $\phi$ goes from $\pi$ to $\pi/2 +\delta$ in this problem. We see in this case that $y$ is always less than $v_o t$ so it doesn't catch up to the other ship. $\\$ In our equation for x above, we can put in for $tan(\phi)=(y-v_o t)/x$ and we can solve for $sec(\phi)$ as well to get an equation with just x and y. Result is $x=x_o|-\sqrt{((y-v_o t)/x)^2+1)}+(y-v_o t)/x|^{-v_o/v_1}$. I still have a "t" in the equations, so I don't quite have just x and y in the equation.

 

[Chestermiller]

PROBLEM 4

I did something very similar to Charles Link.

Parametric equations for coordinates of pirate ship as a function of time:
$$\frac{dx}{dt}=\frac{(x_0-x)}{\sqrt{(x_0-x)^2+(vt-y)^2}}V$$
$$\frac{dy}{dt}=\frac{(vt-y)}{\sqrt{(x_0-x)^2+(vt-y)^2}}V$$

Change of variable:
$$vt-y=r\sin{\theta}$$
$$x_0-x=r\cos{\theta}$$
In these equations, r(t) is the distance between the pirate ship and the merchant ship at time t, and $\theta (t)$ is the direction of the pirate ship velocity relative to the x axis.

If we substitute these variable changes into the differential equations, we obtain:
$$\frac{dr}{dt}=v\sin{\theta}-V\tag{1}$$
$$r\frac{d\theta}{dt}=v\cos{\theta}\tag{2}$$

If we divide Eqn. 1 by Eqn. 2, we get:
$$\frac{d\ln{r}}{d\theta}=\tan{\theta}-\frac{V}{v}\sec{\theta}$$
The solution to this equation, subject to the initial condition $r=x_0$ @ $\theta = 0$, is:
$$r=\frac{\sec{\theta}}{(\sec{\theta} + \tan{\theta})^{V/v}}x_0$$
Substitution of this into Eqn. 2 gives:
$$\frac{\sec^2{\theta}d\theta}{(\sec{\theta} + \tan{\theta})^{V/v}}=\frac{vdt}{x_0}$$

That's where I'm stuck for now. I don't know how to evaluate the integral on the left-hand side.

 

[Chestermiller]

The solution to part (c) of Problem 4 is easy. For, V/v = 1, the distance of the pirate ship from the merchant ship in the end is equal to the distance r evaluated to $\theta = \pi/2$ in the equation: $$r=\frac{\sec{\theta}}{(\sec{\theta} + \tan{\theta})^{V/v}}x_0$$
The value is $$r=\frac{x_0}{2}$$

 

[Charles Link]

The solution to part (c) of Problem 4 is easy. For, V/v = 1, the distance of the pirate ship from the merchant ship in the end is equal to the distance r evaluated to $\theta = \pi/2$ in the equation: $$r=\frac{\sec{\theta}}{(\sec{\theta} + \tan{\theta})^{V/v}}x_0$$
The value is $$r=\frac{x_0}{2}$$

Yes, I think that's what the limit will give also for my equation for y. (as $\phi$ approaches $\pi/2$). For a solution of y vs. x, I think I have a similar problem that you do. I can plug in $dy/dx=tan(\phi)$ into my equation for x, so that I've got a very complicated expression involving $dy/dx$, the solution for which will be the path of the pirate ship.

 

[Chestermiller]

Yes, I think that's what the limit will give also for my equation for y. (as $\phi$ approaches $\pi/2$). For a solution of y vs. x, I think I have a similar problem that you do. I can plug in $dy/dx=tan(\phi)$ into my equation for x, so that I've got a very complicated expression involving $dy/dx$, the solution for which will be the path of the pirate ship.

If I had an expression for the integral, I could get t, which would then give us everything we need. But so far, I've only been able to solve the integral for V/v = 3. In that case, the distance the pirate ship had to travel to catch the merchant ship is (9/8)x₀. The tangential distance the pirate ship travels in any of the cases is just Vt.

 

[Charles Link]

Yes, I think that's what the limit will give also for my equation for y. (as $\phi$ approaches $\pi/2$). For a solution of y vs. x, I think I have a similar problem that you do. I can plug in $dy/dx=tan(\phi)$ into my equation for x, so that I've got a very complicated expression involving $dy/dx$, the solution for which will be the path of the pirate ship.

I think this complicated expression is solvable:
$(x/x_o)^{v_1/v_o}-(dy/dx)=-\sqrt{((dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[ (x/x_o)^{v_1/v_o}-(x/x_o)^{-v_1/v_o}]$. This gives $\\$ $y=(1/2)x_o[(v_o/(v_0+v_1))(x/x_o)^{(v_1/v_o)+1}-(v_o/(v_o-v_1))(x/x_o)^{(-v_1/v_o)+1}]+x_o v_o v_1/(v_1^2-v_o^2)$ where the last term is the constant of integration. I've got a wrong sign or two somewhere, but I get that at x=0, $y=x_o v_0 v_1/(v_1^2-v_o^2)$. This gives distance traveled by the pirate ship is D=x_o v_1^2/(v_1^2-v_o^2) ##.

 

[Twigg]

#7:

Per the given diagram, $X = -b\sin\theta$ defined on the interval $[-\pi,\pi]$.. Normalization of the probability density function yields $\rho = \frac{1}{2*\pi}$, as $\int_{-\pi}^{\pi} \frac{1}{2\pi} d\theta = \frac{\pi - (-\pi)}{2\pi} = 1$. $$\langle X \rangle = \int_{-\pi}^{\pi} \rho X d\theta = \frac{-b}{2\pi}\int_{-\pi}^{\pi}\sin\theta d\theta = 0$$ This last step (=0) can be justified since sine is an odd function, or since the integral of sine is cosine which is periodic with a period of $2\pi$. Either argument is sufficient. $$Var(X) = \langle X^{2} \rangle - \langle X \rangle ^{2} = \langle X^{2} \rangle = \int_{-\pi}^{\pi} \rho X^2 d\theta = \frac{b^2}{2\pi} \int_{-\pi}^{\pi} \sin^{2} \theta d\theta = \frac{b^{2}}{2\pi} (\frac{\theta}{2} - \frac{\sin(2\theta)}{4})|_{-\pi}^{\pi})$$

The integral of sine squared I took from an integral table found here: http://integral-table.com/. The $\sin(2\theta)$ term makes no difference because it is a periodic function with a period of $\pi$, and the interval from $-\pi$ to $\pi$ is a distance of $2\pi$. By its periodicity, $\sin(2\theta)$ takes the same value at both points. That leaves, $$Var(X) = \frac{b^{2}}{2\pi} (\frac{\theta}{2})|_{-\pi}^{\pi} = \frac{b^{2}}{2\pi} (\frac{\pi - (-\pi)}{2}) = \frac{b^{2}}{2} $$

 

[Erland]

Some preliminary work on Advanced Problem 3.

Let $\frak L$ be the set of all generalized limits. We know from previous challenges that there are several distinct generalized limits. Let $L_0$ and $L_1$ be two distinct generalized limits. Then, there is a sequence $(y_n)_n\in X$ such that $L_0((y_n)_n)\neq L_1((y_n)_n)$. For each $t\in[0,1]$, define $L_t$ on $X$ by $L_t((x_n)_n)=(1-t)L_0((x_n)_n)+tL_1((x_n)_n)$ for all $(x_n)_n\in X$. It is easy to prove that $L_t$ is a generalized limit. This holds for all $t\in [0,1]$ ($L_0$ and $L_1$ will be the same as before). We see that the $L_t(y_n)_n$:s will be distinct for distinct $t\in[0,1]$, so the family $\{L_t\,|\,t\in[0,1]\}\subseteq\frak L$ has the same cardinality as $[0,1]$, i.e. $\frak c$ (continuum).
Thus, $\text {card}\,\frak L\ge \frak c$.

On the other hand, $X$ can be viewed as a subset of the set of all functions from $\mathbb N$ to $\mathbb R$, and the latter set has the cardinality $\frak c^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\frak c$. Since $X$ contains all real constant sequences, $\text {card} \,X=\frak c$.
Next, $\frak L$ is a subset of the set of all functions from $X$ to $\mathbb R$, and the latter set has the cardinality $\frak c^\frak c=(2^{\aleph_0})^\frak c =2^{\aleph_0\cdot\frak c}=2^\frak c$.
Thus $\text{card}\,\frak L\le 2^\frak c$.

So, ${\frak c}\le \text{card}\,\frak L\le 2^\frak c$.

Now, if we could prove in ZFC that ${\frak c}< \text{card}\,\frak L< 2^\frak c$, this would contradict the Generalized Continuum Hypothesis (GCH). But this would imply that ZFC is inconsistent (GCH is not disprovable in ZFC if ZFC is consistent). Since ZFC is not known to be inconsistent, and there should be a proof in ZFC that ${\text {card}}\,\frak L$ equals a specific cardinal (otherwise, our problem would be currently unsolved, which it obviously is not), this means that either $\text {card}\,\frak L=\frak c$ or $\text{card}\,\frak L=2^\frak c$ is provable in ZFC.

But which of them? I don't know, for the moment. I am inclined towards the latter one, since I don't think that the restrictons given in the definition of generalized limit are sufficient to shrink the set of all functions from $X$ to $\mathbb R$ down from cardinality $2^\frak c$ to $\frak c$.
But don't know this... I'm working on it...

 

[Chestermiller]

I think this complicated expression is solvable:
$(x/x_o)^{v_1/v_o}-(dy/dx)=-\sqrt{((dy/dx)^2+1}$. A little algebra gives $dy/dx= (x/x_o)^{v_1/v_o}-(x/x_o)^{-v_1/v_o}$. This gives $\\$ $y=(v_o/(v_0+v_1))(x/x_o)^{(v_1/v_o)+1}-(v_o/(v_o-v_1))(x/x_o)^{(-v_1/v_o)+1}-2 v_o v_1/(v_1^2-v_o^2)$ where the last term is the constant of integration.

What does this predict for y at x = x0 when v1/v0 = 3?

 

[micromass]

$X = -b\sin\theta$

Are you sure of that?

Also, note that I made a correction of the problem.

 

[Twigg]

Are you sure of that?

Also, note that I made a correction of the problem.

Whoops! Thanks for the pointer.

 

[micromass]

I think this complicated expression is solvable:
$(x/x_o)^{v_1/v_o}-(dy/dx)=-\sqrt{((dy/dx)^2+1}$. A little algebra gives $dy/dx= (x/x_o)^{v_1/v_o}-(x/x_o)^{-v_1/v_o}$. This gives $\\$ $y=(v_o/(v_0+v_1))(x/x_o)^{(v_1/v_o)+1}-(v_o/(v_o-v_1))(x/x_o)^{(-v_1/v_o)+1}-2 v_o v_1/(v_1^2-v_o^2)$ where the last term is the constant of integration.

Shouldn't $(0,0)$ be on the curve?

 

[Charles Link]

Shouldn't $(0,0)$ be on the curve?

I made a couple of algebraic mistakes. I will go back and edit my original if I still can. I had to go somewhere for about two hours, so I hurried the result and didn't get the chance to finish it.

 

[micromass]

I made a couple of algebraic mistakes. I will go back and edit my original if I still can. I had to go somewhere for about two hours, so I hurried the result and didn't get the chance to finish it.

No please don't edit your post after you have received replies. It makes it much harder to read. Just make a new post with the correct version.

 

[Charles Link]

No please don't edit your post after you have received replies. It makes it much harder to read. Just make a new post with the correct version.

I made too many mistakes in post #48. Here I will start with my equation from post #43 $x/x_o=|sec(\phi)+tan(\phi)|^{-v_o/v_1}$ and proceed with $dy/dx=tan(\phi)$ and $sec(\phi)=-\sqrt{(dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[(x/x_o)^{-v_1/v_o}-(x/x_o)^{v_1/v_o}]$. This integrates to give $\\$ $y=(x_o/2)[-(v_o/(v_1-v_o))(x/x_o)^{-(v_1-v_o)/v_o}-(v_o/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o)} ]+C$ $\\$ where $C=x_o v_o v_1/(v_1^2-v_o)^2$. The constant of integration occurs because at $x=x_o$ , $y=0$. To find where this curve intercepts the good ship, we set x=0. This gives $y_1=x_o v_o v_1/(v_1^2-v_o^2)$. Distance D traveled by the pirate ship is $D=(v_1/v_o)y_1=x_o v_1^2/(v_1^2-v_o^2)$.

 

[micromass]

I made too many mistakes in post #48. Here I will start with my equation from post #43 $x/x_o=|sec(\phi)+tan(\phi)|^{-v_o/v_1}$ and proceed with $dy/dx=tan(\phi)$ and $sec(\phi)=-\sqrt{(dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[(x/x_o)^{-v_1/v_o}-(x/x_o)^{v_1/v_o}]$. This integrates to give $\\$ $y=(x_o/2)[-(v_o/(v_1-v_o))(x/x_o)^{-(v_o-v_1)/v_o}-(v_o/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o)} ]+C$ $\\$ where $C=x_o v_o v_1/(v_1^2-v_o)^2$. The constant of integration occurs because at $x=x_o$ , $y=0$. To find where this curve intercepts the good ship, we set x=0. This gives $y_1=x_o v_o v_1/(v_1^2-v_o^2)$. Distance D traveled by the pirate ship is $D=(v_1/v_o)y_1=x_o v_1^2/(v_1^2-v_o^2)$.

Why would $(x_0,0)$ be on the curve?

 

[Charles Link]

Why would $(x_0,0)$ be on the curve?

I have the good ship sailing up the y-axis (starting at (0,0) and the pirate ship starting at $(x_o,0)$.

 

[micromass]

I have the good ship sailing up the y-axis (starting at (0,0) and the pirate ship starting at $(x_o,0)$.

That is not what I wrote in the OP.

 

[Charles Link]

That is not what I wrote in the OP.

Sorry. I think what happened here is I read the initial problem and then began working on it, but I didn't have the coordinates in precise agreement because I didn't go back and re-read the question thoroughly. In my first couple of posts, I didn't think I was going to come close to solving it... editing... I think my slightly modified coordinates might even simply some of the vector algebra because it always has an x=0 for one of the ships...$\\$ The substitution $x=x_o-x'$ will give $y$ vs. $x'$ where the $x'$ is the correct coordinate given in the OP. Putting this addition in the equation of post 57 hopefully gives the correct answer. My answer of post 57 for D is in agreement with @Chestermiller of post 47 for $v_1=3 v_o$.

 

[person_random_normal]

You cannot link to your private mails. Even if you can see the image (I don't know), no one else can. You can use the "Upload" button to upload images here.

But it's size is greater that 3 MB , so what do I do ??

 

[mfb]

You can also upload it to various websites that host images. A better solution would be to reduce its size, if that doesn't ruin the quality.

 

[Chestermiller]

I made too many mistakes in post #48. Here I will start with my equation from post #43 $x/x_o=|sec(\phi)+tan(\phi)|^{-v_o/v_1}$ and proceed with $dy/dx=tan(\phi)$ and $sec(\phi)=-\sqrt{(dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[(x/x_o)^{-v_1/v_o}-(x/x_o)^{v_1/v_o}]$. This integrates to give $\\$ $y=(x_o/2)[-(v_o/(v_1-v_o))(x/x_o)^{-(v_1-v_o)/v_o}-(v_o/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o)} ]+C$ $\\$ where $C=x_o v_o v_1/(v_1^2-v_o)^2$. The constant of integration occurs because at $x=x_o$ , $y=0$. To find where this curve intercepts the good ship, we set x=0. This gives $y_1=x_o v_o v_1/(v_1^2-v_o^2)$. Distance D traveled by the pirate ship is $D=(v_1/v_o)y_1=x_o v_1^2/(v_1^2-v_o^2)$.

My solution to this problem for x vs y, expressed parametrically in terms of $\theta$ is as follows:
$$x=x_0\left[1-\frac{1}{(\sec{\theta}+\tan{\theta})^{V/v}}\right]$$
$$y=x_0\left[\int_0^{\theta}{\frac{\sec^2{\theta '}d\theta '}{(\sec{\theta '}+\tan{\theta '})^{V/v}}}-\frac{\tan {\theta}}{(\sec{\theta}+\tan{\theta})^{V/v}}\right]$$where $\theta '$ is a dummy variable of integration.

Charles, your analytic solution to this problem should match mine, and should thus somehow provide the result of correctly integrating of my "mystery integral" in the equation for y. Could you please see if you can back out the integral evaluation? Thanks.

Chet

 

[Charles Link]

I made too many mistakes in post #48. Here I will start with my equation from post #43 $\\$ $x/x_o=|sec(\phi)+tan(\phi)|^{-v_o/v_1}$ and proceed with $dy/dx=tan(\phi)$ and $sec(\phi)=-\sqrt{(dy/dx)^2+1}$. A little algebra gives $dy/dx=(1/2)[(x/x_o)^{-v_1/v_o}-(x/x_o)^{v_1/v_o}]$. This integrates to give $\\$ $y=(x_o/2)[-(v_o/(v_1-v_o))(x/x_o)^{-(v_1-v_o)/v_o}-(v_o/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o)} ]+C$ $\\$ where $C=x_o v_o v_1/(v_1^2-v_o)^2$. The constant of integration occurs because at $x=x_o$ , $y=0$. To find where this curve intercepts the good ship, we set x=0. This gives $y_1=x_o v_o v_1/(v_1^2-v_o^2)$. Distance D traveled by the pirate ship is $D=(v_1/v_o)y_1=x_o v_1^2/(v_1^2-v_o^2)$.

@micromass With a little effort I found a couple of algebraic errors in my previous posts. Hopefully this one is now correct. $\\$ Adding on to this and making a couple of corrections, thanks to @Chestermiller post #47, I see I have an error in post 43, and my equation should read (instead of $v_o/v_1$) $\\$ $x=x_o|sec(\phi)+tan(\phi)|^{-v_1/v_o}$. $\\$ My equation above then becomes $\\$ $dy/dx=-(1/2)[(x/x_o)^{-v_o/v_1}-(x/x_o)^{v_o/v_1}]$ $\\$ where the addition of the minus sign (missing in the above quote) in front of the (1/2) comes as a result of the absolute value sign putting a minus sign on the dy/dx which for my choice of coordinates is always negative. Solving for y (integrating), the result is $\\$ $y= (1/2) x_o [(v_1/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o}-(v_1/(v_1-v_o))(x/x_o)^{(v_1-v_o)/v_1}]+C$ $\\$ where $C= x_o v_o v_1/(v_1^2-v_o^2)$. $\\$ The substitution $x=x_o-x'$ is then needed to get in the form of the OP because I did a coordinate transformation where I had the good ship sailing up the y-axis and the pirate ship beginning at $(x_o,0)$... Making these couple of corrections did not affect the distance $D$ that the pirate ship travels which is $D=x_o v_1^2/(v_1^2-v_o^2)$. Note: $v_1=V$ and $v_o=v$ of the OP.

 

[micromass]

Adding on to this and making a couple of corrections, thanks to @Chestermiller post #47, I see I have an error in post 43, and my equation should read (instead of $v_o/v_1$) $\\$ $x=x_o|sec(\phi)+tan(\phi)|^{-v_1/v_o}$. $\\$ My equation above then becomes $\\$ $dy/dx=-(1/2)[(x/x_o)^{-v_o/v_1}-(x/x_o)^{v_o/v_1}]$ $\\$ where the addition of the minus sign (missing in the above quote) in front of the (1/2) comes as a result of the absolute value sign putting a minus sign on the dy/dx which for my choice of coordinates is always negative. Solving for y (integrating), the result is $\\$ $y= (1/2) x_o [(v_1/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_o}-(v_1/(v_1-v_o))(x/x_o)^{(v_1-v_o)/v_1}+C$ $\\$ where $C= x_o v_o v_1/(v_1^2-v_o^2)$. $\\$ The substitution $x=x_o-x'$ is then needed to get in the form of the OP because I did a coordinate transformation where I had the good ship sailing up the y-axis and the pirate ship beginning at $(x_o,0)$... Making these couple of corrections did not affect the distance $D$ that the pirate ship travels which is $D=x_o v_1^2/(v_1^2-v_o^2)$. Note: $v_1=V$ and $v_o=v$ of the OP.

It still doesn't seem to be correct. If I graph the curve you posted, then I notice that if $v1>v0$, then the $y$ becomes negative, meaning that the the pirate ships goes in the negative direction first. This seems wrong.

 

[Charles Link]

It still doesn't seem to be correct. If I graph the curve you posted, then I notice that if $v1>v0$, then the $y$ becomes negative, meaning that the the pirate ships goes in the negative direction first. This seems wrong.

I will need to take another look at it but my x goes from $x_o$ to 0 as the $x'$ (which is the coordinate I should be using) goes from 0 to $x_o$. Did you include the constant of integration C in the graph ? I also missed a " ]" which I will now include.editing... I found the typo: THe denominator of the first exponent should be $v_1$. Hopefully that makes it work.

 

[Charles Link]

@micromass I found a typo: The denominator in the first exponent should be $v_1$. It correctly reads $\\$ $y= (1/2) x_o [(v_1/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_1}-(v_1/(v_1-v_o))(x/x_o)^{(v_1-v_o)/v_1}]+C$ $\\$ where $C= x_o v_o v_1/(v_1^2-v_o^2)$. $\\$ Hopefully this is now finally correct.

 

[micromass]

@micromass I found a typo: The denominator in the first exponent should be $v_1$. It correctly reads $\\$ $y= (1/2) x_o [(v_1/(v_1+v_o))(x/x_o)^{(v_1+v_o)/v_1}-(v_1/(v_1-v_o))(x/x_o)^{(v_1-v_o)/v_1}]+C$ $\\$ where $C= x_o v_o v_1/(v_1^2-v_o^2)$. $\\$ Hopefully this is now finally correct.

That seems ok, but the solution is invalid if $v_1=v_0$.

 

[Charles Link]

That seems ok, but the solution is invalid if $v_1=v_0$.

The $v_1=v_o$ is a special case that has the result $\\$ $y=(1/2)x^2/x_0-(1/2)x_o ln(x/x_o)-(1/2) x_o$ where again the substitution $x=x_o-x'$ is necessary to get it into the form of the original post.