# Ancora Imparo

## 28 February 2014

### Snail Problem, Part 5

Filed under: mathematics — Darmok @ 03:04 UTC
Tags: , ,

I hope no one is tired of snails yet. In the last several posts, I tackled the following problem: a snail crawls along a twig of length L0 at 1 cm/d, and the twig grows (along its entire length) at 2 cm/d. Will the snail reach the end, and when? In parts 1, 2, and 3, I solved this problem, and then in part 4 I turned to the general case, leaving the snail’s velocity and twig growth velocity unspecified.

We found that the snail’s position, x, can be given by the following equation, where g is the growth rate of the twig, s is the (inherent) velocity of the snail, and t is the time:

$\displaystyle x = \frac{s}{g}(gt+L_0)\ln\frac{gt+L_0}{L_0}.$

We want to find the time t where the snail reaches the end of the twig. Since the length of the twig is gt+L0, we set x equal to this.

$\displaystyle x = gt+L_0$

Substituting our equation for x gives

$\displaystyle \frac{s}{g}(gt+L_0)\ln\frac{gt+L_0}{L_0}=gt+L_0$.

To solve this for t, we start by dividing both sides by gt+L0 (note that this is the twig length, and is only applicable for nonzero twig lengths).

$\displaystyle \frac{s}{g}\ln\frac{gt+L_0}{L_0}=1$

We can multiply both sides by g/s (note that the snail velocity cannot be zero).

$\displaystyle \ln\frac{gt+L_0}{L_0}=\frac{g}{s}$

Exponentiating both sides gives

$\displaystyle e^{\ln\frac{gt+L_0}{L_0}}=e^{\frac{g}{s}}$

$\displaystyle \frac{gt+L_0}{L_0}=e^{\frac{g}{s}}$

$\displaystyle gt+L_0 = L_0 e^{\frac{g}{s}}$

$\displaystyle gt = L_0 e^{\frac{g}{s}}-L_0$

$\displaystyle gt = L_0(e^{\frac{g}{s}}-1)$

And finally, the time is

$\displaystyle t_f = \frac{L_0}{g}\left(e^{g/s}-1\right) \blacksquare$

So the time it takes is the initial length of the twig divided by its growth rate (which incidentally is the amount of time it would have taken to grow to its start position) times the quantity e to the power of the ratio of the twig and snail rates minus 1.

Interestingly, this will be finite no matter how small s is. If the snail is extremely slow, then the exponent g/s will be very large, but as long as s is positive, the snail will eventually reach the end of the twig — no matter how fast the twig grows!

Let’s also see how long the twig will be when the snail reaches the end. Recall that the length of the twig is

$\displaystyle L(t) = gt+L_0$

We can substitute our time above in for t:

$\displaystyle L(t_f) = g\left[\frac{L_0}{g}\left(e^{g/s}-1\right)\right] +L_0$

$\displaystyle L(t_f) = L_0\left(e^{g/s}-1\right)+L_0$

$\displaystyle L(t_f) = L_0\left[(e^{g/s}-1)+1\right]$

And finally, the length is

$\displaystyle L(t_f) = L_0 e^{g/s}\, \blacksquare$

## 27 February 2014

### Snail Problem, Part 4

Filed under: mathematics — Darmok @ 06:29 UTC
Tags: , , ,

In the last three posts (see Snail Problem parts 1, 2, and 3), I tackled the following problem: a snail crawls along a twig of length L0 at 1 cm/d, and the twig grows 2 cm/d. Will the snail reach the end, and when?

The answer is yes, and it will reach the end in

$\displaystyle t = (e^2-1)\cdot\frac{L_0}{2}.$

I’m interested in solving this now for the general case, leaving the twig growth and snail velocities unspecified. Fellow blogger Ken Roberts has already solved this, but I still want to try my hand at it. I’ll follow his example and use g for the growth rate of the twig (I’m already using t for time), s for the inherent velocity of the snail, and I’ll continue using L0 for the initial length of the twig (and L for its length at any time).

The length of the twig L is

$\displaystyle L(t)= gt+L_0$,

since it grows at g cm/d and its grown for t days. The velocity at any point x is length is the fraction of the length of the twig,

$\displaystyle \frac{x}{L(t)} = \frac{x}{gt + L_0}$

times the twig’s overall growth rate, g:

$\displaystyle \frac{gx}{gt + L_0}$.

The net velocity v of the snail is the sum of the growth of the twig at that point (which we just calculated) and its own inherent velocity, s:

$\displaystyle v = \frac{gx}{gt + L_0} + s$

Of course, v is the derivative of the snail’s position x with respect to time. Remember that L0, g, and s are constants.

$\displaystyle \frac{dx}{dt} = \frac{gx}{gt + L_0} + s$

This is a linear first-order differential equation.

$\displaystyle \frac{dx}{dt} - \frac{g}{gt + L_0}x = s$

The way to solve this is to multiply by the following integrating factor (see part 2 for an explanation):

$\displaystyle \mu(t) = e^{\int \left(-\frac{g}{gt + L_0}\right) \,dt}$

$\displaystyle \mu(t) = e^{-\int g\cdot\frac{1}{gt + L_0}\,dt}$

$\displaystyle \mu(t) = e^{-\ln|gt+L_0| + c_1}$

I actually might prefer doing this the other way (the longer method from the previous post), since integrating inside an exponent is inconvenient. Using sum and product rules for exponents:

$\displaystyle \mu(t) = e^{c_1}{\left(e^{ln |gt+L_0|}\right)}^{-1}$

$\displaystyle \mu(t) = c_2(gt+L_0)^{-1}$

I got rid of the absolute value because the quantity gt + L0 will always be positive, since it’s the length of the twig. And more generally, we can select a quantity for the constant that would have the appropriate sign. We just need one possible solution for this integrating factor, so I’ll select the one where the constant is equal to one.

$\displaystyle \mu(t) = (gt+L_0)^{-1}$

Now we take this and multiply it by our prior differential equation.

$\displaystyle (gt+L_0)^{-1}\left(\frac{dx}{dt} - \frac{g}{gt + L_0}x\right) = (gt+L_0)^{-1}s$

$\displaystyle (gt+L_0)^{-1}\frac{dx}{dt}-g(gt+L_0)^{-2}x = \frac{s}{gt+L_0}$

The whole point of this diversion was to get the left side of the equation into a form that is the inverse of the product rule.

$\displaystyle \frac{d}{dt}\left[(gt+L_0)^{-1}x\right] = \frac{s}{gt+L_0}$

Integrating both sides with respect to t gives

$\displaystyle (gt+L_0)^{-1}x = \int\frac{s}{gt+L_0}\,dt$

$\displaystyle (gt+L_0)^{-1}x = s\int\frac{1}{gt+L_0}\,dt$

We need to set up the inverse of the chain rule.

$\displaystyle (gt+L_0)^{-1}x = \frac{s}{g}\int\frac{g}{gt+L_0}\,dt$

$\displaystyle (gt+L_0)^{-1}x = \frac{s}{g}\left(\ln|gt+L_0| + c_3\right)$

We can multiply through by s/g, and define a new constant. Also, I’ll remove the absolute value, since I’m only interested in positive t.

$\displaystyle (gt+L_0)^{-1}x=\frac{s}{g}\ln(gt+L_0) + c_4$

$\displaystyle x = \left[\frac{s}{g}\ln(gt+L_0) + c_4\right](gt+L_0)$

We can find the constant by noting that when t=0, x=0:

$\displaystyle 0 = \left(\frac{s}{g}\ln L_0 + c_4\right)(L_0)$

Dividing both sides by L0 gives

$\displaystyle 0 = \frac{s}{g}\ln L_0 + c_4$

$\displaystyle c_4 = -\frac{s\ln L_0}{g}$

Substituting this back into our equation for x:

$\displaystyle x = \left[\frac{s}{g}\ln(gt+L_0) -\frac{s\ln L_0}{g}\right](gt+L_0)$

$\displaystyle x = \frac{s}{g}\left[\ln(gt+L_0)-\ln L_0\right](gt+L_0)$

Let’s combine the natural logarithms and rearrange the terms for clarity:

$\displaystyle x= \frac {s}{g}(gt+L_0)\ln\frac{gt+L_0}{L_0}$

That’s it! Noting that the quantity gt+L0 represents the current twig length, L, we can say that

$\displaystyle x = \frac{s}{g}L\ln\frac{L}{L_0}$.

In other words, at any given time, the position of the snail is equal to the ratio of the snail’s inherent velocity to the twig growth velocity times the current twig length times the natural logarithm of the ratio of the twig’s current length to its original length. Keeping it in this form, using the general case rather than s=1 and g=2, certainly makes the formula more intuitive. I’ll solve for the time it takes to reach the end of the twig in the next post.

## 23 February 2014

### Snail Problem, Part 3

Filed under: mathematics — Darmok @ 20:28 UTC
Tags: , , ,

This is a continuation of my attempt to solve the following problem: A snail crawls along a twig of length L0 at 1 cm/d, and the twig grows (along its entire length) at 2 cm/d. Will the snail reach the end, and when? You can see part 1 where I successfully set up the differential equation and unsuccessfully tried to solve it, and part 2 where I did manage to solve it.

At the end of the last post, I had solved for the equation that describes the position of the snail over time,

$\displaystyle x = \left(t+\frac{L_0}{2}\right)\ln\left(\frac{2t}{L_0}+1\right)$.

Also, recall that the length of the twig is

$\displaystyle L = 2t + L_0$

since it grows at 2 cm/d. We want to find the time t when these two are equal.

$\displaystyle 2t + L_0 = \left(t+\frac{L_0}{2}\right)\ln\left(\frac{2t}{L_0}+1\right)$

[Edit: I missed a very obvious step at this point, so the next few steps will be unhelpful. Free free to skip ahead to the next edit.] We need to isolate the natural logarithm.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2t+L_0}{t+\frac{L_0}{2}}$

Exponentiating both sides gives

$\displaystyle e^{\ln\left(\frac{2t}{L_0}+1\right)} = e^{\frac{2t+L_0}{t+\frac{L_0}{2}}}$

$\displaystyle \frac{2t}{L_0}+1 = e^{(2t+L_0)(t+\frac{L_0}{2})^{-1}}$

Well, the left side is great, but the right side is a mess.

$\displaystyle \frac{2t}{L_0}+1 = \left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{(2t+L_0)}$

If I bring the 2tL0 term out, I could split it up.

$\displaystyle \frac{2t}{L_0}+1 = \left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{(2t+L_0)}$

$\displaystyle \frac{2t}{L_0}+1 = \left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{2t}\left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{L_0}$

This was unhelpful and I have no idea how to simplify this.

[Edit: At this point I get back on track.] Let’s go back to the last simple equation.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2t+L_0}{t+\frac{L_0}{2}}$

I should get rid of this complex fraction, even if it means making the numerator more complex.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2t+L_0}{t+\frac{L_0}{2}}\cdot\frac{2}{2}$

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{4t+2L_0}{2t+L_0}$

Oh, it’s so obvious now. I don’t know how I missed realizing that this fraction would simplify.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2(2t+L_0)}{2t+L_0}$

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = 2$

I feel a bit foolish. Now, let’s exponentiate both sides.

$\displaystyle e^{\ln\left(\frac{2t}{L_0}+1\right)} = e^2$

That’s much simpler.

$\displaystyle \frac{2t}{L_0}+1 = e^2$

$\displaystyle \frac{2t}{L_0} = e^2 - 1$

$\displaystyle t = (e^2-1)\cdot\frac{L_0}{2}$.

Here are my initial observations: It seems “elegant” enough to be a solution to this problem. I like that it depends on e. I also notice that the time it takes is directly proportional to the initial length of the twig. I’m a bit surprised, as the snail’s net speed is not constant but rather increases as it progresses. Also, it’s apparent in retrospect that the first factor, $\left(t+\frac{L_0}{2}\right)$ in the equation for the snail’s position over time,

$\displaystyle x = \left(t+\frac{L_0}{2}\right)\ln\left(\frac{2t}{L_0}+1\right)$,

is half the length of the twig at that time, L(t):

$\displaystyle x(t) = \frac{1}{2}L(t)\ln\left(\frac{2t}{L_0}+1\right)$.

I don’t have an intuitive understanding of what the other quantity could represent, nor do I know if the 2 and 1 have any connection to the twig growth and snail inherent velocity. It would be interesting to repeat this but leave all the quantities unspecified.

## 22 February 2014

### Snail Problem, Part 2

Filed under: mathematics — Darmok @ 23:57 UTC
Tags: , , ,

I read through some more of my college differential equations textbook and I think I am ready to further tackle the snail problem (see yesterday’s post). In short, a snail crawls along a twig of length L0 at 1 cm/d, and the twig grows 2 cm/d. Will the snail reach the end, and when?

I got as far as this differential equation, which I believe is correct:

$\displaystyle \frac{dx}{dt} = \frac{2x}{2t + L_0} + 1$

I read about a technique to solve linear first-order differential equations last night, so I’m going to try it here. First we put this into standard form:

$\displaystyle \frac{dx}{dt} -\frac{2}{2t + L_0}\cdot x = 1.$

Since we already have an x’ and an x, if only they were multiplied by another function μ and μ’ respectively, we would have an inverse of the product rule (μx)’ = (μx’ + μ’x). Of course we can’t multiply different parts of the equation by different functions; we’ll have to multiply the entire equation by the same function μ(t). This will put a μ term in front of the x’, so we have to select a μ that, when multiplied by the function in the second term, will result in μ’. Let’s take a detour to find this function.

$\displaystyle \mu' = \mu\left( -\frac{2}{2t + L_0} \right)$

This is a separable differential equation; we can get the μ terms on one side and the t terms on the other. (Incidentally, I don’t think we could have done that with the original equation, or I would have.)

$\displaystyle \frac{d\mu}{dt} = \mu\left( -\frac{2}{2t + L_0} \right)$

$\displaystyle \frac{1}{\mu}\,d\mu = -\frac{2}{2t+L_0}\,dt$

I just realized we divided both sides by μ. I hope that was “allowed.” Of course we would have lost any solutions where μ = 0, though I’m not sure of the significance of that. In any case, I can’t stop now. Let’s integrate:

$\displaystyle \int \frac{1}{\mu}\,d\mu = \int \left( -\frac{2}{2t+L_0} \right) \,dt$

The left side is easy; it’s a standard rule.

$\displaystyle \ln |\mu|= \int \left( -\frac{2}{2t+L_0} \right) \,dt$

The right side is essentially the same; we need to use the inverse of the chain rule.

$\displaystyle \ln |\mu| = - \int \left( \frac{1}{2t+L_0} \right) (2) \,dt$

$\displaystyle \ln |\mu| = - \ln | 2t+L_0 | + c_1$

Exponentiating both sides yields

$\displaystyle e^{ \ln |\mu| } = e^{- \ln | 2t+L_0 | + c_1}$

$\displaystyle |\mu| = {e^{c_1}} {\left( e^{ln |2t+L_0|} \right)}^{-1}$

Setting $c_2 := e^{c_1}$

$\displaystyle |\mu| = c_2 {|2t+L_0|}^{-1}$

$\displaystyle \mu = \pm \frac{c_2}{2t+L_0}$

I think we can absorb the ± into the constant (though I’m not positive).

$\displaystyle \mu = \frac{c_3}{2t+L_0}$

Actually, we need a solution, not all solutions, so I’ll set the constant equal to 1. So dropping the ± was fine.

$\displaystyle \mu = {(2t+L_0)}^{-1}$

Armed with μ, we should now be ready to return to our earlier work. We’ll take our previous equation, in standard form,

$\displaystyle \frac{dx}{dt} -\frac{2}{2t + L_0}\cdot x = 1$

and multiply both sides by the function μ:

$\displaystyle {(2t+L_0)}^{-1} \cdot \frac{dx}{dt} - 2{(2t + L_0)}^{-2} \cdot x = {(2t+L_0)}^{-1}$

We should now be able to a use the inverse of the product rule.

$\displaystyle \frac{d}{dt} \left[{(2t+L_0)}^{-1} x \right] = {(2t+L_0)}^{-1}$

Integrating both sides gives

$\displaystyle {(2t+L_0)}^{-1} x = \int {(2t+L_0)}^{-1}\,dt$

We’ll need to use the inverse of the chain rule again, so let’s set it up.

$\displaystyle \frac{x}{2t+L_0} = \frac{1}{2}\int 2{(2t+L_0)}^{-1}\,dt$

$\displaystyle \frac{x}{2t+L_0} = \frac{1}{2} \left(\ln |2t+L_0| + c_4\right)$

$\displaystyle \frac{x}{2t+L_0} = \frac{ \ln |2t+L_0|}{2} + c_5$

Multiplying both sides by 2tL0 gives

$\displaystyle x = \left( \frac{\ln |2t+L_0|}{2} + c_5 \right) (2t+L_0)$

Wow, we already solved for x. I wasn’t expecting it so soon. The solution is a little more complex than I’d expected, and I suspect I made a mistake somewhere. However, we should keep going. Let’s find the constant. The snail starts at position x = 0 at time t = 0, so

$\displaystyle 0 = \left( \frac{\ln |0+L_0|}{2} + c_5 \right) (0 + L_0)$

L0 is positive, so we don’t need the absolute value.

$\displaystyle 0 = \left( \frac{\ln L_0}{2} + c_5 \right)L_0$

$\displaystyle 0 = \frac{L_0 \ln L_0}{2} + c_5 L_0$

$\displaystyle c_5 L_0 = - \frac{L_0 \ln L_0}{2}$

$\displaystyle c_5 = - \frac{\ln L_0}{2}$

This is actually promising. Substituting c5 back into the equation we solved for x,

$\displaystyle x = \left( \frac{\ln |2t+L_0|}{2} + c_5 \right) (2t+L_0)$

$\displaystyle x = \left( \frac{\ln |2t+L_0|}{2} - \frac{\ln L_0}{2} \right) (2t+L_0)$

$\displaystyle x = \left( \frac{\ln |2t+L_0| - \ln L_0}{2} \right) (2t+L_0)$

I want to get rid of the absolute value. I think we can. The natural logarithm is only defined for positive numbers (at least without using complex numbers), so the absolute value allows our function to be defined for all nonzero arguments (just like $1/x$). If I remove the absolute value function, then the logarithm and therefore the equation will not be defined for 2t + L0 < 0, or when t < –L0/2. Since negative time is meaningless in this example (the snail started at time t = 0), I don’t mind discarding those times. (As an aside, I haven’t kept the units, but the 2 in the denominator is actually 2 cm/d — the speed at which the twig grows. The time –L0/2 corresponds to the time at which the twig had length zero. Clearly it would be meaningless in this problem to consider times before this.)

Back to the problem:

$\displaystyle x = \left[ \frac{\ln (2t+L_0) - \ln L_0}{2} \right] (2t+L_0)$

$\displaystyle x = [\ln (2t+L_0) - \ln L_0] \cdot \frac{2t+L_0}{2}$

$\displaystyle x = \ln \frac{2t+L_0}{L_0} \cdot \left(t + \frac{L_0}{2}\right)$

$\displaystyle x = \left(t+\frac{L_0}{2}\right) \ln \left(\frac{2t}{L_0} + 1\right)$

And that’s it! That is the equation for the distance the snail has crawled, x, at any given time t. We haven’t actually answered the question of if the snail will reach the end and when that would be, but we can save that for the next post.

### Snail Problem

Filed under: mathematics — Darmok @ 01:44 UTC
Tags: , , ,

A friend mentioned this problem a while back, and while I was able to use a computer to get an accurate numerical solution, I really want to find an analytical solution using math. The problem is that it will require differential equations, which I no longer remember. Edit: I haven’t yet been able to solve this problem, but I’m preserving my initial attempts below. See Part 2 where I made more progress.

The problem is this: a snail starts crawling along a twig, from one end towards the other. The snail can crawl at 1 cm/d. But the twig is growing at 2 cm/d. The twist is that the twig is adding material along its entire length, so the snail will be carried forward a bit, too, depending on where it is. Will the snail ever reach the end? Let’s let L0 be the initial length of the twig.

Let’s say the snail crawls at 1 cm/d and the twig grows at 2 cm/d. The length of the twig at time t would be

$\displaystyle L(t) = L_0 + 2t$.

For any point along the twig, the velocity will be proportional to the distance along the length of the twig, so at a distance x

$\displaystyle v_t(x, t) = \frac{x}{L_0 +2t} \cdot 2 = \frac{2x}{L_0 + 2t}$

Let’s double-check this: the beginning of the twig is $x = 0$:

$\displaystyle v_t(0,t) = \frac{2 \cdot 0}{L_0 + 2t} = 0$.

And at the end of the twig, $x = L_0 + 2t$:

$\displaystyle v_t(L_0+2t,t) = \frac{2(L_0+2t)}{L_0+2t} = 2$.

The velocity of the snail will include both its movement along the twig (at 1 cm/d) and the growth of the twig. It will be therefore

$\displaystyle v_s(x,t) = \frac{2x}{L_0+2t} + 1$

Since the velocity is the derivative of its position,

$\displaystyle \frac{dx}{dt} = \frac{2x}{L_0 + 2t} + 1$

and we want to know if and when $x(t) = L(t)$. In other words, solve for t:

$\displaystyle x(t) = L_0 + 2t$

Taking the derivative with respect to t,

$\displaystyle \frac{d}{dt} x(t) = \frac{d}{dt} (L_0 + 2t)$

$\displaystyle \frac{dx}{dt} = 2$

Note that this makes sense, at the time t when the snail reaches the end, … wait, no it doesn’t, that’s not correct. Since the functions are only equal at a specific time t, I can’t differentiate them and expect them to be equal.

Going back:

$\displaystyle \frac{dx}{dt} = \frac{2x}{L_0 + 2t} + 1$

Is this an linear differential equation? I hope so, since I’m hoping that will make it easier to solve (though at this point, I don’t remember how to solve linear or nonlinear differential equations.

$\displaystyle \frac{dx}{dt} - \frac{2}{L_0 + 2t} \cdot x = 1$.

I think I got it into the correct form. I don’t yet know what to do from here, so I’ll read on and also refresh my memory of differentiation rules.

I actually don’t think we’ll need complex differential equation techniques. We should be able to just integrate this entire equation with respect to t. I think I see why linear differential equations are easier to solve.

$\displaystyle \int ( \frac{dx}{dt} - \frac{2}{L_0 + 2t} \cdot x )\,dt = \int 1\,dt$

By the sum rule,

$\displaystyle \int \frac{dx}{dt} \,dt - \int \frac{2}{L_0+2t} \cdot x\,dt = \int 1\,dt$.

I spoke too soon. The first term is easy to integrate and the last is trivial, but I don’t know how to integrate the second.

$\displaystyle x + c_1 - \int \frac{2}{L_0+2t} \cdot x\,dt = t + c_2$

I can rearrange the terms and combine the constants

$\displaystyle \int \frac{2}{L_0+2t} \cdot x\,dt = x - t + c_3$

and I’m now stuck.

OK, I think we might be able to use integration by parts, using the formula

$\displaystyle \int u\,dv = uv - \int v\,du$.

If I set

$\displaystyle u =\frac{2}{L_0+2t}$

and set

$\displaystyle dv = x\,dt$

then to find v, we need to integrate:

$\displaystyle v = \int x\,dt$

$\displaystyle v = \frac{x^2}{2} + c_4$.

And to find du, we need to take the derivative of u. I’m just going to use the product rule, since I’m rusty at this and I don’t remember the quotient rule.

$\displaystyle du = \frac{d}{dt}\frac{2}{2t+L_0} = \frac{d}{dt}[2\cdot (2t+L_0)^{-1}]$

$\displaystyle du = \left(\frac{d}{dt}2\right) \cdot \frac{1}{2t+L_0} + 2 \cdot \frac{d}{dt}(2t+L_0)^{-1}$

$\displaystyle du = 0 + 2 \cdot (-1)\cdot(2t+L_0)^{-2}\cdot 2$

I’m going to keep going, but I just realized that I am not going to be able to integrate $\int v\,du$.

$\displaystyle du = -\frac{4}{(2t+L_0)^2}$

Shoot.

$\displaystyle uv - \int v\,du = \frac{2}{2t+L_0}\cdot \left(\frac{x^2}{4}+c_4 \right) + \int \left(\frac{x^2}{4}+c_4 \right) \left[\frac{4}{(2t+L_0)^2}\right]\,dt$.

This is a disaster.

I will have to study this more and return to this problem later.

## 8 January 2009

### Individual Action is Not Enough

Filed under: environment,Uncategorized,video — Darmok @ 06:25 UTC
Tags: , ,

The Canadian chapter of the World Wildlife Federation produced this cool video/commercial:

Most people try, to at least some degree, to take steps to help the environment. And these small changes, when summed across the whole population, are significant. But still, the collective action of individuals can only do so much — government and industry need to be on board, too. Unfortunately, in the United States, leadership from the federal government has been lacking (and at times, actively impedes) so state and local governments and industry have had to take their own steps. There is some hope, though, that this will change when the Obama administration takes office (I hope to write more on this in a later post).

## 1 December 2008

### Questions on How to Live a Green Lifestyle

Filed under: environment — Darmok @ 05:25 UTC
Tags:

Most Americans are interested in living more environmentally friendly lifestyles, yet it’s not always easy to know which practices really are best. New Scientist takes a look at these “Dumb Eco-questions You Were Afraid to Ask”. Hope this helps clear up some doubts or misconceptions.

## 20 November 2008

### President-elect Obama Delivers Strong Statement on Climate Change

As we prepare to move past President Bush’s disastrous environmental policies, I’ve been interested to see what President-elect Obama plans to do for the environment. The economy has garnered the most attention, and in the short term, is more important. But continued neglect of the environment will, in the long-term, lead to crises both in the economy and in other sectors.

President-elect Obama addressed the attendees of the Governor’s Global Climate Summit in a four-minute video (high-resolution version is available at change.gov; full text of speech at the end of this post).

He thanked the governors for their work (Governor Schwarzenegger of California along with governors of other U. S. states are hosting the Governor’s Global Climate Summit; leaders of key nations around the world are attending) and also thanked businesses for their efforts, going on to remark “But too often, Washington has failed to show the same kind of leadership. That will change when I take office. My presidency will mark a new chapter in America’s leadership on climate change that will strengthen our security and create millions of new jobs in the process.”

President-elect Obama went on to deliver more specific goals: “That will start with a federal cap-and-trade system. We’ll establish strong annual targets that set us on a course to reduce emissions to their 1990 levels by 2020 and reduce them an additional 80 percent by 2050. Further, we’ll invest $15 billion each year to catalyze private sector efforts to build a clean energy future”, indicating plans to invest in renewable resources as well as nuclear power and clean coal technology. He intends for this to help the economy as well, creating jobs and helping industry. Mr. Obama also indicated a change in the way the U. S. has participated on the international stage, stating that the U. S. would work with and depend on other nations: “And once I take office, you can be sure that the United States will once again engage vigorously in these negotiations, and help lead the world toward a new era of global cooperation on climate change.” Perhaps the most significant statement is the strong importance Mr. Obama still places on environmental problems, despite the problems with the economy. As John Broder writes in the New York Times, “State officials and environmental advocates were cheered that Mr. Obama choose to address climate change as only the second major policy area [after the economy] he has discussed as president-elect.” Reaction from environmental groups appears quite favorable. The CEO of the World Wildlife Fund (WWF) praised President-elect Obama’s remarks: “Today President-elect Obama gave us his first official statements on climate and without a doubt he nailed it. He sees clearly the huge risk that climate change poses to our economy and our future, and he understands that solving climate change is a foundation for a global economic recovery. Writing in the Sierra Club blog, Heather Moyer called the speech “very enjoyable”. And Peter Miller, in the National Resources Defense Council blog, wrote “Looking very presidential, Obama enunciated an unambiguous commitment to enacting a federal cap and trade program with tight annual caps leading to an 80% reduction in emissions by 2050. The contrast with President Bush’s stance on climate change was abundantly evident to everyone. It was the first time I’ve ever seen a standing ovation for a video.” I look forward to more. Below is a transcript of the speech, taken from Grist with slight editing. Let me begin by thanking the bipartisan group of U.S. governors who convened this meeting. Few challenges facing America — and the world — are more urgent than combating climate change. The science is beyond dispute and the facts are clear. Sea levels are rising. Coastlines are shrinking. We’ve seen record drought, spreading famine, and storms that are growing stronger with each passing hurricane season. Climate change and our dependence on foreign oil, if left unaddressed, will continue to weaken our economy and threaten our national security. I know many of you are working to confront this challenge. In particular, I want to commend Governor Sebelius, Governor Doyle, Governor Crist, Governor Blagojevich and your host, Governor Schwarzenegger — all of you have shown true leadership in the fight to combat global warming. And we’ve also seen a number of businesses doing their part by investing in clean energy technologies. But too often, Washington has failed to show the same kind of leadership. That will change when I take office. My presidency will mark a new chapter in America’s leadership on climate change that will strengthen our security and create millions of new jobs in the process. That will start with a federal cap-and-trade system. We’ll establish strong annual targets that set us on a course to reduce emissions to their 1990 levels by 2020 and reduce them an additional 80 percent by 2050. Further, we’ll invest$15 billion each year to catalyze private sector efforts to build a clean energy future. We’ll invest in solar power, wind power, and next generation biofuels. We’ll tap nuclear power, while making sure it’s safe. And we will develop clean coal technologies.

This investment will not only help us reduce our dependence on foreign oil, making the United States more secure. And it will not only help us bring about a clean energy future, saving the planet. It will also help us transform our industries and steer our country out of this economic crisis by generating five million new green jobs that pay well and can’t be outsourced.

But the truth is, the United States can’t meet this challenge alone. Solving this problem will require all of us working together. I understand that your meeting is being attended by government officials from over a dozen countries, including the U.K., Canada, Mexico, Brazil and Chile, Poland and Australia, India and Indonesia. And I look forward to working with all nations to meet this challenge in the coming years.

Let me also say a special word to the delegates from around the world who will gather at Poland next month: your work is vital to the planet. While I won’t be president at the time of your meeting and while the United States has only one president at a time, I’ve asked members of Congress who are attending the conference as observers to report back to me on what they learn there.

And once I take office, you can be sure that the United States will once again engage vigorously in these negotiations, and help lead the world toward a new era of global cooperation on climate change. Now is the time to confront this challenge once and for all. Delay is no longer an option. Denial is no longer an acceptable response. The stakes are too high. The consequences, too serious.

Stopping climate change won’t be easy. It won’t happen overnight. But I promise you this: When I am president, any governor who’s willing to promote clean energy will have a partner in the White House. Any company that’s willing to invest in clean energy will have an ally in Washington. And any nation that’s willing to join the cause of combating climate change will have an ally in the United States of America. Thank you.

## 2 September 2008

### Sarah Palin’s Anti-Science and Anti-Environment Policies Are Worrisome

Filed under: environment,global warming,politics,science — Darmok @ 06:37 UTC
Tags: ,

Senator John McCain, the presumptive Republican presidential nominee, just announced Alaska governor Sarah Palin as his running mate. She was a surprise pick and is relatively unknown, but what I’ve found so far is somewhat disturbing. While I haven’t made my final electoral decision, what I do know is that I don’t want another George W. Bush.

Wired Science, part of the Wired blog network, discusses her views on teaching creationism in public school science classes. (Merriam-Webster defines “creationism” as “a doctrine or theory holding that matter, the various forms of life, and the world were created by God out of nothing and usually in the way described in Genesis [the first book of the Judeo-Christian Bible].”) They refer to an article in the Anchorage Daily News covering a 2006 Alaska gubernatorial debate:

The volatile issue of teaching creation science in public schools popped up in the Alaska governor’s race this week when Republican Sarah Palin said she thinks creationism should be taught alongside evolution in the state’s public classrooms.

Palin was answering a question from the moderator near the conclusion of Wednesday night’s televised debate on KAKM Channel 7 when she said, “Teach both. You know, don’t be afraid of information. Healthy debate is so important, and it’s so valuable in our schools. I am a proponent of teaching both.”

The article goes on to point out:

The Republican Party of Alaska platform says, in its section on education: “We support giving Creation Science equal representation with other theories of the origin of life. If evolution is taught, it should be presented as only a theory.”

This stance alone is a significant strike against her. However, her anti-environment policies are also troubling. For instance, she told NewsMax, “I’m not one though who would attribute [global warming] to being man-made.” As I discussed in a previous post, all major scientific societies concur that humans are responsible for climate change. Senator McCain, as well as Democratic nominee Senator Barack Obama and his running mate Senator Joe Biden, all agree that climate change is a real threat and have proposed plans to combat it.

It’s not surprising, therefore, that her policies appear to show general disregard for the environment, especially with regards to her strong advocacy for oil drilling. For instance, she stated, “I beg to disagree with any candidate who would say we can’t drill our way out of our problem…”, as quoted in Investor’s Business Daily (IBD) and “When I look every day, the big oil company’s building is right out there next to me, and it’s quite a reminder that we should have mutually beneficial relationships with the oil industry” as quoted in Roll Call. She supports opening up the Arctic National Wildlife Refuge (ANWR, commonly pronounced “AN-war”) for drilling, a move generally opposed by environmentalists as well as Congress. Expressing her frustation, she stated to IBD, “But these lands [ANWR] are locked up by Congress, and we are not allowed to drill to the degree America needs the development…”; to Lawrence Kudlow on CNBC, “Very, very disappointed in Congress though [for not voting on drilling in ANWR]”; and so on. Both Senators Obama and McCain opposing drilling in ANWR, and she has attacked Senator McCain for this stance: “I have not talked him into ANWR yet…I think we need McCain in that White House despite, still, the close-mindedness on ANWR” (Lawrence Kudlow, CNBC).

Nor has Alaska, under Mrs. Palin’s governorship, promoted environmental issues. In Massachussets v. Environmental Protection Agency, when twelve states as well as several cities and environmental organizations sued the EPA to regulate carbon dioxide and other greenhouse gases, Alaska argued against them. (In a split decision, the Supreme Court largely agreed with Massachussets et al; see my previous post.)

Earlier this year, the Interior Department listed the polar bear as a threatened species under the Endangered Species Act (ESA). Somewhat bizzarrely, Governor Palin claims that polar bears are not threatened (“In fact, the number of polar bears has risen dramatically over the past 30 years” she states). She opposed the ESA listing and Alaska now plans to sue the Interior Department. Similarly, Governor Palin is opposing plans to list beluga whales as endangered, as it could damage Alaska’s economy.

Eight years of disregard for science and for the environment is enough; I don’t think I want to see someone like this in high office, certainly not in a position where she could become president. If anyone has any examples of Governor Palin promoting science or the environment, please let me know.

## 29 May 2008

### Incredible Photograph of Phoenix landing on Mars

Filed under: astronomy,science,visualization — Darmok @ 05:55 UTC
Tags: , , ,

Phoenix, a NASA robotic probe, landed successfully on Mars on May 25. It landed in the north polar region of Mars, at around the equivalent latitude of northern Alaska, and it will study Mars’ soil to look for clues of past water patterns and if it was ever hospitable to life.

Incredibly, as it parachuted down towards the surface, its picture was taken by a satellite orbiting Mars, the Mars Reconnaissance Orbiter (MRO)! From an amazing  distance of 750 kilometers (470 miles), it snapped this photograph of Phoenix parachuting towards Mars. This is the first time one probe has photographed another landing on a planet.

See full-sized version. Credit: NASA/JPL-Caltech/University of Arizona

To see how this fits in to the landing, take a look at this cool animation of Phoenix landing, produced by MAAS Digital and NASA.

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