January 14

It says that the financial cost of WWII in 1944 daollars is one trillion. How much would that be today? —Preceding unsigned comment added by Jwking (talk • contribs) 18:31, 14 January 2008 (UTC)[reply]

I humbly suggest Science or Humanities as more apropriate desks for this question. Anyway, according to this, such amount will correspond to 11.49 trillions in 2006 dollars. Pallida  Mors 19:06, 14 January 2008 (UTC)[reply]

It really depends on how you are calculating the historical exchange rate. This site discusses a variety of ways to calculate it; depending on what index you use, it can be anywhere between $9.6 trillion and $60 trillion. The numbers are only meaningful if you are considering things correctly; I'd lean towards the relative share of GDP (the $60 trillion figure) because that makes more sense in terms of big government spending than does the CPI (the $11.4 trillion figure above), which compares things like how much a loaf of bread costs at any given time. --24.147.69.31 (talk) 00:08, 16 January 2008 (UTC)[reply]

24 is right in pointing at the fact that there are many ways (or indices) to translate a sum of money from one year to another. 11.49 is the amount (of 2006 dollar trillions) one trillion in 1944 represents in purchase power. In other words, what could an average consumer adquire with such amount, translated to present money?

The $60 trillion figure 24 mentions is built like this: What percentage of America's GDP in 1944 did a trillion represent? Then, what's the current value of such a fraction of 2006's GDP?

I leave to the OP the task of choosing the right answer according with her/his needs. Pallida  Mors 21:29, 16 January 2008 (UTC)[reply]

A country's GDP grows partially because its economy is expanding, and not just because of inflation. Suppose inflation did not exist, and the current cost of all products are identical to the cost of the products in 1945. Calculating the cost of World War II in today's dollars, according to the 1945 percentages of their GDPs countries used, would be misleading. It would imply a much higher cost when expressed in today's dollars, even though the cost of the resources used is the same. In other words, the same resources will "cost" more simply because the country's economy is growing. --Bowlhover (talk) 08:10, 18 January 2008 (UTC)[reply]

I have just been introduced to parametric equations and I'm having a bit of a hard time finding the derivative of one.

Suppose I am using ${\displaystyle x=f(t)}$ and ${\displaystyle y=g(t)}$.

I am using the formula ${\displaystyle y-f(a)=f'(a)(x-a)\,\!}$ to find the tangent to a curve. For ${\displaystyle f(a)}$, am I right in saying that I want to find ${\displaystyle g(a)}$? 172.142.94.249 (talk) 20:20, 14 January 2008 (UTC)[reply]

The formula ${\displaystyle y-f(a)=f'(a)(x-a)\,\!}$ is for "normal" functions, not for parametric curves. The formula you need is ${\displaystyle f'(a)(y-g(a))=g'(a)(x-f(a))\;\!}$. -- Meni Rosenfeld (talk) 20:34, 14 January 2008 (UTC)[reply]

Thank you; everything is working out fine now. I wonder though, since I haven't been taught this and I can't find it in my textbook, is there a way of finding a tangent to a parametric curve at a particular point without using the formula you gave me? 172.142.94.249 (talk) 20:44, 14 January 2008 (UTC)[reply]

Depends on what you are allowed to use. The tangent is the line which gives a first order approximation to the curve. Suppose you have ${\displaystyle t\approx 0}$. Then your first order approximations for the coordinates of the point corresponding to ${\displaystyle a+t}$ are ${\displaystyle x=f(a+t)\approx f(a)+f'(a)t}$ and ${\displaystyle y=g(a+t)\approx g(a)+g'(a)t}$, so the point on the tangent corresponding to t satisfies ${\displaystyle x=f(a)+f'(a)t\;\!}$ and ${\displaystyle y=g(a)+g'(a)t\;\!}$. If you eliminate t from these equations, you will have the equation of the tangent. -- Meni Rosenfeld (talk) 20:53, 14 January 2008 (UTC)[reply]