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Please accept my present, which has been brought to you, for God has favored me, and I have plenty.
求你收下我带来给你的礼物，因为神恩待我，使我充足。
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and literally beg you for whatever they want like, if it's money, something to drink, food.
求你给他们一些东西，钱、喝的、食物，什么都可以。
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But then having found one answer, you can add to it anything that gets killed by the act of taking derivatives.
一旦你算出了一个答案,你就可以往式子里加入任何,在求完导后消失的项
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Well I'm relying on the contract, if you like, that the manufacturer of square root put together, which is, if I know I'm giving it two floats, which I do because I make sure they're floats, the contract, if you like, of square root says I'll give you back a float.
好吧我能靠一些类似合约来保证，如果这么说你喜欢的话,其实就是当求平方根的方法,获得了两个我已经确认了是浮点数的数后,这个类似合约的东西就会保证，返回一个浮点数。
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He greets his former rival and enemy with these words 1 This is in Genesis 33:10-11 If you would do me this favor, accept for me this gift, for to see your face is like seeing the face of God, And you have received me favorably.
他用以下的话来迎接往昔的仇敌,这段话出现在《创世纪》10章第10到11节中1,不然，我若在你眼前蒙恩，就求你从我手里收下这礼物，因为我见了你的面，如同见了神的面，并且你接纳了我。
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Now, you've already seen how entropy depends on temperature.
如何来求熵的表达式,现在你已经知道熵对温度的依赖关系。
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It's useful to have the average acceleration, which you can find by taking similar differences of velocities.
平均加速度也是很有用的,你可以用类似方法,取速度的变化量求出来
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Normally, I will give you a function and tell you to take any number of derivatives.
通常情况下,我会给你一个函数,然后让你求任意阶的导数
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you're looking for that equal and opposite forces acting at two ends of the rope.
你就得求出作用在绳子两端的,大小相等方向相反的力
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Of course, you can take a function and take derivatives any number of times.
当然,你可以随意拿一个函数,对它求任意阶的导数
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Another thing you can do is you can find the speed here.
你能做的另一件事就是你可以求出这里的速度
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The basic idea in solving these equations and integrating is you find one answer, so then when you take enough derivatives, the function does what it's supposed to do.
解决这类方程以及积分的基本思想就是,你求出一个解,然后进行多次求导,求导的结果就满足条件
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Everyone knows from calculus that if you're trying to find a function about which you know only the derivative, you can always add a constant to one person's answer without changing anything.
学过微积分的人都知道,如果你想根据已知的导数,求出其原函数,你总是可以给某人的答案,随便加一个常数,且不影响结果
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Once you can take one derivative, you can take any number of derivatives and the derivative of the velocity is called the acceleration, and we write it as the second derivative of position.
只要你能求一阶导数,你就能求任意阶的导数,速度的导数被称为"加速度",我们把它写成位移的二阶导数
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At any point on the graph you can take the derivative, which will be tangent to the curve at each point, and its numerical value will be what you can call the instantaneous velocity of that point and you can take the derivative over the derivative and call it the acceleration.
在图上的任意一点,你可以进行求导,得到曲线上每一点的切线斜率,所得到的数值,即为该点处的瞬时速度,然后你再求一次导,得出它的加速度
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You're going to get a chance to explore them, and you'll see not only are there the standard numerics for strings, there are things like length or plus or other things you can do with them.
以后你们还会学到更多，你会学到字符串,不仅仅能运用标准数值运算，你还可以对他们求长度,或者加上其他的东西。
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You can actually take derivatives of a vector with time.
你可以直接求位矢对时间的导数
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In some sense, strictly speaking they shouldn't be necessary because the fact that my specification starts with an assumption, says, hey you, who might call square root make sure that the things you call me with obey the assumption.
在某种意义上严格的来讲,它们不是必要的,因为我已经对假设做了声明,也就是说，如果你要调用我的，求平方根的函数,请确保你传递给我的,参数满足假设条件。
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You can easily check by taking two derivatives that this particle does have the acceleration a.
你可以很容易地通过求两次导来验证,这个质点的加速度确实为 a
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If you want to find the speed there, you put the equation v^=v0^-2g.
如果你想求出那一点的速度,可利用方程v^=v0^-2g
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Now, we are done because now we can ask how high does it go, and you go back to your y of 1 is 15+10-5, which is what?
现在问题解决了,因为我们可以算出,最高点的高度,你回到这个式子,求出y=15+10-5,是多少
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So, that is a tricky problem to begin with because if you take this formula here, it tells you y if you know t, but no, we're not saying that.
以这个问题刚开始的时候有点难度,因为如果你要用这个公式,求y的时候就一定要知道t,但是,我们不知道
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Mathematically complete problem is that you can find the function x of t by saying that the second derivative of the function is equal to -k over m times the function.
这个数学问题就是,你可以求出函数 x,只要令函数的二阶导数,等于 -k 除以 m 再乘以函数
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That means, when I take two derivatives, I want to get a, then you should know enough calculus to know it has to be something like at^, and half comes from taking two derivatives.
也就是说,当我想求二阶导时,得到了a,你应该有足够的微积分知识,才能知道必须有类似at^的项,而这个1/2则是因为求了两次导数
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The most important result from last time was that if you took this r, and you took two derivatives of this to find the acceleration, d^2 r over dt^2, try to do this in your head.
上节课最重要的结论,就是如果你把 r 写成这样,对 r 求两次导就能得到加速度,d^2 r / dt^2,心算一下
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You can add something else to the answer without invalidating it, and that is anything with one power of t in it, because if you take one derivative it'll survive, but if you take two derivatives, it'll get wiped out.
你可以往答案后面加上其它东西,而且不影响结果,也就是任何t的一次项,因为你求一次导数之后它还在,但如果你求两次导数,这一项就不存在了
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