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You've got to juggle the two equations, multiply that by 4, multiply that by 3, add and subtract and so on.
同时处理这两个方程,这个两边乘以4,另一个两边乘以3,将两方程相加或相减求出答案
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Take the difference of the two velocities and divide by the difference of the two times, and you've got the acceleration.
求出速度的差值,除以时间的差值,这样就能得到加速度
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First, it's pointing to the beginning of the list, which initially might be down here at but after a while, might be part way through. And to that, I simply add a halfway point, and then I check.
列表中间的一个部分了，然后我求出列表的中值点,然后看看该点的值是不是等于目标值，如果是的话就完成了，如果不是的话，如果中位值大于我要找的目标值。
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So I differentiated this object, this is my first derivative and I set it equal to 0 Now in a second I'm going to work with that, but I want to make sure i'm going to find a maximum and not a minimum, so how do I make sure I'm finding a maximum and not a minimum?
这样我就对它求出导数了,这是一阶导数,令它等于0,一会我们就要计算了,但我先确定一下是最大值还是最小值,我怎么确定是最大值还是最小值呢
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In my days, they would say, "Find out when you want an object to go through that point."
我读书的时候,老师要求我,"求出物体何时经过该点"
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You add them all up algebraically, keeping track of their sign, and that's the total force.
你再对这些力做代数加法,注意符号,就求出合力了
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I want you to find the speed so it will land here.
我希望你们求出初速度,使它落在这儿
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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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That's how by either balancing the unknown force with a known force or by simply measuring the acceleration as I fall towards this podium and multiplying by mass, you can find the force that exerts on me.
我们既可以让未知力与已知力平衡,也可以只测量我冲向讲台时加速度的大小,再与我的质量相乘,都能求出作用在我身上的力的大小
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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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No one has a clue on how I can go back from these to that?
难道没人能给出倒求出它们的方法吗
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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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Do this in your head and ask, "What can I possibly get?"
心算一下然后问问自己,"我能求出什么"
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I've got to find them by experimenting, by putting other bodies and seeing how they react and then finding out what's the force that acts on a body when it's placed in this or that situation.
我得通过实验找出这些力,我会增加其它物体,观察其反应,求出当它处在这种或者那种情况下时,作用在物体上的力
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Then go to the x equation and demand that this be equal to the desired x value and find the time.
然后列出关于 x 的方程,令这个式子等于给定的 x,并求出时间
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Once you got a = 2, you can go back and realize here that f = 4, so you got really 4 Newtons.
一旦你算出 a = 2,你可以回到上面来,求出 f = 4,所以实际上这个力有 4 牛
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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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If you give me a pair of numbers, Ax and Ay, that's as good as giving me this arrow, because I can find the length of the arrow by Pythagoras' theorem.
如果给我一组数字,Ax 和 Ay,就相当于给了我这个箭头示意图,因为我可以利用毕达哥拉斯定理定理求出模长
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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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Take a derivative of S1.
求出S1的导数
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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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We know it's going around in a circle because if I find the length of this vector, which is the x-square part, plus the y-square part, I just get r square at all times, because sine square plus cosine square is one.
我们之所以知道它做圆周运动,是因为我求出了这个矢量的模长,也就是 x 的平方加上 y 的平方,我就得到了它在任意时刻的模长平方,因为正弦平方加余弦平方始终等于1
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