VOA: standard.other
If it's in polar form I passed in a radius and angle and I'll compute what the x- and y- value is.
以及半径和角度,但是现在是这样的,不管我是以哪种形式。
OK, as I said, I want equality in the case of points to be, are the x- and y- coordinates the same?
好,正如我所说,我想要这个例子中的,点相等的意思是?
If you solve for that, you find y-y0=v0^/, and if you put in the v_0 I gave you, which was what, 10?
如果你解这个式子,你会得到y-y0=v0^/,如果再把我给你的v0代入,那个数是多少 10
This actually I think derives from a Greek word, "Synkope, " S-y-n-k-o-p-e, synkope.
这个术语我觉得来源于希腊语,Synkope,S-y-n-k-o-p-e
Right now it's a simple template, but it's a template for creating what a class looks like, and I now have an x- and y- value associated with each instance of this.
那么大家明白了为什么,我说类是一个模板了,对不对?,现在它只是一个简单的模板,但是它是一个用来-,创建形成一个类的模板。
I was expecting to compare x- and y- values, not radius and angle.
噢,发生了什么?,好吧有错误了。
I said both the x- and y- coordinates are bigger, then I'm going to return something to it.
我说过如果x和y坐标,都是更大的。
I take the difference in the x-values, squared, the difference in the y-values, squared, add them up, take the square root of that.
好,是毕达哥拉斯定力对不对?,求x坐标的差,然后平方,求y坐标的差,然后平方,把它们加起来,开平方。
Then, I gave you one other very important example of a particle moving in the x-y plane.
下面我再拿一个重要的例子,质点在 x-y 平面内的运动
It says, if I want to print out something I built in Cartesian form up here, says, again, I'm going to pass it in a pointer to the instance, that self thing, and then I'm going to return a string that I combine together with an open self and close paren, a comma in the middle, and getting the x-value and the y-value and converting them into strings before I put the whole thing together.
这不仅仅是个列表,它是怎么来做的?,流程是:如果我想要返回,一些已经在笛卡尔模式下建好的值,好,再说一遍,我首先要传入一个,指向实例的指针,也就是,然后我会返回一个,由开括号,闭括号,中间的一个逗号,以及提前转换为字符串格式的。
So I don't know don't, John, I would argue if I'd written this better, I would have had a method that returned the x- and the y- value, and it would be cleaner to go after it that way.
我会去辩论这么写是不是更好,我也可以写一个,返回x坐标和y坐标的方法,这么做可能会更清楚一点,这是很棒的缩写,好。
I ask when y=0, then I say 0=15+10t-5t^.
我问何时y=0,然后得到0=15+10t-5t^
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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