![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
留学考试中数学的答题格式(用英语答题!!!)
本人最近准备参加留学考试,要考数学,英语试题,英语作答,听考过的人说,一定要注意格式问题,因为外国人跟咱的书写格式不兼容...比如:“因为所以”的3个点外国人是不认的.....
本人最近准备参加留学考试,要考数学,英语试题,英语作答,听考过的人说,一定要注意格式问题,因为外国人跟咱的书写格式不兼容...比如:“因为所以”的3个点外国人是不认的...
但是,外国数学答题的格式是怎么样的啊?谁能给个例子?还有介绍下格式上的差异
最好是题目和解答都有,越详细越好...
请参加过类似考试的同学
或者是比较了解这方面的人
多帮帮忙
谢谢,谢谢 展开
但是,外国数学答题的格式是怎么样的啊?谁能给个例子?还有介绍下格式上的差异
最好是题目和解答都有,越详细越好...
请参加过类似考试的同学
或者是比较了解这方面的人
多帮帮忙
谢谢,谢谢 展开
1个回答
展开全部
Here is an example:
"Call the two rays R (contains P) and S (contains A,B) and let the angle between them be α.
I tried this in Cartesian form, but geometry is probably better.
WLOG, choose origin O and R as the +x-axis.
The points are A(a,0) B(b,0)
Let's scale by 1/a such that A' is mapped to (1,0) and B' is mapped to (b/a,0). Let's call that (k,0). k is some constant.)
Let P be distance r along R (strictly, that distance has been scaled by 1/a too).
We are trying to find r (r is going to depend on α and k)
Coords of APB are thus A(1,0) P(r cos α, r sin α) B(k,0)
Let angle APB = θ, and apply the Cosine Rule.
Sidelengths are
|PA| = √((r cos α -k)² + (r sin α)²) = √(r² +k² -2kr cos α)
|PB| = √((r cos α -1)² + (r sin α)²) = √(r² +1 -2r cos α)
|AB| = (k-1)
You can get a (nasty) expression for cos θ in terms of parameter r and constants k, α.
Then find the r which maximizes θ, by minimizing cos θ (regardless whether 0 < θ < π/2 or π/2 < θ < π (cosine goes -ve)).
This will get the right answer, it'll just be awkward.
A change of notation might make it clearer.
Anyway, for what it's worth:
cos θ = |PA|²+|PB|²-|AB|² / 2|PA||PB|
= ((r² +k² -2kr cos α) +(r² +1 -2r cos α) -(k-1)²) / (r² +k² -2kr cos α)*(r² +1 -2r cos α)
and so on"
"Call the two rays R (contains P) and S (contains A,B) and let the angle between them be α.
I tried this in Cartesian form, but geometry is probably better.
WLOG, choose origin O and R as the +x-axis.
The points are A(a,0) B(b,0)
Let's scale by 1/a such that A' is mapped to (1,0) and B' is mapped to (b/a,0). Let's call that (k,0). k is some constant.)
Let P be distance r along R (strictly, that distance has been scaled by 1/a too).
We are trying to find r (r is going to depend on α and k)
Coords of APB are thus A(1,0) P(r cos α, r sin α) B(k,0)
Let angle APB = θ, and apply the Cosine Rule.
Sidelengths are
|PA| = √((r cos α -k)² + (r sin α)²) = √(r² +k² -2kr cos α)
|PB| = √((r cos α -1)² + (r sin α)²) = √(r² +1 -2r cos α)
|AB| = (k-1)
You can get a (nasty) expression for cos θ in terms of parameter r and constants k, α.
Then find the r which maximizes θ, by minimizing cos θ (regardless whether 0 < θ < π/2 or π/2 < θ < π (cosine goes -ve)).
This will get the right answer, it'll just be awkward.
A change of notation might make it clearer.
Anyway, for what it's worth:
cos θ = |PA|²+|PB|²-|AB|² / 2|PA||PB|
= ((r² +k² -2kr cos α) +(r² +1 -2r cos α) -(k-1)²) / (r² +k² -2kr cos α)*(r² +1 -2r cos α)
and so on"
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询