STPM 数学练习题

蝎杰

the normal at P(at^2,2at)to the parabola y^2=4ax is y+tx=2at+at^3.find the equation of the normal to the curve which is perpendicular to the normal at P.
不是很明白，没有头绪找另一个的equation。。。。

express 1/(r^3-r)in terms of partial fraction .hence find the sum of the series 1/(r^3-r) which is from r=1 to r=n。
不知道我是否可以要求大家给一个漂亮的解答呢？
在此先谢谢了！

灰羊

原帖由 蝎杰 于 13-10-2005 02:57 PM 发表
the normal at P(at^2,2at)to the parabola y^2=4ax is y+tx=2at+at^3.find the equation of the normal to the curve which is perpendicular to the normal at P.
不是很明白，没有头绪找另一个的equation。。 ...

r應該從2開始因為r=1,1/(r^3-r)沒有意義

蝎杰

如果遇到这种问题，即sum of the 3 different terms which is from r=1 to n的话就必须把它分成２part 来做，是吗？我之前是 这么做的，我不知道为何我总是得不到答案....
∑ [{1/2(r-1)-1/2r}-{1/2r-1/2(r+1)}]
=1/2∑ [{1/(r-1)-1/r}-{1/r-1/(r+1)}]
=1/2[∑ f(r)-f(r+1)]
=1/2[f (1)-f (n+1)]
=1/2[(-1-0)-1/n+1/ (n+1)]
=-1/2 -1/ [2n(n+1)]

蝎杰

这种问题是用什么原理做呢？？
prove that x^2+x-10
             x-3
can take all values except those between 7±2√2

point P(a cos k,b sin k)lies on the curve x^2/a^2 +y^2/b^2=1,where a and bare constants.find dy/dx in terms of k and hence find the equation of the normal to the curve at the point P.

the area enclosed between the curve y=4-x^2 and the line y=4-2x is rotated through 360 about the x-axis,find the exact volume of the solid generated.

find the area enclosed between the 2 graphs y=1-x^2 and y=x(1-x^2)
所谓的enclosed是指这２个graphs的交叉点吗？

dunwan2tellu

如果遇到这种问题，即sum of the 3 different terms which is from r=1 to n的话就必须把它分成２part 来做，是吗？我之前是 这么做的，我不知道为何我总是得不到答案....
∑ [{1/2(r-1)-1/2r}-{1/2r-1/2(r+1)}]
=1/2∑ [{1/(r-1)-1/r}-{1/r-1/(r+1)}]
=1/2[∑ f(r)-f(r+1)]
=1/2[f (1)-f (n+1)]
=1/2[(-1-0)-1/n+1/ (n+1)]
=-1/2 -1/ [2n(n+1)]

也可以分成两个部分

Sum 1/2 {1/r(r-1) - 1/r(r+1)} = 1/2 {1/2 - 1/n(n+1)}
=(n-1)(n+2) / 4n(n+1)

至于你上面写的，有些乱。若照你写的来看的话，你的第二行就factor错了。

dunwan2tellu

这种问题是用什么原理做呢？？
prove that x^2+x-10
             x-3
can take all values except those between 7±2√2

把它设成 y 过后交叉成，得到quadratic equation。过后再用b^2-4ac >= 0 表示(in terms of y) 。你就会得到 y的 range 。那么，y 的range 以外的就是你所要找的答案了。

point P(a cos k,b sin k)lies on the curve x^2/a^2 +y^2/b^2=1,where a and bare constants.find dy/dx in terms of k and hence find the equation of the normal to the curve at the point P.

2x/a^2 + 2y/b^2 *dy/dx = 0 --->dy/dx = -xb^2 / ya^2

所以substitute P点进去就可以得到dy/dx = cot^2 k

之后应该不难了吧

the area enclosed between the curve y=4-x^2 and the line y=4-2x is rotated through 360 about the x-axis,find the exact volume of the solid generated.

先找intersecting point (得到 x=0 ,x=2)。之后就用找体积的方法来integrate ，从x=0 到x=2 。

find the area enclosed between the 2 graphs y=1-x^2 and y=x(1-x^2)
所谓的enclosed是指这２个graphs的交叉点吗？

你必须先找交叉点。之后大概sketch出两个graph来看。enclosed就是这两个graph围住的范围。而你就是要找那围起来的area 。

蝎杰

原帖由 dunwan2tellu 于 15-10-2005 11:11 AM 发表


也可以分成两个部分

Sum 1/2 {1/r(r-1) - 1/r(r+1)} = 1/2 {1/2 - 1/n(n+1)}
=(n-1)(n+2) / 4n(n+1)

至于你上面写的，有些乱。若照你写的来看的话，你的第二行就factor错了。

恕我愚笨，我真的还不明白我到底做错了什么？我看不见我的factor哪里错了，可以清楚地告诉我吗？谢谢！还有这一题你的solution我也是看不明白。为何1/r不见了的呢？

dunwan2tellu

你把1/2 factor 出来，但是 其中一个是1/2r-1 ,你不可以就酱把１／２拉出来变成 1/2 ( 1/r-1)

我的 r并没有不见啊。我只是用partial fraction 把他变成

1/r(r+1)(r-1) = A/r(r-1) + B/r(r+1)

得到 A=1/2 , B=-1/2

蝎杰

抱歉那个不是1/2r-1,(1/2r) - 1/2(r+1)，我的factor没有问题吧？
the area enclosed between the curve y=4-x^2 and the line y=4-2x is rotated through 360 about the x-axis,find the exact volume of the solid generated.用什么体积的formula?是否要画了graph才知道呢？

灰羊

原帖由 蝎杰 于 16-10-2005 12:17 PM 发表
抱歉那个不是1/2r-1,(1/2r) - 1/2(r+1)，我的factor没有问题吧？
the area enclosed between the curve y=4-x^2 and the line y=4-2x is rotated through 360 about the x-axis,find the exact volume of the so ...

$(pi)*y^2 dx面積公式
不過要正確還是要畫圖出來看

QooLuo

A solid circular cylinder has a given volume. Show that its total surface area will be least when its height is equal to the diameter of the base.

可否告诉我问题到底在说什么吗？我一点头绪都没有。。请教我如何解答。。

dunwan2tellu

A solid circular cylinder has a given volume. Show that its total surface area will be least when its height is equal to the diameter of the base.

意思是你知道一个cylinder的体积，那么那cylinder 的total surface area 的minimum 是当它的高等于它的diameter .试试用
diffenrentiation 来找它的minimum 就可以了。

V = pi x r^2 x h  , S = 2 pi x r(r+h)

找dS/dr 或 ， dS/dh 都可以 ....

QooLuo

不是很明白。 我要differenciate来做什么？ pi 是什么？
ds/dh 了之后，我要和什么compare? 题目到底是要我找什么？

dunwan2tellu

pi = 3.142

题目等于要你找 S的 minimum . 还记得form 4学的pembezaan吗？当中不是有用pembezaan 来找minimum的吗？

已知体积,V＝pi x r^2 x h , 所以  h = V/(pi x r^2)

又因为surface area ,
S = 2 x pi x r^2 + 2 x pi x r x h =2pixr^2+2V/r

所以S 的minimum 就是 dS/dr = 0 . 酱你就可以得到 r (in term of V )了。之后再把 r 带入 h = V/(pi x r^2) ， 你就可以得到 h。比较一下 diameter (=2r) 和 height (=h) 看他们是否相同。

蝎杰

show that x^2-x+1 is positive for all values of x.hence show that intergrate |(2x-1)/(x^2-x+1)|dx from 0 to 2 is equal to ln 16/3

我的做法是:
x^2-x+1=(x-1/2)^2+(3/4)>0 for all values of x
intergrate |(2x-1)/(x^2-x+1)|dx from 0 to 2
=[ln |x^2-x+1|](from 0 to 2)
=[ln{(x-1/2)^2+3/4}]
=ln3-ln4
ln3/4                                          
我不知道我哪里出错了....

dunwan2tellu

我的做法是:
x^2-x+1=(x-1/2)^2+(3/4)>0 for all values of x
intergrate |(2x-1)/(x^2-x+1)|dx from 0 to 2
=[ln |x^2-x+1|](from 0 to 2)
=[ln{(x-1/2)^2+3/4}]
=ln3-ln4
ln3/4                                          
我不知道我哪里出错了....

第三行出错。看到modulus 的符号吗？ 这意味着当0=<x=<1/2 时，那expression 是 negative 的，所以要乘个负号。反之，当1/2=<x=< 2时，那expression 是positive 的。所以你integrate 时要分cases ,那就是从0 到1/2 (这里expression 用 negative)  和 从1/2到2 (这里expression 用positive) .

蝎杰

given that a and b are real roots of the aquation 2x^2 -kx+c=0,a>b. if the equation x^2-qx+r=0 has roots (a+2)and (b+1),find q and r in terms of k and c.in the case a=b,show that q^2-4r-1=0.
我只会做到
k/2+3=q
但是r in terms of k & c我就做不到了，我记得是用某个方法做的，可是我却记不起。。。。

dunwan2tellu

given that a and b are real roots of the aquation 2x^2 -kx+c=0,a>b. if the equation x^2-qx+r=0 has roots (a+2)and (b+1),find q and r in terms of k and c.in the case a=b,show that q^2-4r-1=0.
我只会做到
k/2+3=q
但是r in terms of k & c我就做不到了，我记得是用某个方法做的，可是我却记不起。。。。

用quadratic 的formula ... 2x^2 -kx +c=0的两个root 是

x= [k+- sqrt{k^2-8c}]/4 .因为a>b ,所以取 a 为正号的根， b 为负号的根. 所以 r = [2c+3k+8-sqrt{k^2-8c}]/4 。

[ 本帖最后由 dunwan2tellu 于 7-11-2005 06:55 PM 编辑 ]

QooLuo

有几题数学又不会做了。。请大家教教我。。

1. Jack cycled from point A to point B at an average speed of 20 km/h. He then cycled back from B to A at an average speed of 18 km/h. If he took a total od 1 h 25 min, find the distance between points A and B.

2. Alex can either walk or cycle to school. He walks at an average speed of 10 km/h and cycles at an average speed of 20 km/h. If he can save 15 minutes by cycling, find the distance between Alex's home and school.

3. Joanne takes 7 hours to complete a puzzle. Her sister, Jeslin takes only 5 hours to complete the same puzzle. If they work together, how long will they take? Give your answer in h and min.

dunwan2tellu

这应该不是STPM的题目吧？

1. Jack cycled from point A to point B at an average speed of 20 km/h. He then cycled back from B to A at an average speed of 18 km/h. If he took a total od 1 h 25 min, find the distance between points A and B.

distance = d , 用 v=d/t 有
d/18 + d/20 = 17/12 ....

2. Alex can either walk or cycle to school. He walks at an average speed of 10 km/h and cycles at an average speed of 20 km/h. If he can save 15 minutes by cycling, find the distance between Alex's home and school.

同样 distance =d ,
d/10 - d/15 = 1/4 ....

3. Joanne takes 7 hours to complete a puzzle. Her sister, Jeslin takes only 5 hours to complete the same puzzle. If they work together, how long will they take? Give your answer in h and min.

Joanne 每小时完成1/7 ,她妹妹每小时完成1/5 . 两人每小时完成1/7 + 1/5 = 12/35 . 所以总共需要 35/12 小时＝２小时５５分钟