Some of the questions I thought of to train my students at times... feel free to use it or add on :D
Shall post whenever remember some or got inspiration to create.
Complex Numbers:
Evaluate square root of i in the form a + bi in exact form, showing your workings clearly.
Integration:
Integrate x / (x4+4) dx
Enjoy :D
Originally posted by eagle:
I am here.... to spoil market.
square root i = (i)^1/2 = [ ( i X i )^ 1/2 ] ^1/2 = [ (-1)^1/2 ] ^1/2 = (-1)^1/4
x / ( x^4 + 4 )
= (x/4) X ( 4 / ( 4 + x^4 )
= (x/4) X ( 1 / [ ( 4 + x^4 ) / 4 ] )
= (x/4) X ( 1 / [ 1 + (1/4)(x^4)] )
= (x/4) X ( 1 / [ 1 + [(1/2)(x^2)]^2 ] )
Here is when students will say "Ya Hor! Why I never see this?"
After integrating the above, it will become,
(1/4)(tan^-1 [(1/2)(x^2)]) + constant
Eh......
arctan in A level syllabus meh?
Originally posted by Forbiddensinner:I am here.... to spoil market.
square root i = (i)^1/2 = [ ( i X i )^ 1/2 ] ^1/2 = [ (-1)^1/2 ] ^1/2 = (-1)^1/4
x / ( x^4 + 4 )
= (x/4) X ( 4 / ( 4 + x^4 )
= (x/4) X ( 1 / [ ( 4 + x^4 ) / 4 ] )
= (x/4) X ( 1 / [ 1 + (1/4)(x^4)] )
= (x/4) X ( 1 / [ 1 + [(1/2)(x^2)]^2 ] )
Here is when students will say "Ya Hor! Why I never see this?"
After integrating the above, it will become,
(1/4)(tan^-1 [(1/2)(x^2)]) + constant
first one wrong.. but I think is because I never phrase the question properly... Changed it
2nd one correct
Originally posted by deepak.c:
Eh......
arctan in A level syllabus meh?
yup, in calculus, both differentiation and integration
They can use their MF15 sheet to do as well
Originally posted by eagle:first one wrong.. but I think is because I never phrase the question properly... Changed it
2nd one correct
sq rt i
= i^(1/2)
= (2i)^(1/2) / 2^(1/2)
= (1+2i-1)^(1/2) / 2^(1/2)
= [(1+i)^2]^(1/2) / 2^(1/2)
= (1+i) / 2^(1/2)
= 1 / 2^(1/2) + [1 / 2^(1/2)] i
yup :D
a shorter method would be to convert
i = e^(i pi/2)
sqrt(i) = e^(i pi/4)
expand Euler's formula to cos + i sin
sqrt (i) = 1 /
2^(1/2) + [1 / 2^(1/2)] i
Originally posted by eagle:yup, in calculus, both differentiation and integration
They can use their MF15 sheet to do as well
Ooops.....looks like I was daydreaming during lectures.
Hi,
Let me add a question that connects three topics (i.e. graphing techniques, applications of differentiation and applications of integration).
The curve C has equation given by x^2 + 3xy = 4.
(i) Sketch C, indicating its asymptotes, stationary points and axial intercepts, where appropriate.
(ii) Prove, algebraically, that there are no points on C at which the tangent is parallel to the x-axis.
(iii) Find the exact area of the region bounded by C, the line x = 6 and the x-axis.
(iv) The region bounded by C, the line y = 1 and the axes is rotated through 4 right angles about the y-axis. Find the volume of revolution.
Thanks!
Cheers,
Wen Shih
Okay off topic here, but calculus is useful in which area of study?
Originally posted by Ah Chia:Okay off topic here, but calculus is useful in which area of study?
Engineering?
Physics?
Any others?
Hi,
Business, economics, biology and medicine too. Thanks!
Cheers,
Wen Shih
Hi Mr Wee,
Mr Chia is only interested in politics.
He will not be interested on the applications of calculus to business, economics, biology, medicine, engineering or physics.
He is only interested if calculus can be applied to politics.
The url link for the application of differentiation and integration to politics is
http://www.jstor.org/pss/2778814
But it might be too mathematical for Mr Chia.
Originally posted by wee_ws:The curve C has equation given by x^2 + 3xy = 4.
(iv) The region bounded by C, the line y = 1 and the axes is rotated through 4 right angles about the y-axis. Find the volume of revolution.
Hi,
I have written an article on my website:
http://www.freewebs.com/weews/solidsofrevolution.htm
Enjoy!
Cheers,
Wen Shih
Originally posted by eagle:
Complex Numbers:
Evaluate square root of i in the form a + bi in exact form, showing your workings clearly.
root of i, wouldnt there be 2 answers?
let i^1/2 be a+bi
i = (a+bi)^2
i = a^2 - b^2 + 2abi
comparing coeff of real part,
a^2-b^2 = 0
=> a = b -(1)
comparing coeff of imaginary,
1 = 2ab
since a=b by (1), a^2 = 0.5 and b^2 = 0.5
=> a = plus minus 1/sqrt2 and b = plus minus 1/sqrt 2.
therefore i^1/2 = 1/sqrt2 + 1/sqrt2 i and -1/sqrt2 - 1/sqrt2 i...?