past exam question!!!!!! (1 Viewer)

[fush]

sorry for being a bitch and asking questions but i cant figure this out and its pissing me off

question 3cii on the 2004 math1001 paper

check attachment for question and suggested answer


i just dont get the last line, what ta hell is happening there

 

[withoutaface]

The second last line indicates that dV (the change in volume) is approximately equal to 2dh (change in height). Since the height changes by 2%, then the volume changes by 4%.

 

[fush]

yea but how do u know that dv is approx 2dh?

 

[tennille]

yea but how do u know that dv is approx 2dh?

Yeah, I don't get that 2dh/h thing. Where does the 2 come from (not the 2%)?

 

[acmilan]

Answer attached

 

[fush]

ah very good

thanx man

 

[fush]

hey for the dV = differnetial,

why is the da part a negative?

 

[acmilan]

hey for the dV = differnetial,

why is the da part a negative?

Because you are differentiating with respect to a, holding everything else constant ie (pi*h²)/2 = k

V = k/a when holding h constant, when differentiating that you get -k/a², or -(pi*h²)/(2a²)

 

[acmilan]

Question 5(b) from the 1901 2004 paper anyone? Im no good with complex numbers

 

[xiao1985]

post q here acmilian?!

*shit, just realise i forgot everything there is to first yr maths =S

 

[Xayma]

5)b)

Consider the complex function z->w=z+1/z (z maps to w which equals z+1/z), whose domain is the set of all complex numbers except zero.

i) Show that the image (under this function) of a circle of radius 2, centered at the origin, is an ellipse.

ii) Find the image of a circle of radius 1 centered at the origin.

i) Let z=x+iy

Now |z|=2 (circle at origin, radius of 2)

w=z+1/z
=z+z(bar)/[z*z(bar)]
=z+z(bar)/|z|²

Now the image of z=x+iy+x/4-iy/4
=5/4x+3/4iy

Hence w, the image of z, is an ellipse.

Similar for ii) except you should end up with a line (z maps to w=z+1/z=2x)

...I hope

Edit: You probably should sub in the equation at some stage in there (opps, thats what I get for doing it at 11pm)

 

[nit]

yep that's pretty much correct. I used z=x+iy to get at it. As for the second part, it's the x-axis. (I stand corrected)

 

[SeDaTeD]

i) z = 2e^itheta, 1/z = 1/2 e^-itheta , theta is in all reals
z + 1/z = 2e^itheta + 1/2 e^-itheta
= 2cos(theta) + 2isin(theta) + 1/2 cos(-theta) + 1/2 isin(-theta)
= 5/2 cos(theta) + 3/2 isin(theta)
which is an ellipse
x = 5/2 cos(theta) , y = 3/2 sin(theta)
(2x/5)^2 + (2y/3)^2 = 1

ii) z = e^itheta, 1/z = e^-itheta
z + 1/z = e^itheta + e^-itheta
= cos(theta) + isin(theta) + cos(-theta) + isin(-theta)
= 2cos(theta)
x = 2cos(theta), y =0
interval on x-axis [-2,2]

Oh and nit, you can map to the origin,
z = i, 1/z = -i
z + 1/z = i - i = 0

 

[acmilan]

Thanks guys