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Preliminary mathematics marathon (2 Viewers)

hscishard

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Use the trapazodial rule with 4 function values to estimate the volume when the region between y= 1 +lnx, the x axis and the lines x=2 and x=5 , is rotated about the x axis
I don't get it.

Are you meant to find the area first to find the volume? Because the area doesn't look very helpful...
 

random-1005

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I don't get it.

Are you meant to find the area first to find the volume? Because the area doesn't look very helpful...
you do the normal thing, ie V= pi integral y^2 dx

V= pi integral (1 +ln(x)) ^2 dx

then use trapazodial rule to find integral of (1 +ln(x) ) ^2 then multiply answer by pie, also if i put y=1+sin(x) {or any trig function}, you would need to use RADIANS, a common mistake is to use degrees, thats what most people dont know is that the trapazodial/ simpsons rule doesnt find the "area" so to speak, it finds the integral

ie

V= pi/2 * [ ( (1+ln(2))^2 + (1+ln(5) )^2 +2( (1+ln(3))^2 +(1+ln(4))^2) ) ]

whatever that evaluates to, provided i have remembered trapazodial rule correctly = h/2 ( first +last +2(others))
 
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random-1005

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Which book gives the shortcut for the trapeziodal rule?
One of my books has the simpsons rule one but not the trap :(

i dnt think the one i had the trapazodial rule either , see last post for the formula, i never remembered the formula from the book, teacher just said, this is a way , remember it and you will be ok, if u try to do simpsons rule the way it taught in the books you will make a mistake
 

hscishard

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i dnt think the one i had the trapazodial rule either , see last post for the formula, i never remembered the formula from the book, teacher just said, this is a way , remember it and you will be ok, if u try to do simpsons rule the way it taught in the books you will make a mistake
You mean the long way? Yea it's tedious. Espec. for more than 3 values
 

random-1005

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NEW QUESTION:

Find the co ordinates of the focus, vertex and equation of the directrix of the parabola :

(y+2)^2 = -4(2x+1)
 
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random-1005

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^^

Vertex (-0.5, -2)
Focus (-2.5, -2)
Directrix: x=1.5

correct

thats like as exciting as prelim questions get, and thats not exciting, just full of tricks, cant think of any good questions to put up that are still in prelim
 
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New Question:

A swimmer is directly 1000 m out in the ocean from point A on the beach, the swimmer has to get to point B which is on the beach 5000 m away from point A. The swimmer can swim at 2 m/s and walk on the beach at 2.5 m/s, at which point from point A does the swimmer get out of the water and start walking for the quickest journey?
 

random-1005

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New Question:

A swimmer is directly 1000 m out in the ocean from point A on the beach, the swimmer has to get to point B which is on the beach 5000 m away from point A. The swimmer can swim at 2 m/s and walk on the beach at 2.5 m/s, at which point from point A does the swimmer get out of the water and start walking for the quickest journey?

lol, this is exactly like the max/min question they had in last yrs 2 unit
 

random-1005

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I'm finding it hard to interpret the question :S

look at the picture given with last yrs 2unit hsc Q9 b, its very similar , the distance between the person and the point on the shore is the hypotenuse of a right angled triangle plus a line along the shore
 
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hscishard

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look at the picture given with last yrs 2unit hsc Q9 b, its very similar , the distance between the person and the point on the shore is the hypotenuse of a right angled triangle plus a line along the shore
That helps heaps. My point B was like a 5km radius.

In 9b) i-iv are easy..
Not sure on v though. I'm sure this part is similar to the question above.
 

hscishard

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New Question:

A swimmer is directly 1000 m out in the ocean from point A on the beach, the swimmer has to get to point B which is on the beach 5000 m away from point A. The swimmer can swim at 2 m/s and walk on the beach at 2.5 m/s, at which point from point A does the swimmer get out of the water and start walking for the quickest journey?
I got it. It's 2km, 2000m
 

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