2u Mathematics Marathon v1.0 (1 Viewer)

[chinkyeye]

wtf..please explain liek simplyer bro

 

[icycloud]

hey i need help do not understand this trig identity.. HSC PAST PPER 2004 Q9a i
"consider the GP 1 - tan^2"theta" + tan^4"theta"...when is da limitngsum exists, find its value in simplest form... i done all da steps, buti don' understand last line of working..their last line is 1/sec^2"theta" and then jumps to da anwser of cos^2"theta" wtf pelase explain

sec(x) = 1/cos(x)

Thus,

sec^2 (x) = sec(x) * sec(x)
= 1/cos(x) * 1/cos(x)
= 1/cos^2 (x)

Thus, 1/sec^2 (x) = 1/(1/cos^2 (x))
= cos^2 (x)

 

[chinkyeye]

yea lol, i got it now jsut couldn't b borhtered pictureng in my mind the doubel division signs, yeh i get it cheers

 

[icycloud]

Spoiler

y=xe^x
y' = e^x(1+x)
y'' = e^x(2+x)
let y'' = 0 to find point of inflexion
0 = e^x(2+x)
therefore x = -2
point of inflexion at (-2,-2e^-2)

next question
find the equation of the curve that is always concave upwards with a stationary point at -1,2 and y-intercept = 3

Hey! You changed the question!
Edit: Oops, just read your edit reason.

 

[praise897]

Spoiler

haha this is the best!!






ever

 

[skyrockets1530]

Hey! You changed the question!
Edit: Oops, just read your edit reason.

haha, yea, sorry bout that- was doing it myself and came out with a negative answer, so either i made a mistake in my working or the question, but by all means post an answer if u did one, the question was: find the volume when the curve y=ln(x^2 + 2x) is rotated about the y-axis between y=2 and y=1 is it actually do-able?

 

[icycloud]

haha, yea, sorry bout that- was doing it myself and came out with a negative answer, so either i made a mistake in my working or the question, but by all means post an answer if u did one, the question was: find the volume when the curve y=ln(x^2 + 2x) is rotated about the y-axis between y=2 and y=1 is it actually do-able?

Yes it's doable, but I don't think it's 2U standard, or even 3U for that matter .

You can either integrate:

pi * e^y + 1 +/- 2 Sqrt(e^y+1) + 1 dx
bounds: y = 1, y = 2

or you can integrate:

pi * (ln(x^2+2x))^2 dx
bounds: x = +/- Sqrt(e+1) - 1 and x = +/- Sqrt(e^2+1) - 1

using integration by parts (taught in 4U)

Either way, don't think it's a 2U question

 

[Riviet]

find the equation of the curve that is always concave upwards with a stationary point at -1,2 and y-intercept = 3

Spoiler

let the equation of curve be y=ax²+bx+c
when x=0, y=3
.: c=3
sub in (-1,2) with c=3 to obtain:
a-b=-1 (1)
now y'=2ax+b
when y'=0, we obtain stat point (-1,2 in this case)
.: 2ax+b=0
sub in x=-1 because it is the x co-ordinate stat point
=>-2a+b=0 (2)
from (1)
a=b-1 (3)
sub (3) in (2)
=>-2(b-1)+b=0
b=2
.:a=b/2
a=1
.: equation of curve is x²+2x+3=0

Next question: find the derivative of (x²+2)(5-x³)

 

[switchblade87]

Next question: find the derivative of (x²+2)(5-x³)

Spoiler

d / dx (x²+2)(5-x³)
= (5 - x³) * 2x + (x² + 2) * -3x²
= 2x(5 - x³) - 3x²(x² + 2)

Next Question: The length of an arc is 8.9cm and the area of the sector is 24.3cm² when an angle of @ is subtended at the centre of the circle. Find the area of the minor segment cut off by @, correct to 1 decimal place.

 

[skyrockets1530]

Yes it's doable, but I don't think it's 2U standard, or even 3U for that matter .

You can either integrate:

pi * e^y + 1 +/- 2 Sqrt(e^y+1) + 1 dx
bounds: y = 1, y = 2

or you can integrate:

pi * (ln(x^2+2x))^2 dx
bounds: x = +/- Sqrt(e+1) - 1 and x = +/- Sqrt(e^2+1) - 1

using integration by parts (taught in 4U)

Either way, don't think it's a 2U question

thanks, just found my mistake though, i think it is 2U do-able... making x the subject then integrating, as you mention in the first way u say to do it produces an answer, whether its correct or not it hard to say, but it seems right 7.1pi is what i ended up with

 

[skyrockets1530]

Next Question: The length of an arc is 8.9cm and the area of the sector is 24.3cm² when an angle of @ is subtended at the centre of the circle. Find the area of the minor segment cut off by @, correct to 1 decimal place.

Spoiler

theta = x for typing purposes
L=rx
8.9 = rx
x = 8.9/r
A=.5r²x
24.3 = .5r²x
48.6 = r²(8.9/r)
48.6 = 8.9r
r = 5.46
x = 8.9 / 5.46
x = 1.63
A = .5r²(x-sinx)
A = 14.906(1.63-sin1.63)
A of minor sector = 9.4u²

next question: find the area enclosed between y=sqrt(x) and y = x³

 

[jake2.0]

next question: find the area enclosed between y=sqrt(x) and y = x³

Spoiler

the curves intersect at x=1 and y=sqrt(x) is on top.
so area = ∫sqrt(x) - x³ dx. with limits 1 and 0

= (⅔x^3/2 - (x⁴)/4). from 1 to 0
= 2/3 - 1/4
= 5/12 units squared

find the range of f(x) = -2tan^-1x

 

[DeanM]

next question: find the area enclosed between y=sqrt(x) and y = x³

Spoiler

alrighty...

Spoiler

let x^0.5 = x^3
x^3 - x^0.5
x^0.5 (x^6 - 1)
therefore x= 0, x=1
integrate x^0.5 - x^3
[x^1.5 / 1.5 - x^4 / 4] between 0, and 1
{1^1.5 / 1.5 - 1^4 / 4} - {0^1.5 / 1.5 - 0^4 / 4}
therefore 5/12 units squared..

bloody hell its hard doing maths on a computer screen...
is that correct ? i dont want to post a question up if i got this answer wrong..

 

[DeanM]

arrrgh you beat me lol... !!!

 

[Slidey]

Try to keep this thread to 2u topics. I have created a thread in the 3u forum for 3u questions

 

[word.]

x^3 - x^0.5
x^0.5 (x^6 - 1)

remember your index laws; xⁿ * x^m = x^m+n
x³ - x^1/2 = x^1/2(x^5/2 - 1)
but you found the same points anyway so it didn't matter -- that's the correct answer

 

[DeanM]

alrighty... jake2.0's question was a 3u question, so ill post a new question...
find where the particile x=1+3cos2t find when the particile come to rest, AND where it is at this time, 0< t <2pi
( sound familiar? it should !!!)

 

[Slidey]

Hopefully I can do this! haha.

Spoiler

x=1+3cos2t
rest at v=0
v=-6sin2t=0
sin2t=0
2t=0, pi, 2pi, 3pi, 4pi
t=0, pi/2, pi, 3pi/2, 2pi
At these times it is:
x(pi/2)=-2
x(pi)=4
x(3pi/2)=-2
x(2pi)=4

Q) What is the locus of the variable point P such that the distance PA is equal to 2 lots of the distance PB, where A is (1,1), B is (3,4) and P is (x,y)?

 

[word.]

Spoiler

PA = 2PB; Using distance formula:
Sqrt[(x - 1)² + (y - 1)²] = 2*Sqrt[(x - 3)² + (y - 4)²]
SBS: (x - 1)² + (y - 1)² = 4(x - 3)² + 4(y - 4)²
x² - 2x + 1 + y² - 2y + 1 = 4(x² - 6x + 9) + 4(y² - 8y + 16)
3x² - 22x + 3y² - 30y + 98 = 0

Question 63
Integrate 3/Sqrt{e^x} dx

 

[icycloud]

Question 63
Integrate 3/Sqrt{e^x} dx

Answer: Warning, my way is a really long and drawn out method. Please see skyrockets way below, which is what you should be doing. I didn't see the simplification when I wrote the answer below...

Spoiler

∫ (3/Sqrt(e^x)) dx
= 3 ∫(1/Sqrt(e^x)) dx

Let u = Sqrt(e^x)
du = e^x / (2*Sqrt(e^x)) dx
dx = 2*Sqrt(e^x) / e^x du

Thus, original integral becomes:

6 ∫ Sqrt(e^x) / (e^x * Sqrt(e^x)) du
= 6 ∫ du / e^x

But, e^x = u^2, thus:

= 6 ∫du/u^2
= -6 / u + C
= -6 / Sqrt(e^x) + C #

Question 64
A glass has a shape obtained by rotating part of the parabola x = y^2 / 30 about the y-axis. The glass is 10 cm deep. Find the volume of liquid which the glass will hold.