HSC 2015 MX2 Marathon (archive) (1 Viewer)

InteGrand · Apr 22, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
Difference of two volumes?

True, probably too easy.

VBN2470 · Apr 22, 2015

Re: HSC 2015 4U Marathon

$$ Alternatively, you can find the volume of the region bounded by $y=x^2$, $x=1$ and the $x$-axis, and find the volume of the the region bounded between the two axes, $y=1$ and $x=1$ (i.e. volume of a cylinder of radius $1$), then take the difference of the two volumes to obtain the desired volume.$

Drsoccerball · Apr 22, 2015

Re: HSC 2015 4U Marathon

screw this... i learn the normal way properly but still dont know how to do washers method and anything with annulie

Ekman · Apr 22, 2015

Re: HSC 2015 4U Marathon

Just a simple question, for questions like:

$$ The region bounded by $ y=\sqrt{x+2} $ and the y axis is rotated around the line $ x=2$

Is the outer radius 2+x or 2-x? Some argue that it is 2-x as x is technically speaking negative in the second quadrant

Drsoccerball · Apr 22, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
^ Do you have a problem visualising the solids of revolution and their slices/shells?

I just don't kno what to do at all...

InteGrand · Apr 22, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Just a simple question, for questions like:

$$ The region bounded by $ y=\sqrt{x+2} $ and the y axis is rotated around the line $ x=2$

Is the outer radius 2+x or 2-x?

$$R(x)=2-x$ (i.e. the distance of 2 from $x$, where $2> x)$.$

InteGrand · Apr 22, 2015

Re: HSC 2015 4U Marathon

(A quick way to see it is to test when x = -2: The outer radius should clearly be 4 here, so we should be using 2 – x (2 – (-2) = 4).)

VBN2470 · Apr 22, 2015

Re: HSC 2015 4U Marathon

^ Do you have Cambridge 4U? That may help since they have pretty solid examples you can follow, if you're struggling, then go back to 2U and revise (and try to understand) the ideas behind basic volumes of solids of revolution. This can help form a basis on which you can strengthen your ability to solve 4U problems (even the basic ones).

InteGrand · Apr 22, 2015

Re: HSC 2015 4U Marathon

(Or, R(x) = 2 + |x|, and since x < 0, |x| = -x, so R(x) = 2 – x.)

Drsoccerball · Apr 22, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
^ Do you have Cambridge 4U? That may help since they have pretty solid examples you can follow, if you're struggling, then go back to 2U and revise (and try to understand) the ideas behind basic volumes of solids of revolution. This can help form a basis on which you can strengthen your ability to solve 4U problems (even the basic ones).

I have new senior maths the examples are so good but i cant to the questions that involve two radii

InteGrand · Apr 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I have new senior maths the examples are so good but i cant to the questions that involve two radii

Maybe check some maths videos, e.g. Khan Academy or PatrickJMT.

E.g. here's one from PatrickJMT: https://www.youtube.com/watch?v=E5OOMbz5jZk

Drsoccerball · Apr 22, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Maybe check some maths videos, e.g. Khan Academy or PatrickJMT.

E.g. here's one from PatrickJMT: https://www.youtube.com/watch?v=E5OOMbz5jZk

NOWW I UNDERSTAND WHAT IT MEANS BY INNER AND OUTER (mind blown)

Drsoccerball · Apr 22, 2015

Re: HSC 2015 4U Marathon

thanks guys understand it now

)) thanks to integrand and VBN

Drsoccerball · Apr 23, 2015

Re: HSC 2015 4U Marathon

now i need to know how you do a question where like this where theres two areas or 2 annuli:
$y=2x-x^2 $ rotated about the y axis $$

InteGrand · Apr 23, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
now i need to know how you do a question where like this where theres two areas or 2 annuli:
$y=2x-x^2 $ rotated about the y axis $$

For this, the inner radius is the distance from the y-axis to the nearer half of the parabola, and the outer radius is the distance from the y-axis to the further half of the parabola.

To find these for a given y, solve the equation $y=2x-x^2$ for x using the quadratic formula (there will be two solutions). The x you get with the smaller value (the one with the negative square root) will be the inner radius r(y). The bigger solution will be the outer radius R(y). After finding these two functions, you can find $R^2 - r^2$ as a function of y (using $R^2 - r^2 = (R+r)(R-r)$ should make this less tedious), and then after simplifying, you integrate $\pi (R^2 - r^2)$ with respect to y between the range of y-values, which will be between y = 0 and y = y_max., the y-value of the vertex of the parabola (which you'll need to find).

Drsoccerball · Apr 24, 2015

Re: HSC 2015 4U Marathon

$$ The shaded region bounded by the curve $ y=(x-1)^3 +1, $ the line $ y=2-x $ and x=2 $ is rotated about the y axis.$
Using the method of cylindrical shells to calculate the volume of this solid is the answer : $\frac{17 \pi}{6}$

Drsoccerball · Apr 24, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ The shaded region bounded by the curve $ y=(x-1)^3 +1, $ the line $ y=2-x $ and x=2 $ is rotated about the y axis.$
Using the method of cylindrical shells to calculate the volume of this solid is the answer : $\frac{17 \pi}{6}$

dw got it

FrankXie · Apr 24, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
dw got it

do u mean the answer IS 17pi/6 ?

Ekman · Apr 24, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
do u mean the answer IS 17pi/6 ?

I got 77pi/30. My integral was:

$2\pi\int_{1}^{2} x((x-1)^3+1 -(2-x))dx = 2\pi\int_{1}^{2} x(x-1)((x-1)^2+1)dx$

Drsoccerball · Apr 24, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
do u mean the answer IS 17pi/6 ?

lol you did the same thing i did interpreted it as $(x-1)^2 not (x-1)^3$

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