HSC 2015 MX2 Marathon (archive) (1 Viewer)

Kaido · Apr 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
solid explanation

that pun doe

Green Yoda · Apr 21, 2015

Re: HSC 2015 4U Marathon

Wow this stuff is looking difficult! Anyone know good yr 10-11 tutor? just post a link if already on bos!

InteGrand · Apr 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
so y=x^2, y=1, x=3 about x axis would be
$pi \int_1^3(3-x)^2dx$ ?

Note quite.

For this one, your typical slice will be a (vertical) washer. At a point x, the outer radius of the washer will be $R(x)=x^2$ . The inner radius will be $r(x)=3$ (constant). The area of an annulus (http://en.wikipedia.org/wiki/Annulus_(mathematics) ) is $A=\pi (R^2 -r^2)$ .

Then the area of a typical washer is $A(x)=\pi((R(x))^2 -(r(x)^2))=\pi(x^4 - 9)$ .

Then integrate this area function between the appropriate bounds (which you have, i.e. x = 1 to x = 3).

Ekman · Apr 21, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Note quite.

For this one, your typical slice will be a (vertical) washer. At a point x, the outer radius of the washer will be $R(x)=x^2$ . The inner radius will be $r(x)=3$ (constant). The area of an annulus (http://en.wikipedia.org/wiki/Annulus_(mathematics) ) is $A=\pi (R^2 -r^2)$ .

Then the area of a typical washer is $A(x)=\pi((R(x))^2 -(r(x)^2))=\pi(x^4 - 9)$ .

Then integrate this area function between the appropriate bounds (which you have, i.e. x = 1 to x = 3).

Shouldn't the inner radius be 1?

Drsoccerball · Apr 21, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Note quite.

For this one, your typical slice will be a (vertical) washer. At a point x, the outer radius of the washer will be $R(x)=x^2$ . The inner radius will be $r(x)=3$ (constant). The area of an annulus (http://en.wikipedia.org/wiki/Annulus_(mathematics) ) is $A=\pi (R^2 -r^2)$ .

Then the area of a typical washer is $A(x)=\pi((R(x))^2 -(r(x)^2))=\pi(x^4 - 9)$ .

Then integrate this area function between the appropriate bounds (which you have, i.e. x = 1 to x = 3).

Can you explain as if youre talking to an idiot i still dont get it

InteGrand · Apr 21, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Shouldn't the inner radius be 1?

Oh yeah, inner radius is 1 (didn't draw a diagram myself, so forgot what the radius was meant to be).

Ekman · Apr 21, 2015

Re: HSC 2015 4U Marathon

Interesting Question:

$$ Find the volume of the solid formed when the region between the circle $ x^2+y^2=16 $ and the ellipse $ 4x^2+y^2=16 $ where the region is rotated about the line $ x=4$

Don't use cylindrical shells to solve this.

InteGrand · Apr 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Can you explain as if youre talking to an idiot i still dont get it

Draw a diagram and shade in the region that will be revolved. Now draw a thin vertical rectangle at a point x (where x is between 1 and 3). This thin vertical rectangle starts at the line y = 1 and goes up to the point where it touches the parabola (which will be at a place where the y-value is x²). (This isn't exactly a rectangle, since at the top, it's not straight, but rather, parabolic. But in the limit as you take very small strips, it becomes rectangular.)

Now, imagine rotating this rectangular strip, which is between y = 1 and y = x², about the line y = 0 (i.e. the x-axis). What you'd get would be an annulus (basically a doughnut shape). You're not getting a full circle because the rectangular strip does not include any area between the x-axis and the line y = 1. Now, the radius of 'hole' of the annulus is the distance from the x-axis to the line y = 1, which is 1. So the inner radius of the annulus is r(x) = 1 (a constant function of x). The outer radius of the annulus is the distance of the furthest end of the strip from the x-axis to the x-axis, which is just x². Hence R(x) = x². (Using R to denote outer radius and r to denote inner radius.)

Then the area of a typical annulus is A(x) = π((R(x))² – (r(x))²), where R(x) = x², r(x) = 1. Now you can find the volume using $V=\int _1 ^3 A(x) \text{ d}x$ .

VBN2470 · Apr 22, 2015

Re: HSC 2015 4U Marathon

VOLUMES QUESTION

$(I) Show that the curves $y = \frac{1}{\sqrt{x^2+1}}$ and $y = \frac{\sqrt{2}}{\sqrt{x^2+4}}$ intersect each other at $x=\pm{\sqrt2}$. \\ (II) The region bounded by the above curves is rotated about the $x$-axis. Show that the volume of the solid so generated is 2\pi\cdot\tan^{-1}\frac{1}{2\sqrt{2}}$.$

swagswagyoloswag · Apr 22, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Interesting Question:

$$ Find the volume of the solid formed when the region between the circle $ x^2+y^2=16 $ and the ellipse $ 4x^2+y^2=16 $ where the region is rotated about the line $ x=4$

Don't use cylindrical shells to solve this.

Would you jsut find the volume of the circle rotated around x = 4 subtracted by volume of vertical ellipse rotated around x = 4? Or letting slice have end point P(4cost,4sint) [lies on circle] and Q(2cost,4sint) [lies on ellipse] with height of delta y? Then you have an integral with 't' and 'delta y' so to eliminate delta y I use y = 4sint ==> dy=4cost dt ?

Ekman · Apr 22, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
VOLUMES QUESTION

$(I) Show that the curves $y = \frac{1}{\sqrt{x^2+1}}$ and $y = \frac{\sqrt{2}}{\sqrt{x^2+4}}$ intersect each other at $x=\pm{2}$. \\ (II) The region bounded by the above curves is rotated about the $x$-axis. Show that the volume of the solid so generated is 2\pi\cdot\tan^{-1}\frac{1}{2\sqrt{2}}$.$

$For part (I) $ x=\pm\sqrt{2} $ are the intercepts$

$$(II) Via Washer Method: $ \pi\int_{-\sqrt{2}}^{\sqrt{2}} \frac{1}{x^2+1} - \frac{2}{x^2+4} dx$

$= 2\pi \left[\tan^{-1}x - \tan^{-1}\frac{x}{2} \right]_{0}^{\sqrt{2}}$

$=2\pi \left[\tan^{-1}\sqrt{2} - \tan^{-1}\frac{1}{\sqrt{2}}\right] = 2\pi\cdot\tan^{-1}\frac{1}{2\sqrt{2}}$

Drsoccerball · Apr 22, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
VOLUMES QUESTION

$(I) Show that the curves $y = \frac{1}{\sqrt{x^2+1}}$ and $y = \frac{\sqrt{2}}{\sqrt{x^2+4}}$ intersect each other at $x=\pm{2}$. \\ (II) The region bounded by the above curves is rotated about the $x$-axis. Show that the volume of the solid so generated is 2\pi\cdot\tan^{-1}\frac{1}{2\sqrt{2}}$.$

Are you sure those are the intercepts

VBN2470 · Apr 22, 2015

Re: HSC 2015 4U Marathon

^ Fixed

VBN2470 · Apr 22, 2015

Re: HSC 2015 4U Marathon

$Using the method of cylindrical shells, find the volume of a torus with inner radius 2 and outer radius 4.$

Drsoccerball · Apr 22, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Using the method of cylindrical shells, find the volume of a torus with inner radius 2 and outer radius 4.$

learning cylindrical shells tomorrow </3

InteGrand · Apr 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
learning cylindrical shells tomorrow </3

You can find the volume using washers too.

Ekman · Apr 22, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Using the method of cylindrical shells, find the volume of a torus with inner radius 2 and outer radius 4.$

InteGrand said:
You can find the volume using washers too.

Here is the washer method of solving it:

$$ Let the equation of the circle being rotated about the y axis: $ (x-3)^2 + y^2 =1$

$\therefore $the two x-values of the circle: $ x_1=3-\sqrt{1-y^2} $ and $ x_2=3+\sqrt{1-y^2}$

$V=\pi\int_{-1}^{1} (x_2)^2 - (x_1)^2 dy = 12\pi\int_{-1}^{1} \sqrt{1-y^2} dy$

$V=6\pi^2$

InteGrand · Apr 22, 2015

Re: HSC 2015 4U Marathon

NEW VOLUMES QUESTION

$Find the volume obtained upon rotating the region bound by $y=x^2$, $y=1$ and the $y$-axis about the $x$-axis.$

VBN2470 · Apr 22, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
NEW VOLUMES QUESTION

$Find the volume obtained upon rotating the region bound by $y=x^2$, $y=1$ and the $y$-axis about the $x$-axis.$

Difference of two volumes?

Ekman · Apr 22, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
NEW VOLUMES QUESTION

$Find the volume obtained upon rotating the region bound by $y=x^2$, $y=1$ and the $y$-axis about the $x$-axis.$

$$Outer Radius=$ 1 $ and Inner Radius=$ y$

$$ Hence Via Washer Method: $ V=\pi\int_{0}^{1} 1-y^2 dx =\pi\int_{0}^{1} 1-x^4 dx = \frac{4\pi}{5}$

HSC 2015 MX2 Marathon (archive) (1 Viewer)

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