Author Topic: Mathematics Question Thread (Read 1626086 times) Tweet Share

RuiAce · « **Reply #3210 on:** January 17, 2018, 07:39:40 pm »

Quote from: hsotuhsa on January 17, 2018, 07:34:06 pm

Hey could anyone answer these questions? (any help appreciated)

1) 1) Rotate about the y-axis the region bounded by the curve y = log_e(x-2), y=0 and y=h to create a bowl. Find the exact volume of the bowl.

2) The portion of the curve y = e^x -1 from x=0 to x=1 is rotated about the x=axis. Find the volume of revolution generated

3) Calculate the exact area of the region bounded by the curve y = e^2x, y axis and the line y=e4

$\text{The second one can just be done by formula.}\\ \text{If }y=e^x-1\text{, then expanding a perfect square}\\ \text{gives }y=e^{2x}-2e^x+1$
$\text{So since }V=\pi \int_a^b y^2\,dx\\ \begin{align*}V&=\pi \int_0^1 (e^{2x}-2e^x+1)dx\\ &= \pi \left[\frac{e^{2x}}{2}-2e^x+x\right]_0^1\\&= \pi \left(\frac{e^2}{2}-2e+\frac{5}{2}\right)\end{align*}$
Before we dive into questions 1 and 3, have you drawn a diagram for both of them? You will not see what to do by purely relying on the given information itself.

hsotuhsa · « **Reply #3211 on:** January 17, 2018, 07:56:35 pm »

ThankYou so much for question 2.

And yes, I have drawn up graphs for question 1 and 3

RuiAce · « **Reply #3212 on:** January 17, 2018, 08:03:09 pm »

$\text{When you look at your graph for Q3,}\\ \text{you could be tempted to do something like this.}$

Possible method

$\text{You may consider analysing the question as an area}\\ \text{bound above the }y\text{-axis.}$
$\text{To do this, you would rearrange }y=e^{2x}\text{ to get }x=\frac{1}{2}\ln y\\ \text{and then your area integral is just}\\ A = \int_1^{e^4} \frac{1}{2}\ln y\,dy$
But of course, you don't know how to integrate log in 2U. So you need an alternate approach.

$\text{Consider a rectangle, whose vertices are the points}\\ (0,0), (0,e^4), (2,e^4), (2,0).\\ \text{Draw this rectangle }\textbf{on top of}\text{ the graph you have already sketched.}$
Note: The point $(2,e^4)$ actually lies on the curve $y=e^{2x}$.
$\text{The integral }\int_0^2 e^{2x}\,dx\text{ will give you the area}\\ \textbf{under }y=e^{2x}\text{ between 0 and 2.}\\ \text{But that's not what you want.}$
$\text{Instead, you want the area }\textbf{above }y=e^{2x}\\ \text{but }\textbf{below}\, y=e^4.\\ \text{So this is now an area between curves problem.}$
Note: $e^4$ is a constant. Therefore its primitive is $e^4 x$
$\text{The upper curve will be }y=e^4\\ \text{and the lower curve will be }y=e^{2x}.\\ \text{Hence, your area integral is}\\ \begin{align*}A&=\int_0^2 (e^4-e^{2x})\,dx\\ &= \left[e^4 x - \frac{e^{2x}}{2}\right]_0^2\\ &= 2e^4 - \frac{e^4}{2} + \frac{1}{2}\\ &= \frac{1}{2}(3e^4+1)\end{align*}$
_____________________________________________________________________________
$\text{For Q1, recall that you're rotating about the }y\text{-axis.}\\ \text{Therefore, what you need to make the subject is }x^2.\\ \textbf{NOT }y^2.$
$\begin{align*}y&=\log_e(x-2)\\ e^y&=x-2\\ x&=e^y+2\\ x^2&=(e^y+2)^2\\ &= e^{2y}+4e^y+4\end{align*}$
$\text{So for this one, your volume integral is}\\ V=\pi \int_0^h (e^{2y}+4e^y+4)\,dy$

beeangkah · « **Reply #3213 on:** January 19, 2018, 05:44:47 pm »

Not sure if this is being posted correctly...

Any help would be appreciated!

RuiAce · « **Reply #3214 on:** January 19, 2018, 05:48:33 pm »

beeangkah · « **Reply #3215 on:** January 19, 2018, 05:57:21 pm »

Quote from: RuiAce on January 19, 2018, 05:48:33 pm

$\text{Since we are rotating about the }y\text{-axis}\\ \text{we need to make }\boxed{x^2}\text{ the subject.}$
$y=\log_e x \implies x = e^y\\ \text{so our volume integral is}\\ \begin{align*}V&=\pi\int_1^3 \left(e^x\right)^2\,dx\\ &= \pi\int_1^3 e^{2x}\,dx\\ &=\pi\left( e^6-e^2\right)\end{align*}$

Thanks! But the answers at the back say
$pi/2[e^2 (e^4 - 1)]$

RuiAce · « **Reply #3216 on:** January 19, 2018, 06:03:53 pm »

My bad - dropping the 2 was my mistake and is fixed. (Basically it comes out of the statement $ \int e^{2x}\,dx = \frac{e^{2x}}{2}+C $)

Other than that it's the same. They just factorised: $e^6-e^2 = e^2(e^4-1)$

brooksykait · « **Reply #3217 on:** January 21, 2018, 02:30:59 pm »

Hey Jake, I have a question.
A circle has centre C(-1,3) and radius 5 units
a) Find the equation of the circle
b) The line 3x - y +1 = 0 meets the circle at 2 points, find their coordinates
c) Let the coordinates be X and Y, where Y is the coordinate directly below centre C. Find the coordinates of point Z, where YZ is a diameter of the circle
d) Hence show ZXY = 90 degrees

Specifcally parts c) and d), a) and b) I'm ok with.
Thank youuuu

RuiAce · « **Reply #3218 on:** January 21, 2018, 02:58:11 pm »

Quote from: brooksykait on January 21, 2018, 02:30:59 pm

Hey Jake, I have a question.
A circle has centre C(-1,3) and radius 5 units
a) Find the equation of the circle
b) The line 3x - y +1 = 0 meets the circle at 2 points, find their coordinates
c) Let the coordinates be X and Y, where Y is the coordinate directly below centre C. Find the coordinates of point Z, where YZ is a diameter of the circle
d) Hence show ZXY = 90 degrees

Specifcally parts c) and d), a) and b) I'm ok with.
Thank youuuu

The wording of this question is extremely ambiguous, as the way part c) is written does not make sense. Following some backtracking, it was presumed that X and Y are the coordinates of the points of intersection found in part b), with the remainder information suggesting that Y must be the point (-1,-2) (because it is directly below the circle's center, C).

Assuming no typo was placed, please state the source of this question for future reference.

Making the assumption I suggested above

$\text{So }X\text{ is }(2,7)\text{ and }Y\text{ is }(-1,-2).\\ \text{This would mean }Z\text{ is the point }(-1,8).$
Essentially, because $Y$ is directly below the circle's centre, the diameter of the circle through $Y$ must pass through the point on the circle, but directly above the circle's centre. This can be found by just noting that the circle's radius has length $5$, so its diameter has length $10$.
$\text{Now, }XY\text{ is the line }3x-y+1=0\\ \text{and hence has gradient 3.}$
$\text{The gradient of }XZ\text{ is }m_{XZ} = \frac{8-7}{-1-2} = -\frac13.$
$\text{So clearly, }m_{_{XZ}}m_{_{XY}} = -1\\ \text{suggesting that }XZ \perp XY\text{, and thus }\angle ZXY = 90^\circ.$

as111 · « **Reply #3219 on:** January 21, 2018, 06:37:21 pm »

Hi, Could somebody please help me with this question?

RuiAce · « **Reply #3220 on:** January 21, 2018, 06:45:34 pm »

Quote from: as111 on January 21, 2018, 06:37:21 pm

Hi, Could somebody please help me with this question?

$\text{Write }T_n = a+(n-1)d\text{ in the usual sense.}$
$\text{We are told that }T_{13}=27\text{ and }T_7 = 3T_2.\\ \text{This gives the }\textbf{simultaneous equations}\\ \begin{align*}a+12d&=27\tag{1}\\ a+6d&=3(a+d)\\ 2a-3d&=0\tag{2}\end{align*}$
$\text{Solving this gives }\boxed{a=3, d=2}.\\ \text{From here, all we require is }S_{10} = \frac{10}{2}\left[2(3)+ 9(2)\right] = 120$

as111 · « **Reply #3221 on:** January 21, 2018, 06:58:19 pm »

Thanks heaps

angela23 · « **Reply #3222 on:** January 21, 2018, 07:48:52 pm »

Hi! I'm struggling with this locus question:

A point P(x, y) moves so that the sum of the squares of its distance from each of the points A(-1, 0) and B(3, 0) is equal to 40.
Show that the locus of P(x, y) is a circle, and state its radius and centre.

Thanks!

RuiAce · « **Reply #3223 on:** January 21, 2018, 08:04:59 pm »

Quote from: angela23 on January 21, 2018, 07:48:52 pm

Hi! I'm struggling with this locus question:

A point P(x, y) moves so that the sum of the squares of its distance from each of the points A(-1, 0) and B(3, 0) is equal to 40.
Show that the locus of P(x, y) is a circle, and state its radius and centre.

Thanks!

$\text{The given information is that }\boxed{PA^2+PB^2=40}.$
$\text{So via the distance formula,}\\ \begin{align*}\left[(x+1)^2+y^2\right]+\left[(x-3)^2+y^2\right]&=40\\ 2x^2-4x+10+ 2y^2&=40\\ x^2-2x+y^2&=15\\ (x-1)^2+y^2&=16\end{align*}$
$\text{which is a circle with radius }4\text{ and center }(1,0)$

gumscape · « **Reply #3224 on:** January 22, 2018, 02:47:47 pm »

hellooooooo! can anyone give me a general guideline/tips & tricks when it comes to sketching the primitive function? I'm having a hard time with it

Search the forums now!

We have moved!

Author Topic: Mathematics Question Thread (Read 1626086 times) Tweet Share

RuiAce

Re: Mathematics Question Thread

hsotuhsa

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

beeangkah

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

beeangkah

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

brooksykait

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

as111

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

as111

Re: Mathematics Question Thread

angela23

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

gumscape

Re: Mathematics Question Thread

Recent Posts

User:
Password: