Author Topic: 4U Maths Question Thread (Read 665242 times) Tweet Share

bluecookie · « **Reply #1305 on:** July 12, 2017, 04:34:19 pm »

Sketch:

(x^2+y^2)^2=2(x^2-y^2)
and x^2+y^2=3xy

seventeenboi · « **Reply #1306 on:** July 12, 2017, 04:42:36 pm »

Hi

the ansewr to this question is C - i understand that there must be a vertical tangent at x, but how would you choose between a and c (concavity ??)
thanks

Shadowxo · « **Reply #1307 on:** July 12, 2017, 05:06:59 pm »

Quote from: seventeenboi on July 12, 2017, 04:42:36 pm

Hi

the ansewr to this question is C - i understand that there must be a vertical tangent at x, but how would you choose between a and c (concavity ??)
thanks

For a question like that, I'd probably just test each of the quadrants. Quadrant 1, x and y are both positive so dy/dx is positive. Quadrant 2, y is positive and x is negative, so dy/dx is negative etc.
Then you just have to choose between c and d, when x approaches 0 dy/dx approaches positive/negative infinity so it's c

Hope this helps

seventeenboi · « **Reply #1308 on:** July 12, 2017, 06:12:00 pm »

Quote from: Shadowxo on July 12, 2017, 05:06:59 pm

For a question like that, I'd probably just test each of the quadrants. Quadrant 1, x and y are both positive so dy/dx is positive. Quadrant 2, y is positive and x is negative, so dy/dx is negative etc.
Then you just have to choose between c and d, when x approaches 0 dy/dx approaches positive/negative infinity so it's c

Hope this helps

OHHH TRUE ok thankss :DDDD

RuiAce · « **Reply #1309 on:** July 12, 2017, 07:04:31 pm »

Quote from: bluecookie on July 12, 2017, 04:34:19 pm

Sketch:

(x^2+y^2)^2=2(x^2-y^2)
and x^2+y^2=3xy

These are just random implicit curves that authors throw 4U students when they're bored. Please indicate the source of these questions.

(My honest opinion though - not even worth attempting unless you've done polar curve sketching, which is not in the HSC course. Part b) can possibly be done by using the quadratic formula on $y$ though.)

bluecookie · « **Reply #1310 on:** July 12, 2017, 10:48:01 pm »

Quote from: RuiAce on July 12, 2017, 07:04:31 pm

These are just random implicit curves that authors throw 4U students when they're bored. Please indicate the source of these questions.

(My honest opinion though - not even worth attempting unless you've done polar curve sketching, which is not in the HSC course. Part b) can possibly be done by using the quadratic formula on $y$ though.)

Ohhh thanks, yeah I was wondering why the teacher was setting us so unseenly and obscure questions lol xP Thanks, I feel happier knowing I won't have to cover them at all XD They came from his holiday homework. We don't usually get this much, but since it's the holidays before the last school assessment he decided to give us extra

bluecookie · « **Reply #1311 on:** July 12, 2017, 11:18:57 pm »

a) Find integral from 0 to pi of [dx/(5-4cosx)]
b) Un=integral from 0 to pi of [cosnxdx/(5-4cosx)], show that U(n+1)+U(n-1)-5/2*Un=0.
c) Calculte U0, U1, and hence find U2, U3.

RuiAce · « **Reply #1312 on:** July 13, 2017, 08:33:02 am »

Quote from: bluecookie on July 12, 2017, 11:18:57 pm

a) Find integral from 0 to pi of [dx/(5-4cosx)]
b) Un=integral from 0 to pi of [cosnxdx/(5-4cosx)], show that U(n+1)+U(n-1)-5/2*Un=0.
c) Calculte U0, U1, and hence find U2, U3.

$t=\tan \frac{x}{2} \implies x = 2\tan^{-1}t \implies dx = \frac{2\, dt}{1+t^2}\\ \text{and }\cos x = \frac{1-t^2}{1+t^2}$
$\begin{align*}U_0&=\int_0^\pi \frac{dx}{5-4\cos x}\\ &=\int_0^{\infty} \frac{\frac2{1+t^2}dt}{\frac{5+5t^2}{1+t^2}-\frac{4-4t^2}{1+t^2}} \\ &= \int_0^\infty \frac{2}{1+9t^2}dt\\ &= \lim_{R\to \infty}\left[\frac23 \tan^{-1}3t\right]_0^R\\ &= \frac23\left(\frac{\pi}{2}-0\right)\\ &= \frac{\pi}3\end{align*}$
________________________________
$\textit{Lemma}: \cos A + \cos B = 2\cos \left(\frac{A+B}2\right)\cos \left(\frac{A-B}2\right)\\ \text{Proof: Expanding the compound angles we have }\cos(x-y)+\cos(x+y)=2\cos x \cos y\\ \text{Let }x-y=A\text{ and }x+y=B\text{. The remainder is left as an exercise.}$
$\begin{align*}U_{n+1}+U_{n-1}&=\int_0^\pi \frac{\cos [(n+1)x]+\cos [(n-1)x]}{5-4\cos x}dx\\ &=\int_0^\pi \frac{2\cos nx \cos x}{5-4\cos x}dx \end{align*}\\ \text{from the lemma.}$
$\begin{align*}\therefore U_{n+1}+U_{n-1}-\frac52U_n&= \int_0^\pi \frac{2\cos nx \cos x}{5-4\cos x}dx-\frac52\int_0^\pi \frac{\cos nx}{5-4\cos x}dx\\ &= \frac12 \int_0^\pi \frac{4\cos nx \cos x - 5\cos nx}{5-4\cos x}dx\\&=\frac12\int_0^\pi \frac{\cos nx(4\cos x-5)}{5-4\cos x}dx\\ &= \frac12 \int_0^\pi -\cos nx \, dx\\ &= 0\end{align*}$
________________________________
$\text{Note that}\\ \begin{align*}U_1&=\int_0^\pi \frac{\cos x}{5-4\cos x}dx\\ &= -\frac14 \int_0^\pi \frac{-4\cos x}{5-4\cos x}dx\\ &= -\frac14 \int_0^\pi \frac{5-4\cos x}{5-4\cos x}dx + \frac54 \int_0^\pi \frac{1}{5-4\cos x}dx\\ &= -\frac\pi 4 + \frac54 U_0\\ &= \frac{\pi}6\end{align*}$
You should have no trouble using the recurrence formula to find U2 and U3.

seventeenboi · « **Reply #1313 on:** July 13, 2017, 08:14:33 pm »

hiii

how would you do this question HELPPp
thankkkk

RuiAce · « **Reply #1314 on:** July 13, 2017, 08:22:50 pm »

Quote from: seventeenboi on July 13, 2017, 08:14:33 pm

hiii
how would you do this question HELPPp
thankkkk

$\text{For positive }x,y\text{ we can prove that }\frac{x+y}{2}\ge \sqrt{xy}\\ \text{This is left as your exercise.}$
$\text{Also noting }\frac{w+z}{2}\ge \sqrt{wz}\text{ we have}\\ \begin{align*}\frac{w+x+y+z}{4}&= \frac{\frac{x+y}2+\frac{w+z}2}{2}\\ &\ge \frac{\sqrt{xy}+\sqrt{wz}}{2}\end{align*}$
$\text{Applying the two-variable AM-GM inequality }\textit{again,}\\ \frac{w+x+y+z}{4} \ge \sqrt{\sqrt{xy}\sqrt{wz}}= \sqrt[4]{wxyz}\\ \text{So }\boxed{w+x+y+z=4\sqrt[4]{wxyz}}\\ \text{Substituting }w=a^4, x=b^4, y=c^4\text{ and }z=d^4\text{ proves the desired result.}$

claudiarosaliaa · « **Reply #1315 on:** July 13, 2017, 08:39:30 pm »

Can anyone solve this Volumes questions? Its Questions 15.a) from the 2015 James Ruse Trial Paper:

The diagram shows a segment of the circle X² + Y² = r² which is rotated about the y axis to from a collar. This collar is thus a sphere was a symmetrical hole through it. Let the hole be of the height 2h as shown.

Use the method of cylidical shells to show that the volume of the material in the collar is given by the integral

4pi(integral between r and the square root of (r²-h²) of X multiplied by the square root of (r²-X²)dx)

Evaluate the integral to show that the volume of material in the collar is a function of h only independent of r.

Sidenote: How do I enter maths symbols into my posts?

RuiAce · « **Reply #1316 on:** July 13, 2017, 08:49:18 pm »

Quote from: claudiarosaliaa on July 13, 2017, 08:39:30 pm

Can anyone solve this Volumes questions? Its Questions 15.a) from the 2015 James Ruse Trial Paper:

The diagram shows a segment of the circle X² + Y² = r² which is rotated about the y axis to from a collar. This collar is thus a sphere was a symmetrical hole through it. Let the hole be of the height 2h as shown.

Use the method of cylidical shells to show that the volume of the material in the collar is given by the integral

4pi(integral between r and the square root of (r²-h²) of X multiplied by the square root of (r²-X²)dx)

Evaluate the integral to show that the volume of material in the collar is a function of h only independent of r.

Sidenote: How do I enter maths symbols into my posts?

Assuming a diagram was provided.
$\text{Examine the features of a typical shell}\\ \text{Rotation about the }y\text{-axis means its radius will be }x\\ \text{The height of a typical shell depends on this }x\text{-coordinate}$
$\text{Rearranging the equation of the circle, }y=\pm \sqrt{r^2-x^2}\\ \text{So we may deduce that the height of the shell is }2\sqrt{r^2-x^2}\\ \text{So the volume of a typical shell is }2\pi r h \,\delta x, \text{ i.e.}\\ \boxed{\delta V = 2\pi x (2\sqrt{r^2-x^2})\delta x }$
$\text{Applying the method of cylindrical shells, we have our required integrand}\\ \text{The boundaries of the integral are demonstrated by the diagram}$
$\text{Clearly, the furthest }x\text{ can go is all the way to }r\\ \text{As for the closest it can go}\\ \text{Since we know that }y=h\text{ at our point of interest, upon substitution}\\ x^2+h^2=r^2 \implies x = \sqrt{r^2-h^2}$
It should be clear why I took a positive square root.
$\text{Hence }V=4\pi \int_{\sqrt{r^2-h^2}}^{r}x\sqrt{r^2-x^2}dx$
$\text{All that's left to do is to evaluate this integral}\\ \text{We may (but do not need to) make a substitution }u=r^2-x^2\implies -\frac12du=x\, dx\\ \text{to obtain }V=2\pi \int_{h^2}^0-\sqrt u \, du\text{ and evaluating this it's clear that }r\text{ has vanished}$
Last bit subject to a bit of computational inaccuracy, but the method is still the same.

LaTeX guide

claudiarosaliaa · « **Reply #1317 on:** July 13, 2017, 09:11:16 pm »

Can anyone explain why (D) is the correct answer?

claudiarosaliaa · « **Reply #1318 on:** July 13, 2017, 09:19:35 pm »

Quote from: RuiSmash on July 13, 2017, 08:49:18 pm

Assuming a diagram was provided.
$\text{Examine the features of a typical shell}\\ \text{Rotation about the }y\text{-axis means its radius will be }x\\ \text{The height of a typical shell depends on this }x\text{-coordinate}$
$\text{Rearranging the equation of the circle, }y=\pm \sqrt{r^2-x^2}\\ \text{So we may deduce that the height of the shell is }2\sqrt{r^2-x^2}\\ \text{So the volume of a typical shell is }2\pi r h \,\delta x, \text{ i.e.}\\ \boxed{\delta V = 2\pi x (2\sqrt{r^2-x^2})\delta x }$
$\text{Applying the method of cylindrical shells, we have our required integrand}\\ \text{The boundaries of the integral are demonstrated by the diagram}$
$\text{Clearly, the furthest }x\text{ can go is all the way to }r\\ \text{As for the closest it can go}\\ \text{Since we know that }y=h\text{ at our point of interest, upon substitution}\\ x^2+h^2=r^2 \implies x = \sqrt{r^2-h^2}$
It should be clear why I took a positive square root.
$\text{Hence }V=4\pi \int_{\sqrt{r^2-h^2}}^{r}x\sqrt{r^2-x^2}dx$
$\text{All that's left to do is to evaluate this integral}\\ \text{We may (but do not need to) make a substitution }u=r^2-x^2\implies -\frac12du=x\, dx\\ \text{to obtain }V=2\pi \int_{h^2}^0-\sqrt u \, du\text{ and evaluating this it's clear that }r\text{ has vanished}$
Last bit subject to a bit of computational inaccuracy, but the method is still the same.

LaTeX guide

Thank you!

RuiAce · « **Reply #1319 on:** July 13, 2017, 09:19:48 pm »

Quote from: claudiarosaliaa on July 13, 2017, 09:11:16 pm

Can anyone explain why (D) is the correct answer?

$\text{In every case, the parametrisation reduces to }xy=c^2\\ \text{So consider the }\textbf{range}\text{ restrictions imposed for each parametrisation}\\ \text{as this directly affects the }x\text{ values permitted in the graph.}$
$t^2+1\text{ has a range of }Y\ge 1\\ 1-t^2\text{ has a range of }Y \le 1\\ \sqrt{1-t^2}\text{ has a range of }0\le Y\le 1\\ \sin t\text{ has a range of }-1 \le Y \le 1$
$\text{Carefully observe the graph given}\\ \text{The graph given is the hyperbola }\left(ct, \frac{c}t\right)\text{ but specifically for }-1\le t \le 1\\ \text{Hence, matching ranges, D is the answer}$
As an exercise, you should use either technology or by hand, attempt to sketch out what parts of the hyperbola would be present in the other parametrisations.

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