Author Topic: 3U Maths Question Thread (Read 1465110 times) Tweet Share

jamonwindeyer · « **Reply #1395 on:** February 04, 2017, 11:46:47 am »

Quote from: bsdfjn;lkasn on February 04, 2017, 11:42:17 am

Yes I have! Also it's good to know that you can't perform integrations with respect to the y-axis because I was starting to worry that we hadn't been taught that yet (and we've finished the topic )

Thanks for checking that through for me, could you help me with integrating tan²x? I've just tried to rearrange the tan2x formula to put in as the integrand but am getting very very confused

Try $\tan^2{x}=\sec^2{x}-1$, a little play on the pythagorean identity

armtistic · « **Reply #1396 on:** February 04, 2017, 12:15:18 pm »

Hey guys.

So I'm trying to do this question but idk how to get the variables in terms of each other so I can use calculus to find the largest area

jakesilove · « **Reply #1397 on:** February 04, 2017, 12:26:57 pm »

Quote from: armtistic on February 04, 2017, 12:15:18 pm

Hey guys.

So I'm trying to do this question but idk how to get the variables in terms of each other so I can use calculus to find the largest area

Let's try to find ABCD, noting that AB is straight, and DC is an arc. We know that the area of the triangle OAB is going to be

$A=\frac{1}{2}(2r)(2r)\sin(x)$

As two of the sides are the same (radius), and the middle angle is x. Now, we need to subtract the area of the sector of the circle, ODC. This area is calculated by

$A=\frac{r^2 x}{2}$

As the angle is still x.

So, the total area is going to be

$A=\frac{1}{2}(2r)(2r)\sin(x)-\frac{r^2 x}{2}$

Now, noting that r is some constant, we just differentiate with respect to x in order to find the largest value of A.

This gives us

$\frac{dA}{dx}=\frac{1}{2}r^2(4\cos(x)-1)$

Which has a zero at

$\cos(x)=\frac{1}{4}$

Hopefully you can get out an answer from here!

armtistic · « **Reply #1398 on:** February 04, 2017, 02:29:58 pm »

Quote from: jakesilove on February 04, 2017, 12:26:57 pm

Let's try to find ABCD, noting that AB is straight, and DC is an arc. We know that the area of the triangle OAB is going to be

$A=\frac{1}{2}(2r)(2r)\sin(x)$

As two of the sides are the same (radius), and the middle angle is x. Now, we need to subtract the area of the sector of the circle, ODC. This area is calculated by

$A=\frac{r^2 x}{2}$

As the angle is still x.

So, the total area is going to be

$A=\frac{1}{2}(2r)(2r)\sin(x)-\frac{r^2 x}{2}$

Now, noting that r is some constant, we just differentiate with respect to x in order to find the largest value of A.

This gives us

$\frac{dA}{dx}=\frac{1}{2}r^2(4\cos(x)-1)$

Which has a zero at

$\cos(x)=\frac{1}{4}$

Hopefully you can get out an answer from here!

That makes so much sense! Thanks

Also, could someone explain how to do this queston to me, because idek what a cyclic pentagon is

jakesilove · « **Reply #1399 on:** February 04, 2017, 02:38:20 pm »

Quote from: armtistic on February 04, 2017, 02:29:58 pm

That makes so much sense! Thanks

Also, could someone explain how to do this queston to me, because idek what a cyclic pentagon is

Hey! A cyclic pentagon is a pentagon in which all vertexes (points) lie on a single circle. Basically, it's asking you to prove that it is a regular pentagon (ie. all sides are the same length, angles are the same etc.). Knowing that terminology is definitely outside the syllabus, but why don't you have a crack at proving it and if you're still stuck let us know?

RuiAce · « **Reply #1400 on:** February 04, 2017, 02:44:16 pm »

Quote from: armtistic on February 04, 2017, 02:29:58 pm

That makes so much sense! Thanks

Also, could someone explain how to do this queston to me, because idek what a cyclic pentagon is

Briefly expanding on what Jake said

Cyclic can be titled to any polygon, because it's simply an adjective. Concyclic points are points that lie on the same circle, so a cyclic polygon is a polygon whose vertices all lie on the same circle. So you can have cyclic pentagons, hexagons, etc.

A cyclic quadrilateral is simply a 4-sided shape with all vertices on a circle.

A cyclic pentagon did once appear in the 4U exam though.

RuiAce · « **Reply #1401 on:** February 04, 2017, 03:36:43 pm »

Ignore the bit in blue. The congruent triangles proof is not necessary.

Simply use the theorem "the perpendicular from the center of the chord bisects the chord", or rather a converse of it:
"the line from the center of the circle to the midpoint of a chord, meets the chord perpendicularly"

Anyway, bloody hard. To prove something's a cyclic pentagon, it helps to consider TWO cyclic quadrilaterals... somehow.

Mention anything that's illegible.

armtistic · « **Reply #1402 on:** February 04, 2017, 05:55:26 pm »

Quote from: RuiAce on February 04, 2017, 03:36:43 pm

(Image removed from quote.)Ignore the bit in blue. The congruent triangles proof is not necessary.

Simply use the theorem "the perpendicular from the center of the chord bisects the chord", or rather a converse of it:
"the line from the center of the circle to the midpoint of a chord, meets the chord perpendicularly"

Anyway, bloody hard. To prove something's a cyclic pentagon, it helps to consider TWO cyclic quadrilaterals... somehow.

Mention anything that's illegible.

Ohh that makes sense.

Cos 3 points uniquely define a circle so if a circle shares 3 of the same points with another circle then they are the same circle

Lmao i thought there was some rules about exterior angles or something like in cyclic quads

Thanks for the help

bsdfjnlkasn · « **Reply #1403 on:** February 04, 2017, 06:26:54 pm »

Hey there,

I was wondering if I could get some help with these questions

I was able to do part a) of q28 but wasn't sure if parts b) and c) were 4u (i.e. too hard/irrelevant for 3u). If they're not, could someone please explain how to solve the rest of the question?

For Q13, I could do part a) but was struggling with b) ii. So any help there would also be appreciated

Thank you!!

RuiAce · « **Reply #1404 on:** February 04, 2017, 06:34:13 pm »

Quote from: bsdfjn;lkasn on February 04, 2017, 06:26:54 pm

For Q13, I could do part a) but was struggling with b) ii. So any help there would also be appreciated

(Image removed from quote.)

Thank you!!

$\begin{align*}\int_0^{\frac{\pi}{4}}\sin 2x \cos 2x\, dx&=\int_0^{\frac{\pi}4}\frac{1}{2}\sin 4x\, dx\\ &= \left[-\frac{1}{8}\cos 4x\right]_0^{\frac{\pi}4}\\ &= \frac{1}{8}\end{align*}$

bsdfjnlkasn · « **Reply #1405 on:** February 04, 2017, 06:45:28 pm »

Quote from: RuiAce on February 04, 2017, 06:34:13 pm

$\begin{align*}\int_0^{\frac{\pi}{4}}\sin 2x \cos 2x\, dx&=\int_0^{\frac{\pi}4}\frac{1}{2}\sin 4x\, dx\\ &= \left[-\frac{1}{8}\cos 4x\right]_0^{\frac{\pi}4}\\ &= \frac{1}{8}\end{align*}$

Thank you Rui, I'm not sure how you got the second line of your integration is there an identity for sin2xcos2x?

Could you check my working for this question? The answer is apparently 2/9 but I'm not sure where the sign swaps

bsdfjnlkasn · « **Reply #1406 on:** February 04, 2017, 06:48:02 pm »

Quote from: bsdfjn;lkasn on February 04, 2017, 06:45:28 pm

Thank you Rui, I'm not sure how you got the second line of your integration is there an identity for sin2xcos2x?

Also, could you check my working for this question? The answer is apparently 2/9 but I'm not sure where the sign swaps

(Image removed from quote.)

RuiAce · « **Reply #1407 on:** February 04, 2017, 06:49:25 pm »

Quote from: bsdfjn;lkasn on February 04, 2017, 06:26:54 pm

Hey there,

I was wondering if I could get some help with these questions

I was able to do part a) of q28 but wasn't sure if parts b) and c) were 4u (i.e. too hard/irrelevant for 3u). If they're not, could someone please explain how to solve the rest of the question?

(Image removed from quote.)

Important warning: The Extension questions in Cambridge are designed purely for the sake of self-interest. They are only for you to build superior techniques when you feel necessary. They do not reflect the scope of the 2U, 3U or 4U syllabuses and therefore I will not, in general, do these questions from here on after.
$\text{Anti-differentiating the result of part a) gives}\\ \begin{align*}\int_0^N e^{-x}\sin x\, dx&= \left[-\frac{1}{2}e^{-x}(\sin x + \cos x)\right]_0^N\\ &=\frac{1}{2}-\frac{1}{2}e^{-N}(\sin N + \cos N)\end{align*}$
$\text{Clearly, the behaviour of the exponential will dominate over that of the trigonometric function}\\ \text{The trig functions are going to hang between 0 and 1, but}\\ \text{the exponential will go all the way down.}$
$\text{Hence, limiting }N:\\ \begin{align*}\int_0^\infty e^{-x}\sin x\, dx&=\lim_{N\to \infty}\left(\frac{1}{2}-\frac{1}{2}e^{-N}(\sin N + \cos N)\right)\\&=\frac{1}{2}-0\\&=\frac12\end{align*}$
__________________________________________
$\text{Using the result of part b)}\\ \int_0^\pi e^{-x}\sin x\, dx=\frac{1}{2}-\frac{1}{2}e^{-\pi}(\sin \pi + \cos \pi)=\frac{1}{2}+\frac{1}{2}e^{-\pi}$
$\text{Then}\\ \begin{align*}\int_{2\pi}^{3\pi} e^{-x}\sin x\, dx&= \left[-\frac{1}{2}e^{-x}(\sin x + \cos x)\right]_{2\pi}^{3\pi}\\ &=\frac{1}{2}e^{-2\pi}(\sin 2\pi + \cos 2\pi)-\frac{1}{2}e^{-3\pi}(\sin 3\pi + \cos 3\pi)\\ &=\frac{1}{2}e^{-2\pi}+\frac{1}{2}e^{-3\pi}\end{align*}$
$\text{A pattern forms. One may check that}\\ \int_{4\pi}^{5\pi} e^{-x}\sin x\, dx=\frac{1}{2}e^{-4\pi}+\frac{1}{2}e^{-5\pi}$
$\text{Hence, by splitting the integral apart infinitely}\\ \begin{align*}\text{area of top arches}&= \int_0^\pi e^{-x}\sin x\, dx+\int_{2\pi}^{3\pi} e^{-x}\sin x\, dx+\dots\\ &= \left(\frac{1}{2}+\frac{1}{2}e^{-\pi}\right) + \left(\frac{1}{2}e^{-2\pi}+\frac{1}{2}e^{-3\pi}\right)+\dots\\ &= \frac{1}{2}\underbrace{(1+e^{-\pi}+e^{-2\pi}+e^{-3\pi}+\dots)}_{\text{geometric series}}\\ &= \frac{1}{2(1-e^{-\pi})}\\ &= \frac{e^{\pi}}{2(e^\pi - 1)}\end{align*}$
__________________________________________
$\text{I leave it to you to check that:}\\ \int_{\pi}^{2\pi} e^{-x}\sin x\,dx=-\frac{1}{2}e^{-\pi}-\frac{1}{2}e^{-2\pi}\\ \int_{3\pi}^{4\pi} e^{-x}\sin x\,dx= -\frac{1}{2}e^{-3\pi}-\frac{1}{2}e^{-4\pi}\\ \text{and etc.}$
$\text{So we have}\\ \begin{align*} \int_{\pi}^{2\pi} e^{-x}\sin x\,dx+\int_{3\pi}^{4\pi} e^{-x}\sin x\,dx+\dots &=-\frac{1}{2}e^{-\pi}(1+e^{-\pi}+e^{-2\pi}+e^{-3\pi}+\dots)\\ &=- \frac{1}{2(e^{\pi}-1)} \end{align*}$
$\text{Finally, by tearing apart the original integral in b) infinitely:}\\ \begin{align*}\int_0^\infty e^{-x}\sin x\, dx&= \int_{0}^{\pi} e^{-x}\sin x\,dx+ \int_{\pi}^{2\pi} e^{-x}\sin x\,dx + \int_{2\pi}^{3\pi} e^{-x}\sin x\,dx+ \int_{3\pi}^{4\pi} e^{-x}\sin x\,dx+\dots\\ &=\left(\int_{0}^{\pi} e^{-x}\sin x\,dx\right)+ \int_{2\pi}^{3\pi} e^{-x}\sin x\,dx+\dots)+\left( \int_{\pi}^{2\pi} e^{-x}\sin x\,dx+ \int_{3\pi}^{4\pi} e^{-x}\sin x\,dx+\dots\right)\\&= \frac{e^{\pi}}{2(e^\pi-1)}-\frac{1}{2(e^\pi -1)}\\ &=\frac{e^\pi-1}{2(e^\pi -1)}&= \frac{e^\pi-1}{2(e^\pi-1)}\\&= \frac12 \end{align*}\\ \text{agreeing with the result of b).}$

Quote from: bsdfjn;lkasn on February 04, 2017, 06:45:28 pm

Thank you Rui, could you check my working for this question? The answer is apparently 2/9 but I'm not sure where the sign swaps

(Image removed from quote.)

Integral of cosine is negative sine, not positive sine.

The sign swapped on line 2.

bsdfjnlkasn · « **Reply #1408 on:** February 04, 2017, 07:07:54 pm »

Quote from: RuiAce on February 04, 2017, 06:34:13 pm

$\begin{align*}\int_0^{\frac{\pi}{4}}\sin 2x \cos 2x\, dx&=\int_0^{\frac{\pi}4}\frac{1}{2}\sin 4x\, dx\\ &= \left[-\frac{1}{8}\cos 4x\right]_0^{\frac{\pi}4}\\ &= \frac{1}{8}\end{align*}$

Hey Rui, thanks for taking the time to answer all my questions, especially the extension one

With your solution here, I'm not sure how you got the second line of your integration. Is there an identity for sin2xcos2x?

RuiAce · « **Reply #1409 on:** February 04, 2017, 07:09:04 pm »

Quote from: bsdfjn;lkasn on February 04, 2017, 07:07:54 pm

Hey Rui, thanks for taking the time to answer all my questions, especially the extension one

With your solution here, I'm not sure how you got the second line of your integration. Is there an identity for sin2xcos2x?

$\sin 2u = 2\sin u \cos u\\ \text{Let }u=2x\\ \sin 4x = 2\sin 2x \cos 2x$
$\text{Double angles also work on }2x\text{'s}\\ \text{Not just }x\text{'s}$

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