Author Topic: Mathematics Question Thread (Read 1626918 times) Tweet Share

Calley123 · « **Reply #3435 on:** April 20, 2018, 09:41:11 pm »

Quote from: RuiAce on April 20, 2018, 06:08:48 pm

$\text{If we're rotating about the }y\text{-axis, then we need to make }x\text{ the subject.}\\ y = e^x \implies x = \log_e y$
$\text{So we need to consider}\\ V = \pi \int_3^5 (\ln x)^2\,dx.$
$\text{The relevant Simpson's rule approximation is}\\ V \approx \pi \frac{5-3}{3\times 4}\left[ (\ln 3)^2 + 4(\ln 3.5)^2 + 2 (\ln 4)^2 + 4 (\ln 4.5)^2 + (\ln 5)^2 \right]\\ \text{which you can put into your calculator.}$

Much appreciated !

Calley123 · « **Reply #3436 on:** April 21, 2018, 08:06:36 am »

Lol its me again,
I keep getting the wrong answers for these ! - Question 10 & 12
Thank you in advance

RuiAce · « **Reply #3437 on:** April 21, 2018, 09:59:50 am »

Quote from: Calley123 on April 21, 2018, 08:06:36 am

Lol its me again,
I keep getting the wrong answers for these ! - Question 10 & 12
Thank you in advance

Q12 is just $ \pi \int_0^{\pi/2}\cos x\,dx =\pi [\sin x]_0^{\pi/2}$

$\text{Solving }\sin x = \frac12\text{ for }0\le x \le 2\pi\\ \text{gives }x = \frac\pi6. \frac{5\pi}{6}\\ \text{recalling that sin is positive in the 1st and 2nd quadrant.}$
$\text{Then, this becomes an area between curves problem}\\ \text{where the upper curve is }y = \sin x\\ \text{and the lower curve is }y = \frac12.$
Alternatively, treat it as the area under $ y = \sin x$, minus a rectangle of height $ \frac12 $ and length $ \frac{5\pi}{6} - \frac\pi6$.
$\begin{align*}A&= \int_{\pi/6}^{5\pi/6} \left( \sin x - \frac12\right)\,dx\\ &= \left[-\cos x - \frac{x}{2} \right]_{\pi/6}^{5\pi/6}\\ &= \sqrt{3} - \frac\pi3\end{align*}$

Calley123 · « **Reply #3438 on:** April 21, 2018, 05:36:22 pm »

Quote from: RuiAce on April 21, 2018, 09:59:50 am

Q12 is just $ \pi \int_0^{\pi/2}\cos x\,dx =\pi [\sin x]_0^{\pi/2}$
$\text{Solving }\sin x = \frac12\text{ for }0\le x \le 2\pi\\ \text{gives }x = \frac\pi6. \frac{5\pi}{6}\\ \text{recalling that sin is positive in the 1st and 2nd quadrant.}$
$\text{Then, this becomes an area between curves problem}\\ \text{where the upper curve is }y = \sin x\\ \text{and the lower curve is }y = \frac12.$
Alternatively, treat it as the area under $ y = \sin x$, minus a rectangle of height $ \frac12 $ and length $ \frac{5\pi}{6} - \frac\pi6$.
$\begin{align*}A&= \int_{\pi/6}^{5\pi/6} \left( \sin x - \frac12\right)\,dx\\ &= \left[-\cos x - \frac{x}{2} \right]_{\pi/6}^{5\pi/6}\\ &= \sqrt{3} - \frac\pi3\end{align*}$

Thank you !

Calley123 · « **Reply #3439 on:** April 25, 2018, 12:46:34 pm »

Hey,
Help please..

Thanks in advance

RuiAce · « **Reply #3440 on:** April 25, 2018, 01:10:18 pm »

Quote from: Calley123 on April 25, 2018, 12:46:34 pm

Hey,
Help please..

Thanks in advance

\[ m_{PA} = \frac{y+2}{x-3}\text{ and }m_{PB} = \frac{y-7}{x+1} \]
$\text{If }m_{PA} = 2\,m_{PB}\text{, then}\\ \begin{align*} \frac{y+2}{x-3} &= 2\left( \frac{y-7}{x+1} \right)\\ (y+2)(x+1) &= 2(y-7)(x-3) \\ xy + 2x + y + 1 &= 2xy - 14x - 6y + 42\\ 0 &= xy - 16x - 7y + 40\end{align*}$

aryan.gupta1 · « **Reply #3441 on:** April 25, 2018, 05:45:59 pm »

Hey I have a probability question that im really struggling with !!!

in a litter of seven pups it is known that the first three born are all male. What is the probability that the next pup born will also be male?

Thanks in advance

RuiAce · « **Reply #3442 on:** April 25, 2018, 05:47:24 pm »

Quote from: aryan.gupta1 on April 25, 2018, 05:45:59 pm

Hey I have a probability question that im really struggling with !!!

in a litter of seven pups it is known that the first three born are all male. What is the probability that the next pup born will also be male?

Thanks in advance

The intuitive answer is just 1/2 assuming that both genders are equally likely. What are the answers saying?

aryan.gupta1 · « **Reply #3443 on:** April 25, 2018, 05:48:43 pm »

Lol, my teacher just got these questions and pasted them on a word doc, there aren't any answers :/

jamonwindeyer · « **Reply #3444 on:** April 25, 2018, 05:50:55 pm »

Quote from: aryan.gupta1 on April 25, 2018, 05:48:43 pm

Lol, my teacher just got these questions and pasted them on a word doc, there aren't any answers :/

This is a logic problem more than a maths problem

Definitely 1/2. Once we know the first three pups are male then we've sort of moved along the probability tree - So the next stage just becomes another 50/50 shot

it would be different if we didn't know for sure about the first three!

RuiAce · « **Reply #3445 on:** April 25, 2018, 05:52:02 pm »

Well in general, without any assumptions we should assume that the gender of one puppy is independent of the gender of another. That is to say, whether or not the first puppy is male should not influence whether or not the second is, and then the third, and then the fourth, and so on.

So if we assume that both genders are equally likely (which is just the natural assumption), then it'd just be 1/2. Otherwise, the question lacks information by a lot.

(Assuming that the probability that a puppy is male is 3/7 just using the information given is invalid. This is because we have no information about the fifth, sixth or seventh puppy to base our assumption on. If your teacher does that, I'd be questioning it a lot.)

Calley123 · « **Reply #3446 on:** May 03, 2018, 05:41:22 pm »

Quote from: RuiAce on April 25, 2018, 01:10:18 pm

\[ m_{PA} = \frac{y+2}{x-3}\text{ and }m_{PB} = \frac{y-7}{x+1} \]
$\text{If }m_{PA} = 2\,m_{PB}\text{, then}\\ \begin{align*} \frac{y+2}{x-3} &= 2\left( \frac{y-7}{x+1} \right)\\ (y+2)(x+1) &= 2(y-7)(x-3) \\ xy + 2x + y + 1 &= 2xy - 14x - 6y + 42\\ 0 &= xy - 16x - 7y + 40\end{align*}$

THANK YOUUUUUUUU!

Please help!

Mod Edit: Post merge

Calley123 · « **Reply #3447 on:** May 03, 2018, 06:14:19 pm »

This one too please

Quote from: RuiAce on April 20, 2018, 06:08:48 pm

$\text{If we're rotating about the }y\text{-axis, then we need to make }x\text{ the subject.}\\ y = e^x \implies x = \log_e y$
$\text{So we need to consider}\\ V = \pi \int_3^5 (\ln x)^2\,dx.$
$\text{The relevant Simpson's rule approximation is}\\ V \approx \pi \frac{5-3}{3\times 4}\left[ (\ln 3)^2 + 4(\ln 3.5)^2 + 2 (\ln 4)^2 + 4 (\ln 4.5)^2 + (\ln 5)^2 \right]\\ \text{which you can put into your calculator.}$

Hey RuiAce,

In the second line, did you mean ln(y)^2 dy rather than ln(x)^2 dx ( Since it rotates around y-axis )

Mod Edit: Post merge, use the "Modify" button to add to your previous post

owidjaja · « **Reply #3448 on:** May 03, 2018, 07:03:08 pm »

Hey guys,
So my teacher basically spent the last 5 minutes of the period to explain geometric progression to us and set us homework. Considering he's the type of teacher to just push on and I'm a slow learner, can someone please explain how geometric progression works?

Thanks in advance

RuiAce · « **Reply #3449 on:** May 03, 2018, 07:14:45 pm »

Quote from: Calley123 on May 03, 2018, 06:14:19 pm

This one too please
Hey RuiAce,

In the second line, did you mean ln(y)^2 dy rather than ln(x)^2 dx ( Since it rotates around y-axis )

Mod Edit: Post merge, use the "Modify" button to add to your previous post

Yeah my bad, I'll edit that later. The final answer remains the same but for the working out I should've used $y$'s.
$\text{Recalling the logarithm law }\log a^m = m \log a\\ \frac{d}{dx} \ln \frac1{2x-3} = -\frac{d}{dx} \ln (2x-3) = -\frac{2}{2x-3}$

Quote from: owidjaja on May 03, 2018, 07:03:08 pm

Hey guys,
So my teacher basically spent the last 5 minutes of the period to explain geometric progression to us and set us homework. Considering he's the type of teacher to just push on and I'm a slow learner, can someone please explain how geometric progression works?

Thanks in advance

$\text{They work in contrast to arithmetic progressions, that have a common difference.}$
$\text{A geometric progression features terms having a common ratio.}\\ \text{When put to effect, this means that consecutive terms in the sequence}\\ \text{are always the previous one },\textbf{ multiplied by some number.}$
(As opposed to arithmetic progressions, where you add.)
$\text{Some examples of these include}\\ 1, 2, 4, 8, 16, \dots\\ 1, \frac13, \frac19, \frac1{27}, \frac1{81}, \dots\\ 5, \frac52, \frac 54, \frac58, \dots\\ 5, 1, \frac15, \frac1{25}, \frac1{125}, \dots$
$\text{As with APs, the first term is denoted }a.\\ \text{Instead of }d\text{, we call the common ratio }r\text{ instead.}$
Anything else - provide more details.

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