Author Topic: Mathematics Question Thread (Read 1305987 times) Tweet Share

Sukakadonkadonk · « **Reply #1575 on:** March 26, 2017, 09:08:59 pm »

Quote from: RuiAce on March 26, 2017, 09:04:43 pm

Line 3 is wrong. It should not be there.
$\int x^{-\frac12}dx = \frac{x^{\frac12}}{\frac12}+C=2\sqrt{x}+C$
$\text{In fact, }\int x^{+\frac12}dx = \frac23x^{\frac32}+C$

Oh damn, right. That's my mistake. ok thank you.

katnisschung · « **Reply #1576 on:** March 27, 2017, 09:45:15 am »

hi stuck on this q thanks!

RuiAce · « **Reply #1577 on:** March 27, 2017, 12:50:45 pm »

Quote from: katnisschung on March 27, 2017, 09:45:15 am

hi stuck on this q thanks!

$\sqrt[3]{2^x}=\left(2^x\right)^\frac13=2^\frac{x}3$
$\text{Lemma: }\frac{d}{dx} a^x = a^x \ln a\\ \text{Proof: }\frac{d}{dx}a^x = \frac{d}{dx}e^{\ln a^x}=\frac{d}{dx} e^{x\ln a}= e^{x\ln a}\ln a = a^x \ln a$
$\text{So we know that }\frac{d}{dx}2^x = 2^x \ln 2\\ \text{But we have }2^{\frac{x}3}$
$\text{So by the chain rule}\frac{d}{dx}2^{\frac{x}{3}}=\frac{2^x \ln 2}3$

Fahim486 · « **Reply #1578 on:** March 27, 2017, 01:17:15 pm »

Hi, how would you differentiate this?

Hi again, I've also got another question and I am having trouble understanding trapezoidal rules and can't figure this question out myself
Thanks!

Mod Edit: Post merge

jamonwindeyer · « **Reply #1579 on:** March 27, 2017, 01:25:22 pm »

Quote from: Fahim486 on March 27, 2017, 01:13:08 pm

Hi, how would you differentiate this?

Thanks!

Hey! This tricky way to do this is with product rule:

$\frac{d}{dx}x^2\sqrt{x}=\frac{d}{dx}\left(x^2\right)\sqrt{x}+\frac{d}{dx}\left(\sqrt{x}\right)x^2=2x\sqrt{x}+\frac{1}{2}x^{\frac{3}{2}}=2x\sqrt{x}+\frac{1}{2}x\sqrt{x}\\=\frac{5x}{2}\sqrt{x}$

Buuut, you can rewrite it as $x^\frac{5}{2}$ and get that last line straight away with the normal rule

Fahim486 · « **Reply #1580 on:** March 27, 2017, 01:34:45 pm »

Quote from: jamonwindeyer on March 27, 2017, 01:25:22 pm

Hey! This tricky way to do this is with product rule:

$\frac{d}{dx}x^2\sqrt{x}=\frac{d}{dx}\left(x^2\right)\sqrt{x}+\frac{d}{dx}\left(\sqrt{x}\right)x^2=2x\sqrt{x}+\frac{1}{2}x^{\frac{3}{2}}=2x\sqrt{x}+\frac{1}{2}x\sqrt{x}\\=\frac{5x}{2}\sqrt{x}$

Buuut, you can rewrite it as $x^\frac{5}{2}$ and get that last line straight away with the normal rule

Thank you so much Jamon!!!

jamonwindeyer · « **Reply #1581 on:** March 27, 2017, 01:39:52 pm »

Quote from: Fahim486 on March 27, 2017, 01:17:15 pm

Hi again, I've also got another question and I am having trouble understanding trapezoidal rules and can't figure this question out myself
Thanks!

For this one, just before I lend a hand, does this formula look familiar?

$A=\frac{h}{2}\left(y_0+y_n+2y_1+2y_2+2y_3...\right)$

(There's two formula out there for this, just want to make sure I help you properly)

Fahim486 · « **Reply #1582 on:** March 27, 2017, 01:47:19 pm »

Quote from: jamonwindeyer on March 27, 2017, 01:39:52 pm

For this one, just before I lend a hand, does this formula look familiar?

$A=\frac{h}{2}\left(y_0+y_n+2y_1+2y_2+2y_3...\right)$

(There's two formula out there for this, just want to make sure I help you properly)

Yeah I've seen that formula before but I don't know how to apply it to the question

jamonwindeyer · « **Reply #1583 on:** March 27, 2017, 01:58:18 pm »

Quote from: Fahim486 on March 27, 2017, 01:47:19 pm

Yeah I've seen that formula before but I don't know how to apply it to the question

So $h$ in that formula represents the width of our intervals. For that question, this is 5 metres!

The $y_k$ terms just represent the heights of our area at each value, counting from zero. It is first, plus last, plus double the rest. So, in this case it would be:

$A=\frac{5}{2}\left(4.2+6.3+2(6.1)\right)=56.75\text{m}^2$

The answer is B!

katnisschung · « **Reply #1584 on:** March 27, 2017, 01:58:34 pm »

Quote from: RuiAce on March 27, 2017, 12:50:45 pm

$\sqrt[3]{2^x}=\left(2^x\right)^\frac13=2^\frac{x}3$
$\text{Lemma: }\frac{d}{dx} a^x = a^x \ln a\\ \text{Proof: }\frac{d}{dx}a^x = \frac{d}{dx}e^{\ln a^x}=\frac{d}{dx} e^{x\ln a}= e^{x\ln a}\ln a = a^x \ln a$
$\text{So we know that }\frac{d}{dx}2^x = 2^x \ln 2\\ \text{But we have }2^{\frac{x}3}$
$\text{So by the chain rule }\frac{d}{dx}2^{\frac{x}{3}}}=\frac{2^x \ln 2}3$

Hey ruiace the last bit of the answer didn't come through...
so why chain rule for integration? note i only take 2 unit maths

jamonwindeyer · « **Reply #1585 on:** March 27, 2017, 02:15:16 pm »

Quote from: katnisschung on March 27, 2017, 01:58:34 pm

Hey ruiace the last bit of the answer didn't come through...
so why chain rule for integration? note i only take 2 unit maths

Just fixed his last line

we're differentiating not integrating! Chain rule for differentiation

kiiaaa · « **Reply #1586 on:** March 27, 2017, 06:37:33 pm »

Hello
i was wondering if anyone had any way to get their head around or easier way to solve the loan repayment questions. I just don't get how to do it and often end up more confused than what i was lol

thank you

RuiAce · « **Reply #1587 on:** March 27, 2017, 06:56:17 pm »

Quote from: kiiaaa on March 27, 2017, 06:37:33 pm

Hello
i was wondering if anyone had any way to get their head around or easier way to solve the loan repayment questions. I just don't get how to do it and often end up more confused than what i was lol
thank you

Might wanna read Jamon's guides, in particular the series one.

Whilst the process is the same, there's no way you can get how to do it without knowing WHY your equations are true.

cxmplete · « **Reply #1588 on:** March 27, 2017, 08:01:57 pm »

Hi, so this was a question in our HY exam which i found pretty confusing:

A factory assembles torches. Each torch requires one battery and one bulb. It is known that 6% of all batteries and 4% of all bulbs are defective. Find the probability that, in a torch selected at random, both the battery and the bulb are NOT defective. Give your answer in exact form.

kiiaaa · « **Reply #1589 on:** March 27, 2017, 08:02:45 pm »

hi could anyone please help me answer this question? im super lost

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