Author Topic: 4U Maths Question Thread (Read 665273 times) Tweet Share

RuiAce · « **Reply #1170 on:** May 19, 2017, 11:48:04 am »

Quote from: beau77bro on May 19, 2017, 10:11:18 am

(Image removed from quote.)

Please help

$\text{By considering the shells method}\\ (x-1)^2+y^2 = 4\text{ rearranges to } y = \pm \sqrt{4-(x-1)^2}\\(x+1)^2+y^2 = 4\text{ rearranges to } y = \pm \sqrt{4-(x+1)^2}$
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$\text{The radius of a typical shell is always }r=1+x\\ \text{However, the height depends on the }x\text{ values we're interested at the time}$
$\text{For }-1\le x \le 0\text{ we have }h=2\sqrt{4-(x-1)^2}\\ \text{So the volume of a typical shell is }\delta V_1 = 4\pi (1+x)\sqrt{4-(x-1)^2}\delta x\\ \text{and }V_1 =\int_{-1}^0 4\pi (1+x)\sqrt{4-(x-1)^2}dx$
$\text{Similarly for }0\le x \le 1\text{ we have }V_2 = \int_0^1 4\pi (1+x)\sqrt{4-(x+1)^2}\\ \text{And the total volume is }V = V_1 + V_2$
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$\text{Let }u = -x\implies du = -dx\text{ in }V_1\\ \text{Then }\\ \begin{align*}V_1&=-\int_1^0 4\pi(1 - u)\sqrt{4-(-u-1)^2}du\\ &= \int_0^14\pi (1-u)\sqrt{4-(u+1)^2}du\end{align*}$
$\text{Since }u\text{ and }x\text{ are dummy variables in integration we also have}\\ V_1 =\int_0^1 4\pi (1-x)\sqrt{4-(x+1)^2}dx$
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$\text{So we have}\\ \begin{align*}V&= \int_0^1 4\pi(1-x)\sqrt{4-(x+1)^2}dx +\int_0^1 4\pi (1+x)\sqrt{4-(x+1)^2}\\ &= \int_0^18\pi \sqrt{4-(x+1)^2}dx\end{align*}$

beau77bro · « **Reply #1171 on:** May 19, 2017, 02:19:18 pm »

omg u r a god

Jyrgal · « **Reply #1172 on:** May 20, 2017, 08:16:14 pm »

heyo, i have a question on the below (i think) inequality question. i got up to c)ii), i kinda fudged the induction process (doesnt rly make much sense) can anyone help me with step by step for that? (assume c)i) uve already done)

thanks

RuiAce · « **Reply #1173 on:** May 20, 2017, 08:33:04 pm »

Quote from: Jyrgal on May 20, 2017, 08:16:14 pm

heyo, i have a question on the below (i think) inequality question. i got up to c)ii), i kinda fudged the induction process (doesnt rly make much sense) can anyone help me with step by step for that? (assume c)i) uve already done)

thanks

.
$\text{The truth of the proposition when }n=1\text{ is clear as}\\ \left(\frac{a+b}2\right)^1 = \frac{a+b}{2} \le \frac{a^1 + b^1}{2}$
$\text{Take }k\text{ as an integer for which the proposition holds, so we have}\\ \left(\frac{a+b}{2}\right)^k \le \frac{a^k+b^k}{2}\\ \text{and we wish to prove it is true for }k+1$
$\begin{align*}\left(\frac{a+b}{2}\right)^{k+1}&=\left(\frac{a+b}{2}\right)^k\left(\frac{a+b}{2}\right)\\ &\le \left(\frac{a^k+b^k}{2}\right)\left(\frac{a+b}2\right) \tag{assumption}\\ &= \frac{a^{k+1}+b^{k+1}}4 + \frac{a^kb+b^ka}4\\ &\le \frac{a^{k+1}+b^{k+1}}4+\frac{a^{k+1}+b^{k+1}}4\tag{earlier part}\\ &= \frac{a^{k+1}+b^{k+1}}2\end{align*}$

hanaacdr · « **Reply #1174 on:** May 21, 2017, 01:46:21 pm »

hi could i please get some help on this question thanks

Mahan · « **Reply #1175 on:** May 21, 2017, 03:48:27 pm »

Quote from: hanaacdr on May 21, 2017, 01:46:21 pm

hi could i please get some help on this question thanks

There are two methods:
$\int \frac{1}{1-sinx} dx = \int \frac{1-sinx}{1-sin^{2}x} dx = \int \frac{1-sinx}{cos^{2}x} dx = \int sec^{2}x - tanx \ secx dx = tanx - secx$
Then just evaluate the integral in the interval.

Second method,
$t = tan\frac{x}{2}, \Rightarrow sinx = \frac{2t}{1+t^{2}}, cos^{2} \frac{x}{2} =\frac{1}{1+t^{2}} , \frac{dt}{dx} = \frac{1}{2} sec^{2} \frac{x}{2} = \frac{1}{2} (t^{2}+1)$

That implies,

$x= 0 \Rightarrow t = 0, x=\pi/3 \Rightarrow t= \frac{1}{\sqrt{3}}$

Hence,

$\int \frac{1}{1+sinx} dx = \int \frac{1+t^{2}}{(t+1)^{2}} dx = 2\int \frac{1}{(t+1)^{2}} dt = \frac{-2}{t+1}$

That changes the variable of the integration to t and the limit of the integral is now between 0 and 1/(3^0.5)

evaluating the integral in the new interval will give you the same answer.

itssona · « **Reply #1176 on:** May 21, 2017, 07:16:15 pm »

heey so for this, do we differentiate 3 times?? I got -9/8 is it right

RuiAce · « **Reply #1177 on:** May 21, 2017, 07:22:37 pm »

Quote from: sssona09 on May 21, 2017, 07:16:15 pm

heey so for this, do we differentiate 3 times?? I got -9/8 is it right

$\text{The root has multiplicity 3, so it satisfies}\\ P(x) = P^\prime(x) = P^{\prime\prime}(x) = 0\\ \text{The third derivative is not necessary}$
$\text{By inspection, }P^{\prime\prime}(x) =24x^2+54x+12\\ \text{So to obtain the candidate values, we must consider the solutions to }24x^2+54x+12=0$
$\text{Once we have the two candidate points, we substitute back into }P(x)\\ \text{ to eliminate the incorrect one}$
To finish off the question, the last root is easy to find. Just use the sum of roots formula.

itssona · « **Reply #1178 on:** May 21, 2017, 07:30:27 pm »

Quote from: RuiAce on May 21, 2017, 07:22:37 pm

$\text{The root has multiplicity 3, so it satisfies}\\ P(x) = P^\prime(x) = P^{\prime\prime}(x) = 0\\ \text{The third derivative is not necessary}$
$\text{By inspection, }P^{\prime\prime}(x) =24x^2+54x+12\\ \text{So to obtain the candidate values, we must consider the solutions to }24x^2+54x+12=0$
$\text{Once we have the two candidate points, we substitute back into }P(x)\\ \text{ to eliminate the incorrect one}$
To finish off the question, the last root is easy to find. Just use the sum of roots formula.

So I got -2 after eliminating a candidate point and now I use -b/a in the original thing- so the other root is -9/2?
Just making sure
Thank you so so so much Rui

RuiAce · « **Reply #1179 on:** May 21, 2017, 07:32:55 pm »

The triple root is indeed -2
$\text{By the sum of roots}\\ -2-2-2+\alpha = -\frac{b}{a}=-\frac92 \implies \alpha = \frac32$

itssona · « **Reply #1180 on:** May 21, 2017, 07:35:29 pm »

Quote from: RuiAce on May 21, 2017, 07:32:55 pm

The triple root is indeed 2
$\text{By the sum of roots}\\ -2-2-2+\alpha = -\frac{b}{a}=-\frac92 \implies \alpha = \frac32$

ah thank you

itssona · « **Reply #1181 on:** May 22, 2017, 08:26:28 am »

Having trouble using the sum of roots method for this :/

they compare terms in (x+2)^2 (x^2 +ax +b)

but why did they write x^2 +ax +b instead of ax^2 + bx +c

RuiAce · « **Reply #1182 on:** May 22, 2017, 08:50:26 am »

Quote from: sssona09 on May 22, 2017, 08:26:28 am

Having trouble using the sum of roots method for this :/

they compare terms in (x+2)^2 (x^2 +ax +b)

but why did they write x^2 +ax +b instead of ax^2 + bx +c

You could go for $ax^2+bx+c$ but if you look carefully it should be beyond obvious that a=1 as the polynomial is monic (leading coefficient 1)

itssona · « **Reply #1183 on:** May 22, 2017, 09:41:52 am »

Quote from: RuiAce on May 22, 2017, 08:50:26 am

You could go for $ax^2+bx+c$ but if you look carefully it should be beyond obvious that a=1 as the polynomial is monic (leading coefficient 1)

ohhhhhhhhhhhhh

itssona · « **Reply #1184 on:** May 22, 2017, 05:04:46 pm »

at y=f^2 (x)

why is the turning point squared of f(x)?? i dont get it :/

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