Author Topic: 3U Maths Question Thread (Read 1238147 times) Tweet Share

Shadowxo · « **Reply #1560 on:** February 25, 2017, 10:53:50 am »

I think your mistake was the derivative part
u²=x+2
x=u²-2
dx/du=2u
dx=2u du
Then sub that in. That's the easier way. And I don't know what happened to the x when you differentiated yours.
Also, you need to simplify before antidiffing, you can only have one x term per section, you can't antidiff u²/u for example, only u (one u not two)

Try again and if you can't get the answer out I can post a solution for you

michaelalt · « **Reply #1561 on:** February 25, 2017, 11:15:53 am »

Quote from: Shadowxo on February 25, 2017, 10:53:50 am

I think your mistake was the derivative part
u²=x+2
x=u²-2
dx/du=2u
dx=2u du
Then sub that in. That's the easier way. And I don't know what happened to the x when you differentiated yours.
Also, you need to simplify before antidiffing, you can only have one x term per section, you can't antidiff u²/u for example, only u (one u not two)

Try again and if you can't get the answer out I can post a solution for you

Thanks so much! Your explanation made so much more sense than the way I was taught to do it. However, I still get the -4 as a co-efficient, and not -8...

Shadowxo · « **Reply #1562 on:** February 25, 2017, 11:30:05 am »

You just wrote x-2 = u²-2 instead of u²-4

michaelalt · « **Reply #1563 on:** February 25, 2017, 11:36:09 am »

Quote from: Shadowxo on February 25, 2017, 11:30:05 am

You just wrote x-2 = u²-2 instead of u²-4

Hhahah ahh, thanks

michaelalt · « **Reply #1564 on:** February 25, 2017, 12:25:23 pm »

Quote from: Shadowxo on February 25, 2017, 10:53:50 am

I think your mistake was the derivative part
u²=x+2
x=u²-2
dx/du=2u
dx=2u du
Then sub that in. That's the easier way. And I don't know what happened to the x when you differentiated yours.
Also, you need to simplify before antidiffing, you can only have one x term per section, you can't antidiff u²/u for example, only u (one u not two)

Try again and if you can't get the answer out I can post a solution for you

When you said "you can only have one x term per section", how do I simplify this so that there's only one u?

jakesilove · « **Reply #1565 on:** February 25, 2017, 12:37:33 pm »

Quote from: michaelalt on February 25, 2017, 12:25:23 pm

When you said "you can only have one x term per section", how do I simplify this so that there's only one u?

First correction; du=2dx

Now, in terms of your integral

$\int \frac{u+5}{2u^{\frac{1}{2}}} du$

You want to break up the fraction into composite parts.

$\int \frac{u}{2u^{\frac{1}{2}}}+\frac{5}{2u^{\frac{1}{2}}} du$

Then, use indice laws to simplify, and use the standard power rule to integrate!

hanaacdr · « **Reply #1566 on:** February 28, 2017, 05:53:10 am »

Morning to the lovely people at ATARNOTES!
Could i get some help on the induction question
thanks
Hanaa

RuiAce · « **Reply #1567 on:** February 28, 2017, 10:34:05 am »

Quote from: hanaacdr on February 28, 2017, 05:53:10 am

Morning to the lovely people at ATARNOTES!
Could i get some help on the induction question
thanks
Hanaa

$\text{The statement clearly holds when }n=1\\ \text{ as }2^1\ge 2^1(1!) \text{ is true}$
$\text{Assuming the truth of the statement when }n=k\text{ gives}\\ \boxed{(k+1)^k\ge 2^k k!}\\ \text{and we must use this to verify the truth when }n=k+1\\ \text{i.e. }(k+2)^{k+1}\ge 2^{k+1}(k+1)!$
$\text{Nothing stands to being obvious at ALL with how the hypothesis is applied}\\ \text{Hence, commence (obscure) algebraic tactics}$
$\text{Think of the following proposition intuitively with a numeric example}$
____________________________
$\text{It is obvious that }3 \ge 2\\ \text{So if }\frac{a}{3} \ge \frac{b}{2}\\ \text{Then certainly }a \ge b\\ \text{This is because rearranging we have }\frac{2a}3\ge b\\ \text{So we have }a \ge \frac{2a}{3} \ge b \implies a \ge b$
$\text{Note that the converse is not necessarily true: }a \ge b\text{ does not necessarily imply }\frac{a}3\ge \frac{b}2$
____________________________
$\text{Using this analogy, we see that the truth of}\\ \frac{(k+2)^{k+1}}{(k+1)^n}\ge \frac{2^{k+1}(k+1)!}{2^k k!}\\ \text{certainly implies the truth of }(k+2)^{k+1} > 2^{k+1}(k+1)!\\ \text{as from the assumption, we know that } (k+1)^k\ge 2^k k\\\text{We shall prove this now.}$
$\text{Note that the RHS simplifies to }2(k+1)\\\text{The LHS simplifies to this:}\\ \begin{align*}<br />\frac{(k+2)^{k+1}}{(k+1)^k}&= (k+2)\left(\frac{k+2}{k+1}\right)^{k+1}\\ &= (k+2)\left(1+\frac{1}{k+1}\right)^{k+1}<br />\end{align*}\\ \text{So after all the manipulation, we ultimately need to prove this}\\ \boxed{(k+2)\left(1+\frac{1}{k+1}\right)^{k+1}\ge 2(k+1)}\\ \text{Bring in the obvious fact that }k+2\ge k+1\text{ and the binomial theorem}$
$\begin{align*}LHS&=(k+2)\left(1+\frac{1}{k+1}\right)^{k+1}\\&\ge (k+1)\left(1+\frac{1}{k+1}\right)^{k+1}\\ &= (k+1)\left(\binom{k+1}{0}+\binom{k+1}{1}\frac{1}{k+1}+\dots\right)\\ &= (k+1)\left(1 + \frac{k+1}{k+1} +\dots\right) \\ &= (k+1)(2+\dots)\\ &\ge(k+1)(2)\tag{*}\\ &= RHS\\ \therefore LHS&\ge RHS\end{align*}\\(*) - \text{note, from the binomial expansion, all other terms are positive}\\ \text{2 + some positive number will be larger than 2}$
$\text{Recombine everything to finish the proof.}$

Dragomistress · « **Reply #1568 on:** February 28, 2017, 07:19:03 pm »

May someone please help me with this question.

kiwiberry · « **Reply #1569 on:** February 28, 2017, 08:22:37 pm »

Quote from: JKFlacka on February 28, 2017, 07:19:03 pm

May someone please help me with this question.

Here we have to use the angle between two lines formula:
$\tan \theta = \left| \frac{m_1 - m_2}{1+m_1 m_2} \right|$ The question is saying that $\tan \theta = \frac{2}{5}$, and our two gradient values ($m_1, m_2$) are m and 5. Subbing into the formula:
$\begin{align*}\left|\frac{m-5}{1+5m}\right| &= \frac{2}{5} \\ \frac{m-5}{1+5m}&= \pm\frac{2}{5}\end{align*}$
$\begin{align*}\text{Case 1: } \frac{m-5}{1+5m}&= \frac{2}{5} \\ 5m-25&=2+10m \\ 5m &=-27 \\ m&= -\frac{27}{5}= -5.4\end{align*}$
$\begin{align*}\text{Case 2: } \frac{m-5}{1+5m}&= -\frac{2}{5} \\ 5m-25&=-2-10m \\ 15m &=23 \\ m&= -\frac{23}{15}= 1.53\end{align*}$ So m=-5.4 or 1.53

Shadowxo · « **Reply #1570 on:** February 28, 2017, 10:26:52 pm »

Alternatively, if you don't know that formula you could go, θ= ±(θ2-θ1), knowing tan(θ)=2/5, tan(θ2)=5 and tan(θ1)=m, and solve it the long way.
$\arctan \left(\frac{2}{5}\right)=\pm \left(\arctan \left(5\right)-\arctan \left(m\right)\right) \\ \arctan \left(m\right)=\arctan \left(5\right)\pm \arctan \left(\frac{2}{5}\right) \\ m=\tan \left(\arctan \left(5\right)\pm \arctan \left(\frac{2}{5}\right)\right) = 1.53, -5.4$

RuiAce · « **Reply #1571 on:** February 28, 2017, 10:38:50 pm »

Quote from: Shadowxo on February 28, 2017, 10:26:52 pm

Alternatively, if you don't know that formula

The formula is taught in the course. I think they're expected to use it.

Shadowxo · « **Reply #1572 on:** February 28, 2017, 11:02:54 pm »

Quote from: RuiAce on February 28, 2017, 10:38:50 pm

The formula is taught in the course. I think they're expected to use it.

Ah, it's not in VCE but I've seen VCE exam questions testing it, intending for you not to use the formula but instead use a method similar to the one I used.

K9810 · « **Reply #1573 on:** March 01, 2017, 09:12:36 pm »

How do you differentiate this: y=log log secx

RuiAce · « **Reply #1574 on:** March 01, 2017, 09:21:20 pm »

Quote from: K9810 on March 01, 2017, 09:12:36 pm

How do you differentiate this: y=log log secx

$\text{Be careful applying the chain rule, as we need it twice.}\\ \text{Note that }\frac{d}{dx}\log (\sec x) = \frac{\sec x\tan x}{\sec x}=\tan x$
$\text{Hence}\\ \begin{align*}y&=\log (\log (\sec x))\\ \implies \frac{dy}{dx} &= \frac{\tan x}{\log (\sec x)}\end{align*}\\ \text{having, twice, used the rule }\frac{d}{dx}\ln [f(x)]=\frac{f^\prime(x)}{f(x)}$
Note: As an exercise, you should prove that d/dx sec(x) = sec(x)tan(x). Hint: sec(x) = (cos(x))^-1

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