Author Topic: Mathematics Question Thread (Read 1627053 times) Tweet Share

kiwiberry · « **Reply #1155 on:** February 09, 2017, 11:42:51 pm »

Quote from: anotherworld2b on February 09, 2017, 11:24:30 pm

Could i have some help with this question please? I am a bit confused about how i differentiated it wrong

$\text{Remember that }(a+b)^{\frac{1}{2}} \neq a^{\frac{1}{2}} +b^{\frac{1}{2}} \\ \text{So }(x+2)^{\frac{1}{2}} \neq 2^{\frac{1}{2}} + x^{\frac{1}{2}} \\ \text{Instead, } y=\frac{3}{2(x+2)^{\frac{1}{2}}}=\frac{3}{2}(x+2)^{-\frac{1}{2}}$
Use the chain rule to differentiate from here

laurenf58 · « **Reply #1156 on:** February 10, 2017, 04:22:56 pm »

How do I do this question?

Thanks!

jamonwindeyer · « **Reply #1157 on:** February 10, 2017, 04:30:50 pm »

Quote from: laurenf58 on February 10, 2017, 04:22:56 pm

How do I do this question?

Thanks!

Hey! This is a quadratic identity; for something to be equal for ALL values of $x$ then the coefficients need to be the same:

$ax^2+bx\equiv3x(x+2)\\ax^2+bx\equiv3x^2+6x$

So, $a=3$ and $b=6$. This makes the two quadratics the same and is the only way the condition can be satisfied

does that make sense?

Hplovers · « **Reply #1158 on:** February 10, 2017, 05:10:22 pm »

Ah this is confusing me so much, please help

asd987 · « **Reply #1159 on:** February 10, 2017, 05:26:14 pm »

Hi, I have a couple of questions if some can please help me with them.
Thanks

kiwiberry · « **Reply #1160 on:** February 10, 2017, 05:46:11 pm »

Quote from: Hplovers on February 10, 2017, 05:10:22 pm

Ah this is confusing me so much, please help

Hey! If you split up the fraction like this:
$\frac{x^{2}+x+3}{3x^{5}}=\frac{1}{3x^{3}}+\frac{1}{3x^{4}}+\frac{1}{x^{5}}$
You can integrate easily

Hplovers · « **Reply #1161 on:** February 10, 2017, 06:01:31 pm »

Quote from: kiwiberry on February 10, 2017, 05:46:11 pm

Hey! If you split up the fraction like this:
$\frac{x^{2}+x+3}{3x^{5}}=\frac{1}{3x^{3}}+\frac{1}{3x^{4}}+\frac{1}{x^{5}}$
You can integrate easily

I have gotten to that point but still end up with the wrong answer at the end

, could you please finish the integration for me

After that step I got rid of the fractions to make them 1/3 . x^-3 + ... etc. but i'm not sure if that's correct. I thought that to use the integration( ax^n) rule, the function could not be a fraction. What is the rule for when you can integrate?

kiwiberry · « **Reply #1162 on:** February 10, 2017, 06:16:58 pm »

Quote from: RuiAce on February 10, 2017, 06:10:15 pm

$\int \left(\frac{1}{3}x^{-3}+\frac13x^{-4}+x^{-5}\right)dx=-\frac{1}{12}x^{-4}-\frac{1}{15}x^{-5}-\frac{1}{6}x^{-6}+C$
$\int x^n\, dx = \frac{x^{n+1}}{n+1}+C$

Isn't it this?!!
$\int \left(\frac{1}{3}x^{-3}+\frac13x^{-4}+x^{-5}\right)dx=-\frac{1}{6}x^{-2}-\frac{1}{9}x^{-3}-\frac{1}{4}x^{-4}+C$

RuiAce · « **Reply #1163 on:** February 10, 2017, 06:18:29 pm »

Quote from: kiwiberry on February 10, 2017, 06:16:58 pm

Isn't it this?!!
$\int \left(\frac{1}{3}x^{-3}+\frac13x^{-4}+x^{-5}\right)dx=-\frac{1}{6}x^{-2}-\frac{1}{9}x^{-3}-\frac{1}{4}x^{-4}+C$

Aiya. I'll just blame the heat for my dead brain - mistake deleted

RuiAce · « **Reply #1164 on:** February 10, 2017, 06:23:58 pm »

Quote from: asd987 on February 10, 2017, 05:26:14 pm

Hi, I have a couple of questions if some can please help me with them.
Thanks

$\text{Use the fact that for all positive }x, \, x=e^{\ln x}\\ \begin{align*}\frac{d}{dx}2^{3x-4}&=\frac{d}{dx}e^{\ln 2^{3x-4}}\\ &= \frac{d}{dx}e^{(3x-4)\ln 2}\tag{log law}\\ &= 3\ln 2 \cdot e^{(3x-4)\ln 2}\tag{standard e^(ax+b)}\\ &= 3\ln 2 \cdot 2^{3x-4}\tag{reversing above steps}\end{align*}$
_________________________________

I'm only going to do the first one. You can post up your working for the other if you get stuck because the process is the SAME; the only difference being the product rule used.
$y=(\ln x -1)^3 \implies \frac{dy}{dx}=3(\ln x - 1)^2\times \frac{1}{x}$
$\text{Let }\frac{dy}{dx}=0\text{ for stationary points:}\\ \begin{align*}3(\ln x - 1)^2\times \frac{1}{x}&=0\\ (\ln x -1)^2&=0\\ \ln x-1&=0\\ \ln x &=1 \\ x&=e\end{align*}\\ \text{So there is a stationary point at }(e,0)$
$\text{Force the quotient rule in for the second derivative}\\ \begin{align*}\frac{dy}{dx}&= 3\times\frac{(\ln x-1)^2}{x}\\ \implies \frac{d^2y}{dx^2}&=3\times\frac{x\cdot \frac{2}{x}(\ln -1)^1-(\ln x-1)^2\cdot 1}{x^2}\\ &= 3\times \frac{2(\ln x-1)^2-(\ln x - 1)}{x^2}\\ &= 3\times \frac{(\ln x-1)(2-(\ln x -1))}{x^2}\tag{careful factorisation}\\ &= \frac{(\ln x-1)(\ln x + 3)}{x^2}\end{align*}$
$\text{Possible points of inflection are when }\frac{d^2y}{dx^2}=0\\ \text{Solving } \frac{(\ln x-1)(\ln x + 3)}{x^2}=0\text{ gives}\\ x=e\text{ or }x=e^{-3}$
$\text{I leave you to draw the table of values to check that an inflexion DOES happen}\\ \textbf{You should be familiar with this procedure from earlier in the course}\\ \text{So the inflexion points are at }(e,0)\text{ and }(e^3,8)$

sophiegmaher · « **Reply #1165 on:** February 11, 2017, 10:31:05 am »

So I'm stuck on this question, I feel it's a simple answer but I'm not sure how to get it!
"Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space"

RuiAce · « **Reply #1166 on:** February 11, 2017, 10:43:58 am »

Quote from: sophiegmaher on February 11, 2017, 10:31:05 am

So I'm stuck on this question, I feel it's a simple answer but I'm not sure how to get it!
"Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space"

This question is a bit unspecific in how it was worded. I'd assume that when they say "corner" they mean that the corner is right-angled, but they tell us nothing at all about that.

Where did this question come from, and does it come with a diagram?

sophiegmaher · « **Reply #1167 on:** February 11, 2017, 10:46:17 am »

Yes it does! The diagram is of a right angled triangle with the hypotenuse being the "straight 8m length of fencing"

RuiAce · « **Reply #1168 on:** February 11, 2017, 11:01:28 am »

sophiegmaher · « **Reply #1169 on:** February 11, 2017, 12:09:30 pm »

Thank you!!!

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