Author Topic: 3U Maths Question Thread (Read 1470028 times) Tweet Share

anotherworld2b · « **Reply #1275 on:** January 24, 2017, 08:53:52 pm »

thank you very much for your help

I really appreciate it

Quote from: kiwiberry on January 24, 2017, 08:23:21 pm

When you factorise the √2 out, remember to keep the square root sign on the x+h and x as well
$\frac{\sqrt{2(x+h)} - \sqrt{2x}}{h}<br />= \frac{\sqrt{2}\sqrt{x+h} - \sqrt{2}\sqrt{x}}{h}<br />= \frac{\sqrt{2}(\sqrt{x+h}-\sqrt{x})}{h}$
and then from here you can multiply top and bottom by √(x+h)+√x to cancel the h on the bottom!

anotherworld2b · « **Reply #1276 on:** January 25, 2017, 12:00:16 am »

I was wondering how did it go from
√(2)/(√(x) + √(x)) to 1/√(2x)

Quote from: Shadowxo on January 24, 2017, 06:11:39 pm

No worries
I'm not entirely sure how they want you to answer it. You could say that that equation = lim_h->0 (f(x+h)-f(x))/h = dy/dx where y = √(2x)
therefore it = 1/√(2x). If they say use an appropriate derivative maybe it's that (anyone else know?) Edit: RuiAce did this way so probably this
Another way is just working through it.
Multiply both top and bottom by (√(x+h)+√(x)) so you end up (including the lim with all of these) with √2*(x+h-x)/(h(√(x+h)+√(x)))
= √2*h/(h*(√(x+h)+√(x))) = √(2)/(√(x+h)+√(x))
then remove the lim and sub h = 0
=√(2)/(√(x) + √(x)) = 1/√(2x)

RuiAce · « **Reply #1277 on:** January 25, 2017, 12:02:26 am »

Quote from: anotherworld2b on January 25, 2017, 12:00:16 am

I was wondering how did it go from
√(2)/(√(x) + √(x)) to 1/√(2x)

$\frac{\sqrt{2}}{\sqrt{x}+\sqrt{x}}=\frac{\sqrt{2}}{2\sqrt{x}}=\frac{\sqrt2}{\sqrt{2}\sqrt2\sqrt{x}}=\frac{1}{\sqrt{2x}}$

anotherworld2b · « **Reply #1278 on:** January 25, 2017, 12:30:45 am »

I was looking at the answer to this question but i don't really understand it

kiwiberry · « **Reply #1279 on:** January 25, 2017, 04:30:36 pm »

Quote from: anotherworld2b on January 25, 2017, 12:30:45 am

I was looking at the answer to this question but i don't really understand it

Basically, the question wants us to solve the limit by recognising that it is simply a function plugged into the formula for differentiation from first principles
$\text{Remember that } f^{'}(x)=\lim_{h \to 0}{\frac{f(x+h)-f(x)}{h}} \\<br />\text{If we replace the 5 in the question with an }x: \lim_{h \to 0}{\frac{(1+\sqrt{x+h})^{2}-(1+\sqrt{x})^{2}}{h}} \text{ it now matches the formula exactly, with } f(x)=(1+\sqrt{x})^{2} \\ \text{Therefore } \lim_{h \to 0}{\frac{(1+\sqrt{x+h})^{2}-(1+\sqrt{x})^{2}}{h}}=f^{'}[(1+\sqrt{x})^{2}]=2(1+\sqrt{x})\times\frac{1}{2\sqrt{x}}\\ \text{From here, we sub x=5 in to get back to the original question: } \lim_{h \to 0}{\frac{(1+\sqrt{5+h})^{2}-(1+\sqrt{5})^{2}}{h}}=2(1+\sqrt{5})\times\frac{1}{2\sqrt{5}}=...$
And that's it

Rathin · « **Reply #1280 on:** January 25, 2017, 04:42:19 pm »

Just learned Log Integration today at the lectures and I am kinda stuck on this question..

kiwiberry · « **Reply #1281 on:** January 25, 2017, 05:12:34 pm »

Quote from: Rathin on January 25, 2017, 04:42:19 pm

Just learned Log Integration today at the lectures and I am kinda stuck on this question..

$\begin{align*}u&=\ln{x} \\ \frac{du}{dx}&=\frac{1}{x} &\\ du&=\frac{dx}{x} \\ \text{When }x&=e, u=1 \\ \text{When }x&=e^{2}, u=2\end{align*}\\ \therefore \int_{e}^{e^{2}}{\frac{1}{x(\ln{x})^{2}}\,dx}=\int_{1}^{2}{\frac{du}{u^{2}}}=\left[-\frac{1}{u}\right]_{1}^{2}=-\frac{1}{2}+1=\frac{1}{2}$

Rathin · « **Reply #1282 on:** January 25, 2017, 05:52:55 pm »

Quote from: kiwiberry on January 25, 2017, 05:12:34 pm

$\begin{align*}u&=\ln{x} \\ \frac{du}{dx}&=\frac{1}{x} &\\ du&=\frac{dx}{x} \\ \text{When }x&=e, u=1 \\ \text{When }x&=e^{2}, u=2\end{align*}\\ \therefore \int_{e}^{e^{2}}{\frac{1}{x(\ln{x})^{2}}\,dx}=\int_{1}^{2}{\frac{du}{u^{2}}}=\left[-\frac{1}{u}\right]_{1}^{2}=-\frac{1}{2}+1=\frac{1}{2}$

OMG I am actually so stupid...I didn't change the boundaries for the substitution......no wonder I was getting a weird answer
Thanks for the post though

anotherworld2b · « **Reply #1283 on:** January 25, 2017, 07:12:37 pm »

I was a bit confused when it got to to the third line. Could someone please explain it a bit more please?

Shadowxo · « **Reply #1284 on:** January 25, 2017, 08:09:34 pm »

Quote from: anotherworld2b on January 25, 2017, 07:12:37 pm

I was a bit confused when it got to to the third line. Could someone please explain it a bit more please?

Is this about your question? If so, from line 2->3 they just substituted in x = 5 and cancelled the two 2s

If it's about kiwi's explanation, the derivative (f'(x)) of something can be written as lim h->0 (f(x+h)-f(x))/h (don't know how to properly set out formulas here)
So the lim h->0 of that equation is equal to the derivative of f(x), in this case f(x) = (1+√x)². f'(x) = d((1+√x)²)/dx then find the derivative and sub in

anotherworld2b · « **Reply #1285 on:** January 25, 2017, 08:46:13 pm »

Oh okay thanks for your help

I was wondering how to sketch the quation for this question

Quote from: Shadowxo on January 25, 2017, 08:09:34 pm

Is this about your question? If so, from line 2->3 they just substituted in x = 5 and cancelled the two 2s

If it's about kiwi's explanation, the derivative (f'(x)) of something can be written as lim h->0 (f(x+h)-f(x))/h (don't know how to properly set out formulas here)
So the lim h->0 of that equation is equal to the derivative of f(x), in this case f(x) = (1+√x)². f'(x) = d((1+√x)²)/dx then find the derivative and sub in

Shadowxo · « **Reply #1286 on:** January 25, 2017, 09:11:40 pm »

I think you were on the right track but k=0 not 7 as root is x=1 indicating a point of inflection there and no vertical translation.

kiwiberry · « **Reply #1287 on:** January 25, 2017, 09:13:44 pm »

Quote from: anotherworld2b on January 25, 2017, 08:46:13 pm

Oh okay thanks for your help
I was wondering how to sketch the quation for this question

If the graph has a root at x=1, it will also pass through (1,0)!
$\begin{align*}y&=a(x-1)^{3}+k\\\text{Subbing in (1,0), } 0&=a(1-1)^{3}+k \\ \therefore k&=0\\\text{Now }y&=a(x-1)^{3} \\ \text{Subbing in (0,7), }7&=a(0-1)^{3} \, \\\therefore a&=-7 \end{align*}\\ \text{So the equation of the cubic is } y=-7(x-1)^{3}$

anotherworld2b · « **Reply #1288 on:** January 25, 2017, 09:57:23 pm »

ah that makes sense now

I want to say thank you to everyone who has helped answer my numerous questions.
I really appreciate all your help in giving up your precious time to assist me

Quote from: kiwiberry on January 25, 2017, 09:13:44 pm

If the graph has a root at x=1, it will also pass through (1,0)!
$\begin{align*}y&=a(x-1)^{3}+k\\\text{Subbing in (1,0), } 0&=a(1-1)^{3}+k \\ \therefore k&=0\\\text{Now }y&=a(x-1)^{3} \\ \text{Subbing in (0,7), }7&=a(0-1)^{3} \, \\\therefore a&=-7 \end{align*}\\ \text{So the equation of the cubic is } y=-7(x-1)^{3}$

Shadowxo · « **Reply #1289 on:** January 25, 2017, 09:59:03 pm »

Quote from: anotherworld2b on January 25, 2017, 09:57:23 pm

ah that makes sense now
I want to say thank you to everyone who has helped answer my numerous questions.
I really appreciate all your help in giving up your precious time to assist me

No worries, I enjoy helping people out

Glad I've helped (even if just a little

)

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