leehuan's All-Levels-Of-Maths SOS thread (5 Viewers)

leehuan · Mar 19, 2016

InteGrand said:
Try using L'Hôpital's rule or a substitution.

Can't use the former cause haven't actually been taught it yet (or I would've). Will try a substitution later I guess, thanks.

Drongoski · Mar 19, 2016

Drsoccerball said:
hmmm :/ If I pick a different starting point for a vector I get a different answer is it wrong?

Have you done your Linear Algebra yet?

Drongoski · Mar 19, 2016

leehuan said:
$$What elementary trick did I forget (just a hint this time for now please) to be unable to do $\\ \lim_{x\rightarrow64}{\frac{{x}^{1/3}-4}{{x}^{1/2}-8}}$

$= \lim_ {x \rightarrow 64}\frac {(x^{\frac {1}{6}}-2)(x^{\frac {1}{6}}+2)}{(x^{\frac {1}{6}}-2)(x^{\frac {1}{3}}+2x^{\frac {1}{6}}+4)} = \frac {2+2}{4+4+4} = \frac {1}{3}$

cancelling the common factors, as you were taught to do in your HSC - but were not told why you can do this. You would now know the reason, from learning the delta-epsilon definition of a limit.

To use L'Hopital's rule, you simply differentiate the numerator and the denominator wrt 'x', and substitute x=64 in the resultant expression; if still indefinite, you can reapply the rule until no longer indefinite, like the original expression would have given you a so-called indefinite 0/0.

InteGrand · Mar 19, 2016

Drongoski said:
$= \lim_ {x \rightarrow 64}\frac {(x^{\frac {1}{6}}-2)(x^{\frac {1}{6}}+2)}{(x^{\frac {1}{6}}-2)(x^{\frac {1}{3}}+2x^{\frac {1}{6}}+4)} = \frac {2+2}{4+4+4} = \frac {1}{3}$

???

$\noindent Yes. The substitution of $u=x^{\frac{1}{6}}$ $\big{(}$so $u\to 2$, $x^{\frac{1}{3}}=u^2$ and $x^{\frac{1}{2}}=u^3$$\big{)}$ transforms the limit to $\lim _{u \to 2}\frac{u^2 - 4}{u^3 - 8}$, which can make it easier to see what to do.$

leehuan · Mar 19, 2016

I know how to use L'Hopital's; I've used it before. I just haven't been taught it in my course yet to use it in my homework.

But I see. I was looking for something to cancel out so I guess the fractional powers threw me off.

leehuan · Mar 20, 2016

Paradoxica · Mar 20, 2016

leehuan said:
$$Let $f\left(x\right)=\frac{1}{1+\sqrt{x}} \\ $a) Find $L=\lim_{x\rightarrow \infty}{f\left(x\right)} $ (answer was obviously 0 here)$ \\ $b) For each $0<\varepsilon<1, $find an $M$ such that$ \\ \left| f\left(x\right)-L \right|<\varepsilon$ whenever $x>M. \\$

$\\ $Ok so for b) I do need a check up on my answer of $x>{\left(\frac{1}{\varepsilon}-1\right)}^{2}. \\ $I got this by considering the fact that $1+\sqrt{x}$ is always positive and thus as must it's reciprocal so the absolute values get dropped, and by the fact $x$ is positive I reciprocated the inequalities on both sides.$$

$\\ $ If b) is all good, then: $ \\ $c) Is $N=\frac{1}{{\varepsilon}^{2}} $ a correct answer to b)? Give reasons for your answers.$ \\ $I said no, because whilst when $\varepsilon\approx 0 $ this is quite negligible, $\\ $when $\varepsilon\approx 1$ the value of $\frac{1}{{\varepsilon}^{2}}$ and part b)'s answer would be off by about 1, which is much more impacting. To show this I used trial and error. But I'm not confident with my answer at all here. In fact, I don't fully know what I'm doing anymore.$$

I have no idea what's happening, so I'm just gonna say "Geometric Series"

leehuan · Mar 20, 2016

Paradoxica said:
I have no idea what's happening, so I'm just gonna say "Geometric Series"

It has absolutely nothing to do with geometric series; this is all related to nothing more than limits.

Paradoxica · Mar 20, 2016

leehuan said:
It has absolutely nothing to do with geometric series; this is all related to nothing more than limits.

But could the conversion into a geometric series assist your problem?

$\frac{1}{1+\sqrt{x}}\equiv \frac{1}{\sqrt{x}}\sum_{r=0}^{\infty}\left(\frac{-1}{\sqrt{x}} \right)^r$

Which is valid for all x>1

InteGrand · Mar 20, 2016

leehuan said:
$$Let $f\left(x\right)=\frac{1}{1+\sqrt{x}} \\ $a) Find $L=\lim_{x\rightarrow \infty}{f\left(x\right)} $ (answer was obviously 0 here)$ \\ $b) For each $0<\varepsilon<1, $find an $M$ such that$ \\ \left| f\left(x\right)-L \right|<\varepsilon$ whenever $x>M. \\$

$\\ $Ok so for b) I do need a check up on my answer of $x>{\left(\frac{1}{\varepsilon}-1\right)}^{2}. \\ $I got this by considering the fact that $1+\sqrt{x}$ is always positive and thus as must it's reciprocal so the absolute values get dropped, and by the fact $x$ is positive I reciprocated the inequalities on both sides.$$

$\\ $ If b) is all good, then: $ \\ $c) Is $N=\frac{1}{{\varepsilon}^{2}} $ a correct answer to b)? Give reasons for your answers.$ \\ $I said no, because whilst when $\varepsilon\approx 0 $ this is quite negligible, $\\ $when $\varepsilon\approx 1$ the value of $\frac{1}{{\varepsilon}^{2}}$ and part b)'s answer would be off by about 1, which is much more impacting. To show this I used trial and error. But I'm not confident with my answer at all here. In fact, I don't fully know what I'm doing anymore.$$

The answer to b) is a correct one.

The answer in c) is a correct one too. To see this, let eps > 0, N = 1/eps^2, and suppose x > N.

Then x > 1/eps^2, which implies sqrt(x) > 1/eps. This implies that sqrt(x) > (1/eps) – 1, and this is equivalent to 1/(1 + sqrt(x)) < eps, i.e. |1/(1+sqrt(x)) – 0| < eps. Hence if x > N, then |f(x) – L| < eps, so the N is a correct answer.

InteGrand · Mar 20, 2016

An easier way to see that N works is that for any 0 < eps < 1, we have 1/eps > (1/eps – 1) > 0, so (1/eps^2) > (1/eps – 1)^2, that is, N > M. Since we know that M works (by 'works', I mean that x being greater than it will imply that |f(x) – L| < eps), taking any larger value (e.g. N) will also work. This is because if x > N, then x > M (as N > M), implying that |f(x) – L| < eps, so N works too.

leehuan · Mar 20, 2016

Ohhh should've trusted my very first instinct then. Thanks

leehuan · Mar 20, 2016

Lost in my algebra today big time.

$\\ $The concentration of oxygen $x(t)$ is given \\ $x\left(t\right)=\frac{1}{f+E} \left(f+0.21E-0.79f\exp{\left(-(F+E)\frac{t}{V}\right)}\right). \\ $The equilibrium state is achieved as $t\rightarrow \infty$ and this gives us the equilibrium concentration. $ \\ $The question: On the first trial, $f = 18.0, \, E = 17.0$ and$ V = 145. $ What is the equilibrium concentration?$ \\ \textbf{ How long would it take after starting the inflow for the concentration to reach within 0.01 of its equilibrium state?}$

textbf - what I'm stuck on basically

Drongoski · Mar 20, 2016

leehuan said:
Ohhh should've trusted my very first instinct then. Thanks

Nothing to do with instinct. If you had said 'Yes', you'd still have to justify your answer.

$You have already shown that you need: $ x > (\frac{1}{\varepsilon} - 1)^2 \\ $So you need only show that $ \frac {1}{\varepsilon ^2} > (\frac {1}{\varepsilon} - 1)^2$

$\\ $One way to do this is $ \frac{1}{\varepsilon} > 1 $ and $ (\frac{1}{\varepsilon} - 1)^2$ $= \frac{1}{\varepsilon ^2 } - \frac {2}{\varepsilon} + 1 $ $< $ $\frac {1}{\varepsilon ^2} \\$

leehuan · Mar 20, 2016

Drongoski said:
Nothing to do with instinct. If you had said 'Yes', you'd still have to justify your answer.

$You have already shown that you need: $ x > (\frac{1}{\varepsilon} - 1)^2 \\ $So you need only show that $ \frac {1}{\varepsilon ^2} > (\frac {1}{\varepsilon} - 1)^2$

$\\ $One way to do this is $ \frac{1}{\varepsilon} > 1 $ and $ (\frac{1}{\varepsilon} - 1)^2$ $= \frac{1}{\varepsilon ^2 } - \frac {2}{\varepsilon} + 1 $ $< $ $\frac {1}{\varepsilon ^2} \\$

Not sure if you understood what I meant.

My first instinct processed almost the ENTIRE method. But for some reason I had a contradiction at one point in my mind which halted me. And for some reason I went with that.

Drongoski · Mar 20, 2016

leehuan said:
Not sure if you understood what I meant.

My first instinct processed almost the ENTIRE method. But for some reason I had a contradiction at one point in my mind which halted me. And for some reason I went with that.

Ok

Green Yoda · Mar 20, 2016

Question:
Determine the centre and radius of the circle.
x^2+y^2-18x+20y+60=0

Paradoxica · Mar 20, 2016

Rathin said:
Question:
Determine the centre and radius of the circle.
x^2+y^2-18x+20y+60=0

r=11
C(9,-10)

Green Yoda · Mar 20, 2016

Paradoxica said:
r=11
C(9,-10)

Can you tell me how do you do it?

InteGrand · Mar 20, 2016

Rathin said:
Can you tell me how do you do it?

Complete the squares.

leehuan's All-Levels-Of-Maths SOS thread (5 Viewers)

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