First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

InteGrand · May 29, 2017

Re: MATH1131 help thread

matchalolz said:
How do you know what function to pick out for g(x)?

When x is large, f(x) "looks like" x/(2x^3) = 1/(2x^2), because the -1 on the denominator becomes negligible. So we know we want to compare f(x) to 1/(2x^2) (or can just do it to 1/x^2, since all we need is the limit of f(x)/g(x) to be a positive constant), and this is helpful because we know the convergence status of the improper integral of 1/(x^2).

matchalolz · Jun 3, 2017

Re: MATH1131 help thread

I'm not following how we went from the first step to the second step

Screen Shot 2017-06-03 at 11.02.03 am.png

leehuan · Jun 3, 2017

Re: MATH1131 help thread

matchalolz said:
I'm not following how we went from the first step to the second step

View attachment 33967

$\text{Their line one is wrong}\\ \text{It should read }4(z\overline{z}+3i(z-\overline{z})+9)=z\overline{z}\boxed{-3i(z-\overline{z})}+9$

matchalolz · Jun 3, 2017

Re: MATH1131 help thread

leehuan said:
$\text{Their line one is wrong}\\ \text{It should read }4(z\overline{z}+3i(z-\overline{z})+9)=z\overline{z}\boxed{-3i(z-\overline{z})}+9$

Really?

I don't even know how we got to that line in the first place, let alone realise there was a mistake

leehuan · Jun 3, 2017

Re: MATH1131 help thread

$\text{Using the identity they supplied}$
$\begin{align*}|z+3i|^2 &= (z+3i)\overline{(z+3i)}\\ &= (z+3i)(\overline{z}-3i)\\ &= z\overline{z} + 3i\overline{z} - 3iz - 9i^2\\ &= z\overline{z} - 3i(z-\overline{z}) + 9\end{align*}$

matchalolz · Jun 3, 2017

Re: MATH1131 help thread

leehuan said:
$\text{Using the identity they supplied}$
$\begin{align*}|z+3i|^2 &= (z+3i)\overline{(z+3i)}\\ &= (z+3i)(\overline{z}-3i)\\ &= z\overline{z} + 3i\overline{z} - 3iz - 9i^2\\ &= z\overline{z} - 3i(z-\overline{z}) + 9\end{align*}$

oh I see ty c:

matchalolz · Jun 3, 2017

Re: MATH1131 help thread

Why are we multiplying cos(x^6) by 3*x^2?

Screen Shot 2017-06-03 at 3.35.50 pm.png

Rip finals

InteGrand · Jun 3, 2017

Re: MATH1131 help thread

matchalolz said:
Why are we multiplying cos(x^6) by 3*x^2?

View attachment 33968

Rip finals

We're multiplying it by the derivative of the x^3 in the integral upper limit, because of the chain rule.

matchalolz · Jun 3, 2017

Re: MATH1131 help thread

InteGrand said:
We're multiplying it by the derivative of the x^3 in the integral upper limit, because of the chain rule.

why do we need to do this? can you elaborate on this please?

InteGrand · Jun 3, 2017

Re: MATH1131 help thread

matchalolz said:
why do we need to do this? can you elaborate on this please?

$\noindent In other words, let $F(u) = \int_{0}^{u} \cos \left(t^{2}\right)\, \mathrm{d}t$. Then we are asked to find $\frac{\mathrm{d}y}{\mathrm{d}x}$, where $y = \int_{0}^{x^{3}} \cos \left(t^{2}\right) \, \mathrm{d}t = F\left(x^{3}\right)$. You should be able to get the answer via chain rule and Fundamental Theorem of Calculus from here. (Remember in the FTC, the upper limit of the integral is the same as the variable with respect to which we are differentiating.)$

matchalolz · Jun 3, 2017

Re: MATH1131 help thread

is it like y = F(x^3)

and because F(x^3) is a function of a function, to differentiate y with respect to x you go (3*x^2)*F'(x), where F'(x) = f(x) = cos(x^6)

InteGrand · Jun 3, 2017

Re: MATH1131 help thread

matchalolz said:
is it like y = F(x^3)

and because F(x^3) is a function of a function, to differentiate y with respect to x you go (3*x^2)*F'(x), where F'(x) = f(x) = cos(x^6)

$\noindent Right idea, but the chain rule calculation is $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}F'\left(x^{3}\right)$ (rather than $F'(x)$).$

$\noindent Since $F'(u) = \cos \left(u^{2}\right)$ (by the FTC), we have $F'\left(x^{3}\right) = \cos \left(\left(x^{3}\right)^{2}\right) = \cos \left(x^{6}\right)$ (replacing $u$ with $x^{3}$ in the equation $F'(u) = \cos \left(u^{2}\right)$).$

matchalolz · Jun 3, 2017

Re: MATH1131 help thread

InteGrand said:
$\noindent Right idea, but the chain rule calculation is $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}F'\left(x^{3}\right)$ (rather than $F'(x)$).$

$\noindent Since $F'(u) = \cos \left(u^{2}\right)$ (by the FTC), we have $F'\left(x^{3}\right) = \cos \left(\left(x^{3}\right)^{2}\right) = \cos \left(x^{6}\right)$ (replacing $u$ with $x^{3}$ in the equation $F'(u) = \cos u^{2}$).$

Sweet! such a lifesaver

leehuan · Jun 18, 2017

I know that the answer is $\alpha=6$ . But I don't know how to properly set out a logical answer because they wanted us to explicitly state any tests we use and how we used it. So I was just looking for some guidance on how I can set out a response?

$\text{What is the largest value of }\alpha > 0\text{ for which the following improper integral diverges?}\\ I=\int_0^\infty\frac{1+x^2}{\sqrt{x^\alpha+2}}dx$

(I don't remember if I already asked this question either)

InteGrand · Jun 18, 2017

leehuan said:
I know that the answer is $\alpha=6$ . But I don't know how to properly set out a logical answer because they wanted us to explicitly state any tests we use and how we used it. So I was just looking for some guidance on how I can set out a response?

$\text{What is the largest value of }\alpha > 0\text{ for which the following improper integral diverges?}\\ I=\int_0^\infty\frac{1+x^2}{\sqrt{x^\alpha+2}}dx$

(I don't remember if I already asked this question either)

$\noindent We can show this using a limit comparison test and the $p$-test. $\Big{(}$Note that the integrand is continuous on $[0,\infty)$ and is asymptotically equivalent to $\frac{1}{x^{\frac{\alpha}{2}-2}}$ as $x\to \infty$, for $\alpha > 0$.$\Big{)}$$

matchalolz · Jul 29, 2017

I'm back for another semester of struggling through math

Screen Shot 2017-07-29 at 3.57.58 pm.png

Can someone explain this?

Sorry if this is so basic but I legit don't get it

pikachu975 · Jul 29, 2017

matchalolz said:
I'm back for another semester of struggling through math

View attachment 34159

Can someone explain this?

Sorry if this is so basic but I legit don't get it

Basically use chain rule so d(tan^-1 (2/y))/d(2/y) * d(2/y)/dy = (-2/y^2) / (1+(2/y)^2) and simplify that

Sp3ctre · Jul 29, 2017

matchalolz said:
I'm back for another semester of struggling through math

View attachment 34159

Can someone explain this?

Sorry if this is so basic but I legit don't get it

You could also use the the differentiation: d(tan^-1(x/a))/dx = a/(a^2+x^2).

Although this method is easier when you're differentiating something in the form tan^-1(x/a), you can treat a as 1/2 and you will get:

-1/2/(1/2)^2 + x^2

= -1/2/(1/4 + x^2)

Times numerator and denominator by 4:

= -2/(1+4x^2)

Edit: As suggested by He-Mann, although this method can be significantly easier to differentiate certain functions, it is best avoided if you do not have a conceptual understanding of differentiation of inverse trig functions since you are just plugging in values into a formula

He-Mann · Jul 29, 2017

Sp3ctre said:
You could also use the the differentiation: d(tan^-1(x/a))/dx = a/(a^2+x^2).

Although this method is easier when you're differentiating something in the form tan^-1(x/a), you can treat a as 1/2 and you will get:

-1/2/(1/2)^2 + x^2

= -1/2/(1/4 + x^2)

Times numerator and denominator by 4:

= -2/(1+4x^2)

This is just mindlessly plugging into formulas which should be avoided at all costs for this case. matchalolz does not have a good idea of how to do this fundamental question so giving formulas without explanation/proof is worthless for the purposes of learning which is top priority in this case.

Whenever you make a claim, it is best to provide proof because the reader does not necessarily believe whatever you say unless it's trivial.

Sp3ctre · Jul 29, 2017

He-Mann said:
This is just mindlessly plugging into formulas which should be avoided at all costs for this case. matchalolz does not have a good idea of how to do this fundamental question so giving formulas without explanation/proof is worthless for the purposes of learning which is top priority in this case.

Whenever you make a claim, it is best to provide proof because the reader does not necessarily believe whatever you say unless it's trivial.

You're right, sorry. I don't know why but I thought for a second that I could somehow help by giving an alternate solution and thought that it would add to his knowledge of how to tackle those types of questions. Apologies for my stupidity

First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

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