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

#### matchalolz

##### Well-Known Member

$\bg_white \noindent That's not the right answer I'm afraid. Basically, in general, for any u\in [-1,1], \cos^{-1}u is the angle in the range [0,\pi] whose cosine is u (such an angle exists and is unique for any u \in [-1,1]). So \cos^{-1}(\cos x) is that angle in the range [0,\pi] whose cosine is \cos x. In other words, it's an angle like'' x (since the cosine of x is also \cos x), but can't be x itself if \pi < x < \frac{3\pi}{2}, because this is not in the range [0,\pi]. So the answer would be the angle \theta between 0 and \pi that has \cos \theta = \cos x. You can use, for example, general solutions of trig. equations from HSC 3U maths to find this \theta.$
Ok I think I understand what you're saying.

I barely remember anything from 3 unit, I was such a flop back then because I never thought I would be doing a uni math subject

Will try searching up general solutions

#### matchalolz

##### Well-Known Member

so theta = ± cos inverse (cosx)+ 2kπ

How do you know what value to use for k? so lost

#### InteGrand

##### Well-Known Member

so theta = ± cos inverse (cosx)+ 2kπ

How do you know what value to use for k? so lost
$\bg_white \noindent Actually, \theta \emph{is} \cos^{-1}(\cos x), which is what we want to find. This gives \cos \theta = \cos x, so (see e.g. Year 12 3 Unit Pender textbook), we have \theta = x + 2k\pi for some integer k or \theta = - x + 2\ell\pi for some integer \ell. But \theta = \cos^{-1} (\cos x), so \theta is between 0 and \pi. So you need to choose the case (\pm x) and the integer k or \ell that results in \theta being in the interval [0,\pi], given that x is between \pi and \frac{3\pi}{2}.$

#### matchalolz

##### Well-Known Member

$\bg_white \noindent Actually, \theta \emph{is} \cos^{-1}(\cos x), which is what we want to find. This gives \cos \theta = \cos x, so (see e.g. Year 12 3 Unit Pender textbook), we have \theta = x + 2k\pi for some integer k or \theta = - x + 2\ell\pi for some integer \ell. But \theta = \cos^{-1} (\cos x), so \theta is between 0 and \pi. So you need to choose the case (\pm x) and the integer k or \ell that results in \theta being in the interval [0,\pi], given that x is between \pi and \frac{3\pi}{2}.$
I get what we're trying to do but I'm still not getting it, if I sub in like theta = 0 and x = Pi, then I get like l = -1/2 which makes no sense because l is meant to be an integer?

#### InteGrand

##### Well-Known Member

I get what we're trying to do but I'm still not getting it, if I sub in like theta = 0 and x = Pi, then I get like l = -1/2 which makes no sense because l is meant to be an integer?
$\bg_white \noindent Here, think of x as some given number in \left(\pi, \frac{3\pi}{2}\right) and \theta being something depending on x. We can't just sub. arbitrary values for \theta and x. Rather, x is given and we are finding what \theta has to be. We know \theta = x + 2k\pi for some integer k or \theta = -x + 2\ell \pi for some integer \ell (where x is given). We know \theta has to be between 0 and \pi. Since \pi < x < \frac{3\pi}{2}, we have -\frac{3\pi}{2} < - x < -\pi \Rightarrow -\frac{\pi}{2} < 2\pi - x < \pi. So take \ell = 1, then \theta = -x + 2\pi is in the desired range for \theta, so this is what \theta is (remembering there is a \emph{unique} \theta). In other words, \theta = -x + 2\pi = 2\pi - x. Thus \cos^{-1} (\cos x) = 2\pi - x.$

#### matchalolz

##### Well-Known Member

This question is part of the chapter on Gaussian elimination and matrices etc.

Am I correct in thinking that this line would be parallel to the plane if its direction vector <2,3,-1> is a linear combination of <1,1,0> and <0,1,-1>? Or am I totally off? Someone please enlighten me

#### InteGrand

##### Well-Known Member

View attachment 33939

This question is part of the chapter on Gaussian elimination and matrices etc.

Am I correct in thinking that this line would be parallel to the plane if its direction vector <2,3,-1> is a linear combination of <1,1,0> and <0,1,-1>? Or am I totally off? Someone please enlighten me
That is correct.

#### matchalolz

##### Well-Known Member

That is correct.
You know how the line can be expressed as line = point + direction vector? Do I need to do anything with the point at all?

#### InteGrand

##### Well-Known Member

You know how the line can be expressed as line = point + direction vector? Do I need to do anything with the point at all?
No, that point doesn't affect whether the line will be parallel to the plane.

#### matchalolz

##### Well-Known Member

No, that point doesn't affect whether the line will be parallel to the plane.
Thanks so much!

#### matchalolz

##### Well-Known Member

Can someone please explain how you would integrate this integral by inspection? I know how to do it by substitution but the question specifically asks for that specific method.

I tried to differentiate (1+x^2)^3 but I got 6*x(1+x^2)^2 and i don't know what to do afterwards

#### matchalolz

##### Well-Known Member

View attachment 33946

Can someone please explain how you would integrate this integral by inspection? I know how to do it by substitution but the question specifically asks for that specific method.

I tried to differentiate (1+x^2)^3 but I got 6*x(1+x^2)^2 and i don't know what to do afterwards

#### InteGrand

##### Well-Known Member

View attachment 33946

Can someone please explain how you would integrate this integral by inspection? I know how to do it by substitution but the question specifically asks for that specific method.

I tried to differentiate (1+x^2)^3 but I got 6*x(1+x^2)^2 and i don't know what to do afterwards
$\bg_white \noindent The derivative of 1 + x^{2} is sitting there, so an antiderivative is \frac{1}{4}\left(1 + x^{2}\right)^{4}.$

$\bg_white \noindent Use a mental substitution of u = 1 + x^{2}.$

#### matchalolz

##### Well-Known Member

$\bg_white \noindent The derivative of 1 + x^{2} is sitting there, so an antiderivative is \frac{1}{4}\left(1 + x^{2}\right)^{4}.$

$\bg_white \noindent Use a mental substitution of u = 1 + x^{2}.$
Yeah that's similar to what I did, I thought that was the substitution method though :/

#### InteGrand

##### Well-Known Member

Yeah that's similar to what I did, I thought that was the substitution method though :/
The difference between "the substitution method" and inspection is that if you do it by inspection, you don't write out the substitution, you just write down the answer.

#### matchalolz

##### Well-Known Member

The difference between "the substitution method" and inspection is that if you do it by inspection, you don't write out the substitution, you just write down the answer.
Ok thanks for the clarification

I'm probably just going to lose heaps of time in the exam coz I'm too dumb to do it in my head haha

#### leehuan

##### Well-Known Member

Ok thanks for the clarification

I'm probably just going to lose heaps of time in the exam coz I'm too dumb to do it in my head haha
You can always practice doing it the long way but very quickly (and because you're now at uni you can omit a few steps)

#### matchalolz

##### Well-Known Member

wtf is the point of the limit comparison test? I tried reading the theory but it's still not really sinking in

Does the numerator just become f(x) and the denominator become g(x) or do you have to actually figure out a function to compare it to??

#### InteGrand

##### Well-Known Member

wtf is the point of the limit comparison test? I tried reading the theory but it's still not really sinking in

Does the numerator just become f(x) and the denominator become g(x) or do you have to actually figure out a function to compare it to??

View attachment 33953
$\bg_white \noindent The limit comparison test helps us decide whether a given improper integral is convergent or not by comparing the integrand to another function whose improper integral we know is convergent or divergent. For example, in question (a), let f(x) = \frac{x}{2x^{3} - 1} (the integrand) and g(x) = \frac{1}{x^{2}}. Then f and g are positive and continuous on [2,\infty) and \lim_{x\to \infty}\frac{f(x)}{g(x)} = C for some C\in (0,\infty) (you should be able to find what the C is). Then the limit comparison test tells us that \int_{2}^{\infty}f(x)\, \mathrm{d}x has the same convergence status as \int_{2}^{\infty}g(x)\, \mathrm{d}x= \int_{2}^{\infty}\frac{1}{x^{2}}\, \mathrm{d}x. But we know from the p-test that the latter integral converges. Hence by limit comparison test, the integral in (a) is convergent.$

#### matchalolz

##### Well-Known Member
$\bg_white \noindent The limit comparison test helps us decide whether a given improper integral is convergent or not by comparing the integrand to another function whose improper integral we know is convergent or divergent. For example, in question (a), let f(x) = \frac{x}{2x^{3} - 1} (the integrand) and g(x) = \frac{1}{x^{2}}. Then f and g are positive and continuous on [2,\infty) and \lim_{x\to \infty}\frac{f(x)}{g(x)} = C for some C\in (0,\infty) (you should be able to find what the C is). Then the limit comparison test tells us that \int_{2}^{\infty}f(x)\, \mathrm{d}x has the same convergence status as \int_{2}^{\infty}g(x)\, \mathrm{d}x= \int_{2}^{\infty}\frac{1}{x^{2}}\, \mathrm{d}x. But we know from the p-test that the latter integral converges. Hence by limit comparison test, the integral in (a) is convergent.$