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

Flop21 · May 25, 2016

Re: MATH1131 help thread

Flop21 said:
Why does cos simplify to this? Or how do you work it out?

Just bumping this q, hidden in the previous page.

leehuan · May 25, 2016

Re: MATH1131 help thread

Flop21 said:
Just bumping this q, hidden in the previous page.

That's an exact ratio.

$\cos{\frac{\pi}{6}}=\frac{\sqrt3}{2}$

Flop21 · May 25, 2016

Re: MATH1131 help thread

What about say cos[(5*Pi)/6]

leehuan · May 25, 2016

Re: MATH1131 help thread

Flop21 said:
What about say cos[(5*Pi)/6]

Cos is negative in the second quadrant

cos(5π/6)=-√3/2

Flop21 · May 25, 2016

Re: MATH1131 help thread

Why is 3/root(3) = root(3) ??

InteGrand · May 25, 2016

Re: MATH1131 help thread

Flop21 said:
Why is 3/root(3) = root(3) ??

Because sqrt(3)*sqrt(3) = 3, then rearrange.

In general, a/sqrt(a) = sqrt(a) for any a > 0.

RealiseNothing · May 27, 2016

Re: MATH1131 help thread

Flop21 said:
How do you do this with no calculator?? What's going on in the second to third line?

If you want to square a number that ends in 5 you do this as a short cut (i'll show by example):

$35^2$ well 3 times 4 is 12, then just add 25 on the end = 1225.

$45^2$ well 4 times 5 is 20, add 25 to get 2025.

$115^2$ well 11 times 12 is 132, so answer is 13225.

Flop21 · May 27, 2016

Re: MATH1131 help thread

RealiseNothing said:
If you want to square a number that ends in 5 you do this as a short cut (i'll show by example):

$35^2$ well 3 times 4 is 12, then just add 25 on the end = 1225.

$45^2$ well 4 times 5 is 20, add 25 to get 2025.

$115^2$ well 11 times 12 is 132, so answer is 13225.

Yeah or you can just do long multiplication to multiply any two numbers, e.g. 35*35. But then I realised I gotta squareroot it all in the end so getting a big number was useless.

iforgotmyname · May 27, 2016

MATH1131 help thread

Flop21 said:
Yeah or you can just do long multiplication to multiply any two numbers, e.g. 35*35. But then I realised I gotta squareroot it all in the end so getting a big number was useless.

Or you can test it like (35+1)^2 , (36+1)^2 then get the answer

leehuan · May 27, 2016

Re: MATH1131 help thread

There's an infinite number of ways to do questions. Just go with what you find is best

turntaker · May 27, 2016

Re: MATH1131 help thread

Not sure if this is the right place, but I guess the topics are the same across universities.

I have calculated the intersection. Not sure how to find the third plane.

iforgotmyname · May 27, 2016

Re: MATH1131 help thread

turntaker said:
Not sure if this is the right place, but I guess the topics are the same across universities.

I have calculated the intersection. Not sure how to find the third plane.

Put these in a reduced matrix, then row eucholeon form. From there a condition should be clear. Then you can back sub the condition to anyone of the equation

Drongoski · May 27, 2016

Re: MATH1131 help thread

P₃ is: 11x - 10y +18z - 19 = 0

turntaker · Jun 5, 2016

Re: MATH1131 help thread

Not sure how to do this one.

leehuan · Jun 5, 2016

Re: MATH1131 help thread

Have not seen that type of question in my life so my guess would be isn't it just (x+1)²+(y+3)²+(z+5)²≤1

turntaker · Jun 5, 2016

Re: MATH1131 help thread

leehuan said:
Have not seen that type of question in my life so my guess would be isn't it just (x+1)²+(y+3)²+(z+5)²≤1

The answer is

(x-1)^2 + (y-3)^2 + (z-5)^2

Not sure why though

leehuan · Jun 5, 2016

Re: MATH1131 help thread

turntaker said:
The answer is

(x-1)^2 + (y-3)^2 + (z-5)^2

Not sure why though

Oh, then that x²+y²+z²≤1 is the non-hollow sphere that results from after the translation I reckon.

Which makes sense cause you're just adding 1 to the x-component, 3 to the y-component and 5 to the z-component. And because the question says "translate under t".

Only makes sense intuitively though. I still have to have the wording broken down for me

InteGrand · Jun 5, 2016

Re: MATH1131 help thread

turntaker said:
The answer is

(x-1)^2 + (y-3)^2 + (z-5)^2

Not sure why though

$\noindent If we're translating under $\vec{t}$, it means we're moving everything in the direction and amount specified by the vector $\vec{t}$.$

$\noindent I.e. it is the mapping $\vec{x}\mapsto \vec{x} + \vec{t}$. So if we had a point before, we add $\vec{t}$ to it to get it to its new location (which amounts to translating it by $\vec{t}$).$

$\noindent So the vector $\left(x,y,z\right)$ gets moved to $\left(x^\prime,y^\prime,z^\prime \right)=\left(x+1,y+3,z+5\right)$, due to what $\vec{t}$ is. So if we had $x^2 + y^2 + z^2 \leq 1$ originally, we'll now have $\left(x^\prime-1\right)^2 + \left(y^\prime-3\right)^2 + \left(z^\prime-5\right)^2 \leq 1$, and conversely.$

turntaker · Jun 5, 2016

Re: MATH1131 help thread

InteGrand said:
$\noindent If we're translating under $\vec{t}$, it means we're moving everything in the direction and amount specified by the vector $\vec{t}$.$

$\noindent I.e. it is the mapping $\vec{x}\mapsto \vec{x} + \vec{t}$. So if we had a point before, we add $\vec{t}$ to it to get it to its new location (which amounts to translating it by $\vec{t}$).$

$\noindent So the vector $\left(x,y,z\right)$ gets moved to $\left(x^\prime,y^\prime,z^\prime \right)=\left(x+1,y+3,z+5\right)$, due to what $\vec{t}$ is. So if we had $x^2 + y^2 + z^2 \leq 1$ originally, we'll now have $\left(x^\prime-1\right)^2 + \left(y^\prime-3\right)^2 + \left(z^\prime-5\right)^2 \leq 1$, and conversely.$

Thanks fam

turntaker · Jun 5, 2016

Re: MATH1131 help thread

Stuck on this for 30mins. Help would be appreciated

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

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