# MATH1151 HELP (1 Viewer)

#### Voolgdognng

##### New Member
How do i convert from parametric vector form to cartesian form for a plane

#### ExtremelyBoredUser

##### Bored Uni Student
How do i convert from parametric vector form to cartesian form for a plane
pre sure the conventional way is

the eq of plane is given by
l = a + tu + mv where t,m are scalars and u,a,v are vectors

Find the normal of the plane by cross producting the direction vectors so n = u x v in this case
Apply the formula,
n * (x-a) = 0 (where n is the normal, x is (x,y,z) \in R3 and a is the vector and * means dot product)
and then expand and get your cartesian equation

cbf latexing sorry

#### Voolgdognng

##### New Member
pre sure the conventional way is

the eq of plane is given by
l = a + tu + mv where t,m are scalars and u,a,v are vectors

Find the normal of the plane by cross producting the direction vectors so n = u x v in this case
Apply the formula,
n * (x-a) = 0 (where n is the normal, x is (x,y,z) \in R3 and a is the vector and * means dot product)
and then expand and get your cartesian equation

cbf latexing sorry
thanks, understood it now

#### ExtremelyBoredUser

##### Bored Uni Student
np all the best, just make sure to document all this stuff bc its easy to forget maths over the weeks.

#### Voolgdognng

##### New Member

How would you go about doing this

#### Luukas.2

##### Well-Known Member
View attachment 42646

How would you go about doing this
(a) is bounded above but not below as its a negative cubic, and so the region above $\bg_white y > 1$ will be $\bg_white x \in (-\infty,\ a)$ or $\bg_white x \in (-\infty,\ a)\cup(b,\ c)$.

(b) must be bounded below as $\bg_white \ln{x} \in \mathbb{R} \implies x > 0$. It's also bounded above as $\bg_white x\ln{(x + 1) - (x + 1)\ln{x} < 0\ \forall x > 2.3$

(c) must be bounded below as the sum is always at least 1 and it increases without bound, so there must be a largest $\bg_white n$ for which the sum is under 10.

#### scaryshark09

##### ∞∆ who let 'em cook dis long ∆∞
Does anyone know how to do part 3 and 4?

#### chigurh07

##### Member
Does anyone know how to do part 3 and 4?

View attachment 42690
I think you can do polynomial long division for both and keep continuing the dividing process until you have a constant in the quotient, then whatever the quotient is is your answer. So for part 3 I think it would be 4*x-9 then for part 4 you could multiply the numerator by a constant to make the division process easier then you should end up with a linear expression as the quotient after dividing, then divide the quotient by the constant you multiplied the numerator by to get the final answer

#### liamkk112

##### Well-Known Member
Does anyone know how to do part 3 and 4?

View attachment 42690
for iii), i think if u do long division u can write the rational function as (4x-9) + 16/(x+1), and 16/(x+1) is O(x^-1) so the answer should just be 4x-9

for iv) similar process, just note that x^(-2) is O(x^-1)

#### scaryshark09

##### ∞∆ who let 'em cook dis long ∆∞
thanks, but im confused with the 'Big O' and 'Little o' stuff

why would 16/(x+1) be O(1/x). and what does the 1/x even mean? would it be different if it was O(x)

#### liamkk112

##### Well-Known Member
thanks, but im confused with the 'Big O' and 'Little o' stuff

why would 16/(x+1) be O(1/x). and what does the 1/x even mean? would it be different if it was O(x)
dfn of O(f(x)): a function g(x) is O(f(x)) if there exists constants m and k such that for all x >=k, |g(x)|<= m|f(x)|

essentially, it means that the growth of f(x) is greater than (or equal to) the growth of g(x). clearly 16/(x+1) is O(1/x) because their growth rates are the same (or, you can find constants m and k such that the definition is satisfied, in this case k could be any number and m would be 16). similarly you can find that the result in iv) after doing long division is O(1/x).

keep in mind that g(x) being O(f(x)) just means that f(x) is an upper bound on the function. so for example, if g(x) was O(x), then it is clearly O(x^2), because x^2 grows at a greater rate than x. similarly this can be extended so if g(x) is O(x) then it is O(x^3) and so on

#### scaryshark09

##### ∞∆ who let 'em cook dis long ∆∞
Does anyone know how to do part c?

#### kendricklamarlover101

##### Member
u can write det(WM^-1) as
$\bg_white \det (-4MM^{T}M^{-1}) = -4\det(MM^{T}M^{-1})$
$\bg_white =-4\det(M)\det(M^{T}M^{-1})$
$\bg_white =-4\det(M)\det(MM^{-1}) = -4\det(M)$
u could also split the product up at the start to get
$\bg_white \det (-4MM^{T}M^{-1}) = -4\det(M)\det(M^{T})\det(M^{-1})$
$\bg_white =-4\det(M)\det(M)\frac{1}{\det(M)}$
$\bg_white =-4\det(M)$

Last edited:

#### scaryshark09

##### ∞∆ who let 'em cook dis long ∆∞
u can write det(WM^-1) as
$\bg_white \det (-4MM^{T}M^{-1}) = -4\det(MM^{T}M^{-1})$
$\bg_white =-4\det(M)\det(M^{T}M^{-1})$
$\bg_white =-4\det(M)\det(MM^{-1}) = -4\det(M)$
u could also split the product up at the start to get
$\bg_white \det (-4MM^{T}M^{-1}) = -4\det(M)\det(M^{T})\det(M^{-1})$
$\bg_white =-4\det(M)\det(M)\frac{1}{\det(M)}$
$\bg_white =-4\det(M)$
lol i agree, thats what i did but it says im wrong???

#### scaryshark09

##### ∞∆ who let 'em cook dis long ∆∞
Does anyone know what to do??

#### liamkk112

##### Well-Known Member
Does anyone know what to do??
detWM^-1
=det(-4MM^tM^-1)
=16det(M)det(M^t)det(M^-1)
=16det(MM^-1)det(M)
=16det(I)*33
=528 ?

keep in mind that det(kA) = k^2 det(A), you can check this for a 2x2 or 3x3 matrix if u like. i think that’s where the mistake is coming from

#### kendricklamarlover101

##### Member
keep in mind that det(kA) = k^2 det(A), you can check this for a 2x2 or 3x3 matrix if u like. i think that’s where the mistake is coming from
shouldnt it be k^8 det(A) as when a row/column is multiplied by a scalar k then det(A) increases by a factor of k? so since its an 8x8 matrix it would be k^8?