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anonymous
 4 years ago
does anyone want to teach me Multivariable Calculus? :P
anonymous
 4 years ago
does anyone want to teach me Multivariable Calculus? :P

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1you know what you did to x in calc 1 and 2? now do it to y and z

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im good with matrices and all that. it's when it comes a part of vectors that i'm not understanding really well

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1334382554391:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1vector just point in a direction

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1good, casue i aint none to goog at mindreading :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm in the same position! Some teacher failed me along the way in the vector department!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0given vectors a=2xi+2xj+xk, b=xi2xj+2xk and c=2xixj2xk show that {a,b,c} is a negative orthogonal base for x<0

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1an orthogonal base is such that each vector is orthogonal to each other

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1since x is a common factor of everyting; we can ignore it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for wich value(s) of x, will {a,b,c} be an orthornomal base?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1a.b =2xi+2xj+xk, xi2xj+2xk  2  4 + 2 = 0 b.c=2xixj 2xk xi2xj+2xk  2 + 2  4 = 0 a.c =2xi+2xj+xk, xi2xj+2xk  2  4 + 2 = 0 so it is orthonormal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i did that, but when it comes to the second part i get lost.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1and since x is just a common "scalar" id assume x can be anything; but maybe zero since people hate zeros

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1" Moreover, they are all required to have length one" http://mathworld.wolfram.com/OrthonormalBasis.html

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a=\sqrt{4x^2+4x^2+x^2}=\sqrt{9x^2}=3x\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1looks like x needs to be the 1/... yeah, that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i did it for all 3 vectors and they all were 3x

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1x is the scalar that makes them go unit i believe; which means the x = 1 over magnitude

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so since a=b=c=3x=1 > x=1/3

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1x is not part of the length; its factored out and eqauted to the reciprocal of the magnitude

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1each vector has the same basic components so they are going to have the same lengths <2,2,1> ; sqrt(4+4+1) = sqrt(9) = 3 x = 1/3 yep

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there's a third part :X

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1and the negative is just 1/3 right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Find the coordinates to v in the orthornomal base obtained, in wich v in the canonical base has coordinates (1,2,3)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1now your just making up words :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im sorry, its in portuguese, i'm translating :P

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1standard basis perhaps?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1row reduce your vector basis next to this new vector from the standard bases

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1how should we define our vectors in the basis? what should our x y and z parts correlate to?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1it matters; switching columns of a matrix alters things; switching rows doent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for the first vector and i did the same to the other 2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1v1=2,2,1 v2=1,2,2 v3=2,1,2 \[\frac{1}{3}\begin{vmatrix}v_1&v_2&v_3\\2&1&2\\2&2&1\\1&2&2\end{vmatrix}\ \begin{vmatrix}c_1\\c_2\\c_3\end{vmatrix}=\begin{vmatrix}1\\2\\3\end{vmatrix}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1your vectors need to be put in columns, not rows

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1multiply both sides by 3 to get rid of the scalar

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[RREF\begin{vmatrix}2&1&2&&3\\2&2&1&&6\\1&2&2&&9\end{vmatrix}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the left goes to identity and the right goes to the coord vector in the basis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0guess im not as bad as i thought? at least i got the first 2 parts correct :p and thanks for explaining the rest :))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'd like to learn multivariate as well to get a head start on next year, but I'm not a fan of matrices and vectors :(
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