Question

Find the point on the sphere x^2 +y^2 + z^2 = 9 that is closest to the point (2,3,4)

(presumably using lagrange multipliers, since thats what the chapter entails lol)

Answer 1

You have to minimize the distance between a point on the sphere and the point (2,3,4), with the constraint that the point lies on the sphere.

Answer 2

how do i do that? :(

Answer 3

distance function is:
\[\sqrt{(x-a)^{2} +(y-b)^{2} +(z-c)^{2}}\]
let a,b,c be point (2,3,4) and x,y,z point on sphere
\[z = \sqrt{9-x^{2}-y^{2}}\]
\[d = \sqrt{(x-2)^{2}+(y-3)^{2}+(\sqrt{9-x^{2}-y^{2}}-4)^{2}}\]
set partial derivatives equal to 0 and solve for x,y

Answer 4

where did you get the last 4 from? and is there a way to also do this with lagrange?

Answer 5

from the point (2,3,4)
and probably...the Lagrange multipliers may come into play later in the problem, i've forgotten how to use those...something with matrices

Answer 6

so.. how do i set partials when everything is under a root? :\ and the lagrange multiplier is just:
f(x,y,z) - lambda (g(xyz) - k)

Answer 7

you just take the derivative...its a lot of the chain rule :)

Answer 8

would you mind showing me the steps for just one of them, like Fx?
the only thing i can think of is
\[\sqrt{2(x-2)}\]

Answer 9

do you know the solution?
i want to make sure i did it right first...i didn't use langrange multiplier at all so this won't help you for your class too much :(

Answer 10

8/3, according to the book

Answer 11

? it should be a point

Answer 12

just kidding, wrong answer ^.^ the correct one is
\[6/\sqrt{29}, 9/\sqrt{29}, 12/\sqrt{29}\]

Answer 13

ok i got the right answer but doing it a different way

Answer 14

here you go..haha
http://en.wikipedia.org/wiki/Lagrange_multiplier

Answer 15

yeahhh i get how to use them, just not for this problem :(

Answer 16

hmm try
\[f(x,y,z) = \sqrt{(x-2)^{2} +(y-3)^{2}+(z-4)^{2}}\]
\[g(x,y,z)-c = x^{2}+y^{2}+z^{2}-9\]
then
\[\rightarrow \sqrt{(x-2)^{2} +(y-3)^{2}+(z-4)^{2}} +\lambda( x^{2}+y^{2}+z^{2}-9)\]

Answer 17

now you have to take 4 partial derivatives..fun

Answer 18

ohhhhhh okay :o
but.. i still don't understand how to even begin going about getting a partial derivative while everything is under the square root :( :(

Answer 19

\[f(x) = \sqrt{u(x)}\]
\[f'(x) = \frac{1}{2\sqrt{u(x)}}*u'(x)\]
chain rule

Answer 20

i'll do Fx
the derivative inside root is just 2(x-2) right
multiply that by derivative of sqrt(u) ...where u is that whole big function
\[F_x = \frac{2(x-2)}{2\sqrt{u}} +2 \lambda x\]

Answer 21

i'm kind of having a hard time understanding what you're doing, but let me try..is this correct?
\[f_y=2(y-3)/2\sqrt{u}\]

Answer 22

+2lambda y

Answer 23

yes

Answer 24

so then
\[f_z = 2(z-4)/2\sqrt{u} + 2z \lambda \]

Answer 25

do you know why the sqrt(u) is on bottom?

\[d/dx \sqrt{x} = x^{1/2} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\]

Answer 26

OHHHHHHHHHH that totally makes so much more sense now LOL

Answer 27

good...also you can simply all of those by cancelling out the 2's

Answer 28

not in lambda part though

Answer 29

wait another question ... so like lets pretend that i did the chain rule but it WASN'T 2(x-2)/2(u) but it was 2(x-2)/(u) right? and then i had to solve for x... i know you're not supposed to touch the inside when doing the chain rule, but when solving for x, after doing the chain rule.. would i be allowed to distribute then?

Answer 30

yes after differentiating you can distribute if you need to

Answer 31

ohh okay! great! thank you sooooo much!!!! i wish i could like.. give you two medals, or fan you again lol

Answer 32

haha no problem...good luck , you may want to review derivative rules before the test

Answer 33

so the denominator is 2sqrt {(x-2)^2+y-3^2+(z-4)^2, right? do i need to write that out for fx fy and fz? and what can i do with it when i'm solving for stuff? :o

Answer 34

its up to you...since its same for each one you could leave it as "u" or some other variable until you have to break it up
i think for this problem it just ends up cancelling out so it doesn't matter

Answer 35

wait.. is it supposed to be f(x) - lambda g(x) or f(x) + lambda g(x)?

Answer 36

it says "+"
do you use "-" for minimization problems?

Answer 37

hold on let me solve it and i'll find out

Answer 38

i actually have no idea, i just know the formula i've been using is -lambda :o but that's been for finding local max and min, not sure if that counts as minimization =\ and they don't really explain how to do this type of problem in the book -_-"

Answer 39

btw, is this right?
\[2x \lambda = (x-2)/\sqrt{u} , x = (x-2)/ 2 \lambda \sqrt{u}\]

Answer 40

ok using "+" works, i get same answers as before

looks right...change the sign though

Answer 41

okay! thank you!! :D

Answer 42

your welcome

Answer 43

so... i'm kind of stuck again :x i got
x = -x+2 / 2λ√u
y = -y+3 / 2λ√u
z = - -z+4/2λ√u

and am not sure how to proceed =\

Answer 44

sorry :(

Answer 45

haha no problem...well you need to isolate x, right now you have an x on both sides
2 ways to do this:
1. split right side into 2 fractions, then move x term over
2. multiply the denominator back over to left side, then add the x over...factor out the x..

Answer 46

i just tried to do it in my head real quick, and do they both yield the same answer? :o

Answer 47

yes they do...no matter how you do it x can only equal one expression right

Answer 48

haha oh right... -_-" i knew that..

Answer 49

okay so this may be a silly question but ... lets call 2λ√u m?
i split it into two to get x = -x/m + 2/m, moved the x over to get
x + x/m = 2/m so then i factor it? to get
x (1+1/m) = 2/m ?
now.. if i divide by 1+1/m .... ...
i'm not sure how to do this :X

Answer 50

ahh yes, lot of algebra involved here
you need to combine 1+ 1/m into 1 fraction, then to divide, multiply by its reciprocal

Answer 51

then.. would it be... 1m/m + 1/m? o.O

Answer 52

yep

Answer 53

would that make it 2m/m? would the m's cancel out? to just be 2?! :o

Answer 54

uh oh , you are missing something...but yes you will have the m's cancel

m/m + 1/m = (m+1)/m
reciprocal -> m/(m+1)

Answer 55

OH lol -_-" that was silly haha...
so then is
x = 2/m+1?

Answer 56

sorry..yes thats correct

Answer 57

yes! okay :D great! you're my hero.
x = 2/m+1 , y = 3/m+1, and z = 4/m+1?!

Answer 58

good...now plug them into last equation which is just g(x,y,z)-c
and solve for lambda

Answer 59

so i immediately factored out m to get
m(-2(2) - 2(3) - 2(4) which is m(-18) = 9
-18 = 9/m, divide both sides by 9 i get
-2 = 1/m, or -2 = 1/2λ√u+1
i'm not sure what else i can do to try to get lambda alone :(

Answer 60

what equation are you using?
it should be partial wrt lambda
--> x^2 +y^2 +z^2 -9 = 0

Answer 61

and be careful...denominator is 1+m so you can't factor out a m

Answer 62

oh nevermind, i see what you did....m is "m+1" :)

Answer 63

wait i'm using g(x)?? not g'x?!? O.O
i thought i had to find fx, fy, fz, and then f lamdba =\

Answer 64

oh.. i see what i did wrong -_-"

Answer 65

think , what is f lambda ?

Answer 66

yeah sorry, m = 2λ√u and we'll make n = 2λ√u+1 just so it makes more sense haha..
so plugging everything in i get 4/n + 9/n + 16/n = -9, can i multiply both sides by n?

Answer 67

yes...the -9 should be +9 when you move it over
also the denominator is squared...so technically it should be n^2

Answer 68

oh! i totally forgot about the square -_-"
so then is it 4 lambda u? which will make it 28 = 36 lambda^2 u?

Answer 69

lets see...you forgot about the +1 ....n = 1+2lambda*sqrt(u)
28 should be 29....4+9+16

Answer 70

i subtracted 1 from both sides, am i not allowed to do that? :(

Answer 71

btw.. does u ever cancel? :o

Answer 72

29 = 9n^2
take square root
sqrt29 = 3n
sqrt29/3 = n = 1+2lambda*sqrt(u)

Answer 73

then subtract 1

Answer 74

and yes the u will cancel when we plug lambda back in to solve for x,y,z

Answer 75

ohhhhhh okay, order of operations, lol that's right. so then is it:
\[-1+\sqrt{29}/6\sqrt{u} = \lambda\]

Answer 76

is that (-1 +sqrt29) ?
should be (-3+sqrt29)
when you combine fractions...change 1 to 3/3

Answer 77

OH i totally did that at first andthen was like nahh... lol -_-" okay -3+! is the denominator correct? :o

Answer 78

haha yes you got it

Answer 79

YES!!!!! :D :D : D :D omg thank you soooooooooooo much!!!!!!!!! hey, btw, are you a girl or guy? lol

Answer 80

i just realized that sounds creepy lol. i want to write you a testimony and would like to get the sex correct haha

Answer 81

not done yet...still have to plug in lambda to get x,y,z

and im a guy...guess i could add that to my profile

Answer 82

haha with a name like dumbcow that's actually pretty hard to guess.
it's not fair that plugging lambda in makes things even more complicated

Answer 83

its not too bad...the sqrt(u) will cancel , just be careful with combining fractions and simplifying

actually i will say this method turns out to be less messy than my original way

Answer 84

actually, jk, maybe not... is it 1/root(29)? :o

Answer 85

close...haha you have the solutions remember

Answer 86

yeah but i refuse to look at them :o and i'm not plugging into the original equation yet, i'm just trying to solve for x y and z, right?

Answer 87

right
x = 2/(1+2lambda*sqrt(u)) now take the expression for lambda and plug it in

Answer 88

so, i must confess, i'm not exactly sure how to properly plug it in haha but this is what i did:

2/2 x (-3+root(29)/6rootu) x root(u)+1 =
2 / -6 +2root(29)6root(u)) x root(u) +1 , i think the 6 and root u cancel out to get
2 / -1+2root(29)+1, the 1's cancel out to get
2/2root(29) and the 2's reduce to get 1/root(29) :o

Answer 89

ok you're mistake is the 6's do not cancel because the whole numerator is (-6+2root29)
if it was (-6+6root29) then you could factor out a 6, then 6's cancel

in this case, however, the most you can factor out is a 2
--> 2(-3+root29)/6 = (-3+root29)/3

Answer 90

oh...so then it's... but wait.. this is already a denominator.. so what happens with a denominator of a denominator?!

i have 2 / (-2+root(29)/3)
is that right?

Answer 91

careful with those fractions
\[\frac{\sqrt{29}-3}{3} +1 \neq \frac{\sqrt{29}-2}{3}\]

Answer 92

dammit. 2/ (3+root(29)/6) ?!

Answer 93

Note:
when you have to divide by a fraction...multiply by its reciprocal

Answer 94

no thats not it...its alright just think get common denominator right
change the 1 to 3/3

Answer 95

oh lol wait i had it over 6 for some reason...its 0 ^.^
2 / (root(29)/6)
you said to multiply by the reciprocal? so then should it be 2/6root(29)?

Answer 96

haha its still over 6 :)
reciprocal means flip the fraction....root(29)/3 -> 3/root(29)

Answer 97

why.. why is that six.. still there -_-"..... i'm sorry, you must think i'm sooo stupid lol. please don't hate me...

wait so .. am i multplying the entire thing by the reciprocal? so like 2/(root29/3) x 3/root29?

Answer 98

or am i mutiplying 2 /root(29)/3 by 3/root(29)/2? lol