Question

please help me

Using Lagrange's method,find minimum and maximum distance from point (1,2,2) to the sphere x^2+y^2+z^2=36.

Answer 1

which part you having a problem with? or having trouble getting started?

Answer 2

the part about the sphere represents the "constraint".
You have to derive the equation that need to be min/maximised

let A (x,y.z) be an arbitrary point in space, then distance between P and X is given by
\[\sqrt{(x-1)^2+(y-2)^2+(z-2)^2}\]

Answer 3

another helpful piece of info is that you can square the function for distance and then find max/min

Answer 4

*distance between P and A

Answer 5

hmm
im bad at partial
i require a lot of help

Answer 6

@SolomonZelman @aryandecoolest

Answer 7

i need to learn it

pls show me the steps

Answer 8

@sweetburger @Frostbite @shamil98 @ganeshie8 @AravindG @Nishant_Garg

Answer 9

@zepdrix @Loser66 @iambatman pls help :'(

Answer 10

how far have you understood lagranges method?

Answer 11

hmm
we have two functions
F(x,y)=f(x,y)+g(x,y)




sorry for late reply
my laptop got hanged

Answer 12

how did you get that?

Answer 13

to apply lagranges method you need two unconnected functions
w=f(x,y,z)
and
g(x,y,z)=c (constraint)

Answer 14

okay

Answer 15

the function of the sphere represents g(x,y,z) = c (c is 36)

Answer 16

so in this case our sphere will be a constriant

Answer 17

yep

Answer 18

yess

Answer 19

lagranges method states: when you have a function f and a constraint 'g'
then, at max/min points
\[\nabla f = \lambda \nabla g\]

where 'lambda' is an arbitrary scalar quantity

Answer 20

oh i dont know the meaning of that symbol

Answer 21

sorry i m really bad at partial

Answer 22

\[\nabla f =< \partial f/\partial x , \partial f/\partial y, \partial f/\partial z>\]

Answer 23

oh okay

Answer 24

you know how to do partial differentiation? you just have to treat all the other variables as constants

Answer 25

yes i know that

Answer 26

try finding \[\nabla g\]

Answer 27

\[\tt g_x =2x\\ g_y=2y\\ g_z=2z\]

Answer 28

yep :)

Answer 29

yay

Answer 30

you understood the equation for 'f' ?

Answer 31

the one at the top of this thread, under the square root sign.

Answer 32

yea it is the distance equation

Answer 33

what is P there?
centre of sphere?

Answer 34

no, P is a random point somewhere inside the sphere, you need to find which point on the sphere is closest and farthest from P

Answer 35

you understood that part where we said we can square the distance function?

Answer 36

yep
d^2=-----

Answer 37

so from point (1,2,2) we are finding a point which is closest to it
and far from it,correct?

Answer 38

yep

Answer 39

yes...

Answer 40

so, you have understood that the distance function is the "f(x,y,z)" ?

Answer 41

hmmm.. tbh no :(

Answer 42

well for sure we have a function with constant
so the other function has to be the f(x,y,z) function

Answer 43

f(x,y,z) represents the function that needs to minimized/maximized. the question is asking you to minimize or maximize the distance between P and a point on the circle right.

Answer 44

yes @baru

Answer 45

Using Lagrange's method,find minimum and maximum distance from point (1,2,2) to the sphere x^2+y^2+z^2=36.

so lets write our objective as the squared distance from \((1,2,2)\): $$r=(x-1)^2+(y-2)^2+(z-2)^2$$ now we want to extremize \(r\) with respect to the constraint \(x^2+y^2+z^2=36\); we can add a penalty for violating this constraint using a Lagrange multiplier \(\lambda\): $$r=(x-1)^2+(y-2)^2+(z-2)^2-\lambda(36-x^2+y^2+z^2)$$

Answer 46

now extremize that function: $$\frac{\partial r}{\partial x}=2(x-1)+2\lambda x\\\frac{\partial r}{\partial y}=2(y-2)-2\lambda y\\\frac{\partial r}{\partial z}=2(z-2)-2\lambda z$$

Answer 47

oops, it should read \(36-x^2-y^2-z^2\) so the sign of the \(\lambda\) terms in the partial derivatives should be positive

Answer 48

so $$x=1-\lambda\\y=2-\lambda\\z=2-\lambda$$ when \(r_x=r_y=r_z=0\)

Answer 49

and our constraint becomes $$(1-\lambda)^2+(2-\lambda)^2+(2-\lambda)^2=36\\9-10\lambda+3\lambda^2=36\\3\lambda^2-10\lambda-27=0$$

Answer 50

so: $$\lambda=\frac{10\pm\sqrt{100+4(81)}}{6}=\frac{5\pm\sqrt{106}}3$$

Answer 51

plug this in for \(x,y,z\) for the points with minimum and maximum distance, respectively

Answer 52

@rvc followed?

Answer 53

i would love to hear your explainah]tion too @baru

Answer 54

explaination*

Answer 55

are you clear on part about "constraint" and the function "f"?

Answer 56

yepyep

Answer 57

so rather than suffer with an equation under the square root, we convinently choose square of distance for "f". can you find \[\nabla f\]?

Answer 58

\[\rm d^2=(x-1)^2+(y-2)^2+(z-2)^2\]

Answer 59

please reply to my message

Answer 60

now f = d^2
find\[\nabla f\]

Answer 61

\[\rm f_x=2(x-1)\\ f_y=2(y-1)\\ f_z=2(z-1)\]

Answer 62

yep

Answer 63

now try \[\nabla f = \lambda \nabla g\]

Answer 64

the first equation you get would be
\[2(x-1)= \lambda 2x\]

Answer 65

can you do the other two?

Answer 66

\[\tt 2(y-1)=2y\\ 2(z-1)=2z\]

Answer 67

oops i forgot the \(\lambda\)

Answer 68

:) ya

Answer 69

yup :)

Answer 70

well sorry to message you.
I thought you wouldn't mind me :)
its fine
thanks for all the help though

Answer 71

message? you mean pm? i didnt get any

Answer 72

i did message you 3-4 times

Answer 73

oh..lol. didnt get any notifications

Answer 74

now can u please relpyto my messages Sir

Answer 75

yep :)

Answer 76

@rvc are we done with this question? you know how to proceed?

Answer 77

further i have to solve for lambda right

Answer 78

thats upto you, in the end you just need to find values for x,y and z which satisfy those three equations and the constraint equation 'g=c'

Answer 79

oh
i need more help

Answer 80

the best way to do this would be to get rid of lambda
\[2(x-1)/2x = \lambda\]
\[2(y-2)/2y= \lambda\]
\[2(z-2)/2z=\lambda\]

Answer 81

now see how all three equations are equal to lambda and thus equal to each other

Answer 82

yesss