Question

The shortest distance from the curve xy = 4 to the origin is

Answer 1

i did this one early for someone else let me see if i can link you to it

Answer 2

Thanks. anyone know this one:
If f(x) = I(x^2 - 12)(x^2+4)I, how many numbers in the interval [-2,3] satisfy the conclusion of the mean value theorem?

Answer 3

i cant find it lol
anyways y=4/x

\[d=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+(\frac{4}{x})^2}\]
\[d^2=x^2+\frac{16}{x^2}\]

Answer 4

to minimize d all we need to do is minimize d^2

Answer 5

I sort of get confused when minimizing

Answer 6

hey myinin could you use legrange multipliers?

Answer 7

what would your constraint be?

Answer 8

\[(d^2)'=2x-2*16*\frac{1}{x^3}=\frac{2x*x^3}{x^3}-32\frac{1}{x^3}=\frac{2x^4-32}{x^3}\]
\[=2*\frac{x^4-16}{x^3}\]
critical numbers are -2,2,0
except f(0) doesn't exist so we just need to see which gives us shortest distance x=2 or x=-2
and they both give shortest distance

Answer 9

x^2+y^2

Answer 10

that is not a constraint

Answer 11

(2,4/2) and (-2,-4/2)

Answer 12

whats a constraint?

Answer 13

xy=4 would be a constarint

Answer 14

y=4/x

Answer 15

there is a really simple way to do this problem

Answer 16

ok sorry yeah lamda(del(constarint)=x^2+y^2

Answer 17

del(x^2+y^2)

Answer 18

wait zarkon i can't do that way?

Answer 19

your way is fine

Answer 20

I have a quick way to do it

Answer 21

show and tell please

Answer 22

I dont really know anything about constraints... myininaya, how did you get that equation for the derivative of d^2?

Answer 23

i square both sides of the distance formula

Answer 24

do you know how i got d right?

Answer 25

hey zarkon you show me the quick way in a different thread i guess if trap doesn't want to see

Answer 26

min google lagrange multipliers thats an easy way to do it

Answer 27

I'd like to see!

Answer 28

me too
im so excited

Answer 29

\[xy=4\]

so \[y=\frac{4}{x}\]

then \[y'=-\frac{4}{x^2}\]

the shortest distance is the line that is perpendicular to the tangent

this will happen when the the slope of the line from the origin to the function is 45deg...ie has slope 1

so solve \[-1=-\frac{4}{x^2}\]

\[x^2=4\]

staying in the first quadrant x=2

Answer 30

plug the 2 in the distance formula

Answer 31

this works really well when the function is really complicated. This function was fairly simple so it only helps a little.

Answer 32

creative i like it zarkon

Answer 33

this is a stupid question but where does -1 come from
-4/x^2=tan(pi/4) ?

Answer 34

where did the -1 come in?

Answer 35

the slope of a perpendicular line is the negative reciprocal on the original line

Answer 36

our slope was 1 so the perpendicular slope is -1/1=-1

Answer 37

ok!

Answer 38

can anyone do it with Lagrange multipliers :)

Answer 39

nut was talking about lagrange multipliers
i'm not sure

Answer 40

zarkon how can i be smart like you?

Answer 41

\[\nabla F=\lambda \nabla G\]

Answer 42

you are smart

Answer 43

not as smart

Answer 44

I have a lot of experience. I've been doing this stuff for many years

Answer 45

so the older i get hopefully the wiser i will get

Answer 46

so 2 has to be plugged into the distance formula? It is not the answer?

Answer 47

so 2 has to be plugged into the distance formula? It is not the answer?

Answer 48

\[F(x,y)=x^2+y^2\]

\[G(x,y)=xy-4\]

Answer 49

2 is not the answer

Answer 50

you have to plug it into the distance formula

Answer 51

asking for shortest distance
to find distance you need to plug 2 into distance formula since this will give you a distance

Answer 52

what do I use for the y value?

Answer 53

\[\frac{\partial F}{\partial x}=2x\]
\[\frac{\partial F}{\partial y}=2y\]

\[\frac{\partial G}{\partial x}=y\]

\[\frac{\partial G}{\partial y}=x\]

Answer 54

remember i gave you y earlier
f(2)=4/2=2
so use (2,2) and (0,0)
to find distance

Answer 55

\[2x=\lambda y\]
\[2y=\lambda x\]

Answer 56

\[\frac{2x}{y}=\frac{2y}{x}\Rightarrow x^2=y^2\]

Answer 57

restrict to 1st quadrant

Answer 58

xy=4

xx=4
x=2

Answer 59

lol so are there anymore ways?

Answer 60

thanks! would anyone be able to hep me with the other problem i posted in this thread?

Answer 61

good stuff...so many ways to do it

Answer 62

\[f(x) = |(x^2 - 12)(x^2+4)|\]

\[f(x)=(12-x^2)(x^2+4)\]

on [-2,3]

Answer 63

how did you switch around thhose numbers?

Answer 64

on [-2,3]

x^2-12 is negative

Answer 65

so \[|x^2-12|=|-(12-x^2)|=|12-x^2|=12-x^2\]

Answer 66

on [-2,3]

Answer 67

\[x^2+4\] is always positive

Answer 68

was it a product or division?

Answer 69

oops

Answer 70

I get 3 numbers

Answer 71

lol

Answer 72

I cheated to get my answer though ;)

Answer 73

i blame that it is getting later

Answer 74

late*

Answer 75

you on the east coast?

Answer 76

central
its almost 1:00 am

Answer 77

Arkansas?

Answer 78

yes

Answer 79

how did you cheat? I applied the MVT and I'm stuck with an equation I dont know how to solve.

Answer 80

I wrote a program on my calculator a while back that tells me the naswer

Answer 81

hey can someone check this log derivative for me, only check work please?

Answer 82

hey can someone check this log derivative for me, only check work please?

Answer 83

what equation did you get trapp?

Answer 84

-x^4 + 8x^2 + 48 = -5. I played around with it, but...

Answer 85

my calculator could not give me exact answers

Answer 86

it gave me decimal approximations

Answer 87

this section is noncalc.

Answer 88

that is your derivative?

Answer 89

i got f'=-4x^3+16x

Answer 90

that's correct

Answer 91

oooh I forgot to take the deriv after i simplified ok hold on

Answer 92

this doesn't become a pretty equation

Answer 93

no

Answer 94

ok got that. so set equal to -5?

Answer 95

no i set equal to -10/65

Answer 96

how did you get that number?