Question

find critical point and classify

Answer 1

\[\frac{ dx }{ dt }=x+y+7\]
\[\frac{ dy }{ dt }=3x-y-5\]

Answer 2

so you set both of these equal to 0 I suppose?

Answer 3

I tried that, but it's not correct, according to the answer in my book

Answer 4

oh wait,

Answer 5

so hmm we have
x+y=-7
3x-y=5
------ add equations:
4x=-2
you can solve that for x

Answer 6

that is what I did, x=-.5

Answer 7

and then get y
by solving
-.5+y=-7 for y

Answer 8

so y=-6.5

Answer 9

that is not what my book says though, but that is how I solved it

Answer 10

(-1/2,-13/2)

Answer 11

what does the book say?

Answer 12

that should be right

Answer 13

(3,4)

Answer 14

that has to be a mistake
I'm sure you solve the system dx/dt=0 and dy/dt=0 to find the critical points

Answer 15

ok, thank you

Answer 16

http://users.etown.edu/h/hughesjr/ma321/notes/c6s1.html
here is something to back by up

Answer 17

me up*

Answer 18

A critical point of such a system is a point (a,b) such that F(a,b)=G(a,b)=0.
x+y+7
=
-1/2+-13/2+7=-14/2+7=-7+7=0
and
3x-y-5
=
3(-1/2)--13/2-5
=
-3/2+13/2-5
=
10/2-5
=
5-5
=0

just to prove it is a critical point :p

Answer 19

understood, I appreciate the supporting details :p

Answer 20

I know it doesn't make you feel much better since the book still says the critical points to the system you were given was (3,4)

Answer 21

it always feels weird accepting answers that the book doesn't seem to agree with

Answer 22

the book has been wrong before, so its possible again..lol

Answer 23

http://www.wolframalpha.com/input/?i=critical+points+dx%2Fdt%3Dx%2By%2B7+and+dy%2Fdt%3D3x-y-5

there are some 3's and 4's occurring in the differentiation equations solutions here

Answer 24

and also along with our critical point

Answer 25

thank you, we agree so im confident the book is wrong

Answer 26

I'm confident the book is wrong too :)

Answer 27

can i say one more annoying thing

Answer 28

I think I know what went wrong

Answer 29

with the book's solution

Answer 30

ok, im listening..lol

Answer 31

\[\frac{dx}{dt}=x+y-7 \\ \frac{dy}{dt}=3x-y-5 \]
so this would mean
x+y=7
3x-y=5
-------add equations:
4x=12
x=3
and so y=4
(3,4) would be a critical point to \[\frac{dx}{dt}=x+y-7 \\ \frac{dy}{dt}=3x-y-5 \]

Answer 32

this problem looks a lot similar to our problem
\[\frac{dx}{dt}=x+y+7 \\ \frac{dy}{dt}=3x-y-5\]

Answer 33

I think there was just a small type-o in the problem and that is how it got what it got

Answer 34

ahhhh

Answer 35

I think that seems very possible

Answer 36

anyways have fun :)

Answer 37

seems very possible, thanks for your help and insight