Question

Slopefields

Answer 1

need help with generalization of slope fields

Answer 2

I know if you give me a function, example y=x and you ask me to find it slopefields
then I know to take the derivative of it
(dy/dx)=1
and I know all the slope field is made up of the slopes are all + 1
but I am having issues with given a (dy/dx) and matching or finding the slope field.
Is there a generalization for this or do I have to calculate the slope at every point?

Answer 3

may be lets try an example problem directly :)

Answer 4

you're given a differential equation, and asked to match it with the slope field, right ?

Answer 5

yes both ways ie matching differential equations to slopefields and also, given functions and matching to slopefields. Give me a moment and I will try to upload an example.

Answer 6

okie..

Answer 7

my second page tells me to separate the variables and find the general solutions to determine what the slope field should look like for each. Then I am suppose to match the graphs of the slope fields on the page I uploaded.

Answer 8

Okay, where are the equations ?

Answer 9

\[\frac{ dy }{ dx }=e^x\]
\[dy=e^x dx\]
\[\int\limits dy= \int\limits e^x dx\]
\[y=e^x + c\]
so I picked B because I think it looks like e^x

Answer 10

I will attach the equations in a moment

Answer 11

thats right, there is a trick to figure out quickly :

\[\large y' = e^x\]

clearly, the slope can never be 0 or negatve (why ?)
if we look at the graphs, the slope is always positive in only two graphs : B and J

Answer 12

lets work another example equation :)
to figure out slope field, solution is not needed... we just need the differential equation

Answer 13

sorry, my computer is acting strange

Answer 14

ok here are my equations

Answer 15

ok number 4 gives me issues

Answer 16

If I do it the way the instructions tell me I get y=2x+c

Answer 17

the solution is J

Answer 18

again, we interpret only the input differential equation for figuring out slope field, we don't need the solution.


\[\large y' = 2\]

slope = 2, constant always. only J has a constant slope always, so you're right

Answer 19

lets work #8 :

\[\large y' = x\]

so slope is same as x coordinate, that means :
x=-1, slope = -1
x=0, slope = 0
x=1, slope = 1
x=2, slope = 2

looking at the graphs, onlyy `F` meets these

Answer 20

these instructions don't make sense for slope field. I see them as good equations for giving practice for differential equations. I always thought you just take the dy/dx statements and just plot (x,y) points into it to find the slopes
number 8
boy great minds think alike

Answer 21

Oh i missed reading the instructions before, wait..

Answer 22

shouldn't the y axis points look different, isn't x=0

Answer 23

we have negative slope for left of y axis, and positive slope for right of y axis
that means slope is 0 on y axis (assume)

Answer 24

hey the instructions ask us to use the solution directly !
that makes our life easy !!

Answer 25

Maybe I should include what was written on the top of the first one:

Match a slope field to a differential equation.
Since the slope field represents all of the particular solutions to a differential equation, and the solution represents the ANTIDERIVATIVE of a differential equation, then the slope field should take the shape of the antiderivative of dy/dx

Answer 26

thats right !
particular soluton of differential equation = integral curve on slope field

Answer 27

integral curve is just a smooth curve connecting the slope segments

Answer 28

\[\frac{ dy }{ dx}=x\]
\[y=\frac{ 1 }{ 2} x^2+c\]

Answer 29

that is for problem 8

Answer 30

that is why slope field looks like a quadratic

Answer 31

matching `slope field` and `solution of a differential equation` is a algebra problem, not calculus... not very interesting, we just need to match the solution equation to a graph in which the smooth curves through slope segments

Answer 32

yes !

Answer 33

*we just need to match the solution equation to a graph in which the smooth curves through slope segments match the solution equation

Answer 34

thanks, I will post another one I am struggling with. You help me understand this concept a little better

Answer 35

sure, wil try...