Question

Drawing direction field and solution curve passing through a given point

Answer 1

The question says:

Graph a direction field (by a CAS or by hand). In the field graph several solution curves by hand, particularly those passing through the given points (x,y) .

(1) \(y' = 1+y^2\)

Answer 2

Point is \((\cfrac{\pi}{4} , 1\))

Answer 3

Well in a sense, you have a formula for the slope right there! So you can just plug in that point:

y'=1+(1)^2

The slope doesn't even depend on the x-value!

Answer 4

Now, I know that for drawing a direction field, the first step is to put
1+y^2 = 0 which will not give any solution. So, the arrows will never be horizontal.

Now, I also know that 1+y^2 will always be positive, so, arrows will always point upwards.

Now, I assume that 1+y^2 = 1
\(\implies y = 0\)

Which means that arrows lying on x-axis will have a slope 1.

Similarly now I assume 1+y^2 = 5
\(\implies y^2 = 4\)
Which represents a pair of straight lines y = 2 and y = -2.

So, arrows lying on these lines will have a slope of 5.

And so on..

Answer 5

Now I don't know how to plot the solution curve.

Answer 6

Well can you plot the slope field?

Answer 7

Well sorry but I have to leave now. I will draw it and then notify you. Will that work ?

Answer 8

Yeah, sure. Are you going to eat dinner as well?

Answer 9

Yes

Answer 10

I'll be back after I finish breakfast lol.

Answer 11

Lol Bye

Answer 12

This is the direction field.

Answer 13

The corresponding solution curve according to me.

Answer 14

@Kainui Mission Accomplished :)

Answer 15

Is that correct ?

Answer 16

Yeah, exactly! Not bad.

Answer 17

Thanks.

Answer 18

Glad I could help, but you did most of it yourself.

Answer 19

:)