stickyricebuns
Member
- Joined
- Apr 25, 2020
- Messages
- 48
- Gender
- Female
- HSC
- 2021
hi
1. for this equation
the question asks, "based on the slope field, determine the behaviour of y as x approaches infinity". the answer is y approaches plus minus infinity and 0
i'm confused how to find the limits from the slope field? i've drawn it and it just looks like this:
but how do you get y approaches 0 and infinity?? i'm unsure if the solution graph on the left (goes straight up) is the reason for y approaching infinity. because as x approaches infinity, at some point it still has to come back down before x is really "at infinity" no?
for y=0 i asked someone and they explained that the reason y approaches 0 is because when y'=0, y=-e^-x. so along the exponential curve the gradient is 0, so as x approaches 0, y also approaches 0. i only understand it a little so can someone pls explain it more in depth.
2. for this equation
it asks the same thing ("draw slope field, find lim x->inf) (answer is 0)
how do you draw the slope field in the first place ?? my understanding is letting y'=m so you get an equation in terms of y, x and m (y= x/2 * e^-2x - m) then sub in values of m and draw the slopes.
this seems kind of ?? like ?? really tedious? because the graph that u have the draw the slopes across isn't something that's intuitive so u have to take time to draw what it looks like before transforming it and doing the slope field. so is there a diff way to graph this slope field that isn't super time consuming.
i did ask the teacher and they said to "let x=1" and just graph y'=e^-2-2y. but why is this a valid way of drawing the slope graph bc x is a variable and you're just making it a constant so you're overall changing the equation from the question hence not answering the question at all. i put it into geogebra and it also doesn't approach 0 (instead approaching something like 0.1) so like. wat
thank u
1. for this equation

i'm confused how to find the limits from the slope field? i've drawn it and it just looks like this:

for y=0 i asked someone and they explained that the reason y approaches 0 is because when y'=0, y=-e^-x. so along the exponential curve the gradient is 0, so as x approaches 0, y also approaches 0. i only understand it a little so can someone pls explain it more in depth.
2. for this equation

how do you draw the slope field in the first place ?? my understanding is letting y'=m so you get an equation in terms of y, x and m (y= x/2 * e^-2x - m) then sub in values of m and draw the slopes.
this seems kind of ?? like ?? really tedious? because the graph that u have the draw the slopes across isn't something that's intuitive so u have to take time to draw what it looks like before transforming it and doing the slope field. so is there a diff way to graph this slope field that isn't super time consuming.
i did ask the teacher and they said to "let x=1" and just graph y'=e^-2-2y. but why is this a valid way of drawing the slope graph bc x is a variable and you're just making it a constant so you're overall changing the equation from the question hence not answering the question at all. i put it into geogebra and it also doesn't approach 0 (instead approaching something like 0.1) so like. wat
thank u