Slope fields and limits (1 Viewer)

[stickyricebuns]

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

 

[tickboom]

Here's my thoughts on your first question:

 

[tickboom]

... and here's my thoughts on your second question. I may be missing something, but I think what your teacher has suggested is unnecessarily complex. Creating the slope field should be as simple as picking various x and y coordinates, plugging them into the formula for y' and then analysing the resulting diagram.

 

[stickyricebuns]

... and here's my thoughts on your second question. I may be missing something, but I think what your teacher has suggested is unnecessarily complex. Creating the slope field should be as simple as picking various x and y coordinates, plugging them into the formula for y' and then analysing the resulting diagram.

thank you for these explanations ! so from what i gather, we can pick any random x and y points, so it would still be valid to let x=1 and find y and y' for my question?

our teacher did say that in the hsc they usually wouldn't make you graph something as hard as what he gave us and we'd only usually see slope field questions in multiple choice anyway. i think his method is to give us the whole visualisation so when we finish plotting the points we get something that looks like what desmos gives us instead of a couple of lines

 

[tickboom]

thank you for these explanations ! so from what i gather, we can pick any random x and y points, so it would still be valid to let x=1 and find y and y' for my question?

our teacher did say that in the hsc they usually wouldn't make you graph something as hard as what he gave us and we'd only usually see slope field questions in multiple choice anyway. i think his method is to give us the whole visualisation so when we finish plotting the points we get something that looks like what desmos gives us instead of a couple of lines

No worries! Yeah the approach I used was just picking combinations of x and y, and then calculating y' to observe the slope of the tangent at that point. Naturally something like Desmos or Wolfram Alpha can do this over many combinations with little effort, but at the end of the day, the technology is simply doing the same steps (just faster).

I'm not sure I understand what your teacher is saying about just letting x=1 and then considering the resulting equation involving y and y' ... Like are you effectively plotting a chart where the vertical axis is y' and the horizontal axis is y? And if so, what would such a chart represent? It's possible I'm just missing something really basic.

 

[stickyricebuns]

No worries! Yeah the approach I used was just picking combinations of x and y, and then calculating y' to observe the slope of the tangent at that point. Naturally something like Desmos or Wolfram Alpha can do this over many combinations with little effort, but at the end of the day, the technology is simply doing the same steps (just faster).

I'm not sure I understand what your teacher is saying about just letting x=1 and then considering the resulting equation involving y and y' ... Like are you effectively plotting a chart where the vertical axis is y' and the horizontal axis is y? And if so, what would such a chart represent? It's possible I'm just missing something really basic.

the method he taught us was to let y'=m and find an equation in terms of y, and m and x (y=f(x)+m or something like that). then you can either sub in values of m (or y) (first column from my working), then you tan inverse whatever m becomes (third column) which is the angle of the tangent. the second column is just the curves you have to draw using y and you draw little lines in the direction of the angle you get from m. sorry this explanation is kind ass, my graph for the second one was kinda crap but if u look at the first one, i subbed m=-1 hence the angle is -45 and y=-e^-x-1 so u draw the graph of y=-e^-x-1 and draw little lines that are -45 degrees along that curve and for m=0, the angle is 0 and y=-e^-x so all the lines on that curve have 0 degree gradient

i'm also not sure wat subbing x=1 would achieve apart from making the y= second column easier to graph

 

[tickboom]

the method he taught us was to let y'=m and find an equation in terms of y, and m and x (y=f(x)+m or something like that). then you can either sub in values of m (or y) (first column from my working), then you tan inverse whatever m becomes (third column) which is the angle of the tangent. the second column is just the curves you have to draw using y and you draw little lines in the direction of the angle you get from m. sorry this explanation is kind ass, my graph for the second one was kinda crap but if u look at the first one, i subbed m=-1 hence the angle is -45 and y=-e^-x-1 so u draw the graph of y=-e^-x-1 and draw little lines that are -45 degrees along that curve and for m=0, the angle is 0 and y=-e^-x so all the lines on that curve have 0 degree gradient

i'm also not sure wat subbing x=1 would achieve apart from making the y= second column easier to graph

Ahhh yes I think I understand the approach now. It kind of speeds up the process since you can do a whole set of y' values in a single hit. So it's about picking a value for y' (i.e. m), plotting the resulting curve, and then overlaying the gradient lines on that curve. Fair enough.