yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Worked example: range of solution curve from slope field | AP Calculus AB | Khan Academy


2m read
·Nov 11, 2024

If the initial condition is (0, 6), what is the range of the solution curve ( Y = F(x) ) for ( x \geq 0 )?

So, we have a slope field here for a differential equation, and we're saying, okay, if we have a solution where the initial condition is (0, 6), so (0, 6) is part of that solution.

Let's see (0, 6). So this is part of the solution, and we want to know the range of the solution curve. You can eyeball a little bit by looking at the slope field.

So, as ( x ), remember ( x ) is going to be greater than or equal to zero, so it's going to include this point right over here. As ( x ) increases, you can tell from the slope, okay, ( y ) is going to decrease, but it's going to keep decreasing at a slower and slower rate.

It looks like it's asymptoting towards the line ( y = 4 ). So, it's going to get really, as ( x ) gets larger and larger, it's going to get infinitely close to ( y = 4 ) but it's not quite going to get there.

So the range, the ( y ) values that this is going to take on, ( y ) is going to be greater than 4. It's not ever going to be equal to 4. So I'll do, it's going to be greater than 4. That's going to be the bottom end of my range, and at the top end of my range, I will be equal to 6.

Six is the largest value that I am going to take on. Another way I could have written this is ( 4 < y \leq 6 ). Either way, this is a way of describing the range, the ( y ) values that the solution will take on for ( x ) being greater than or equal to zero.

If they said for all ( x )'s, well then you might have been able to go back this way and keep going, but they're saying the range of the solution curve for ( x ) is greater than or equal to zero.

So we won't consider those values of ( x ) less than zero. So there you go, the curve would look something like that, and you can see the highest value it takes on is six, and it actually does take on that value because we're including ( x ) equaling zero, and then it keeps going down, approaching 4, getting very, very close to 4 but never quite equaling 4.

More Articles

View All
Sampling distribution of sample proportion part 2 | AP Statistics | Khan Academy
This right over here is a scratch pad on Khan Academy created by Khan Academy user Charlotte Allen. What you see here is a simulation that allows us to keep sampling from our gumball machine and start approximating the sampling distribution of the sample …
Unlocking the Eyes | Explorer
[Music] What boggles my mind about the eye is everything. But I’m really, really excited by the advances in technology made possible by research, not just into the eye, but into how natural selection caused it to be what it is. The next few decades are go…
How Many Photos Have Been Taken?
Hey, Vsauce. Michael here. In 1826, this became the very first photograph ever taken. And in 1992, this became the very first image ever uploaded to the web. But how many photographs have we all taken, altogether, throughout all of history? Well, 1000memo…
Worked example: coefficient in Taylor polynomial | Series | AP Calculus BC | Khan Academy
Given an f of x, and they say, what is the coefficient for the term containing x plus 2 to the 4th power in the Taylor polynomial centered at x equals negative 2? So, like always, take a see if you can take a stab at this video on your own before we work…
Proof: perpendicular radius bisects chord
So we have this circle called circle O based on the point at its center, and we have the segment OD, and we’re told that segment OD is a radius of circle O. Fair enough! We’re also told that segment OD is perpendicular to this chord, to chord AC, or to se…
Post-Truth: Why Facts Don't Matter Anymore
This is the challenge of a YouTuber, which is, you know, pushing the record button and actually filming something. Because you never know: “Are people going to hate it?” Or “Is it good enough?” Have you thought through what you’re going to say. I’ve not t…