Vector Functions: Problems Plus

Transcript

Hi, it's time for another math, easy solution, and today I'm going to go over further into vector functions and now look at the Problems Plus, which is basically advanced math problems at the end of the chapter in my calculus. and now before we go further, as always, if you don't have time to watch the whole video, well, there's timestamps, sections, playlist, PDF and Hive article notes in the video description, and as always, you could play this video at a faster speed and see my tutorial on that, and basically it helps your brain speed up. Alright, so let's dive right in. So Calculus, book reference, note that I mainly follow along the book Calculates Early Transcendental Seventh Edition by James Stewart. There's a link to that. And then I use the following solution manual for this chapter. There's a link to that. And there's, I also used, this is actually for the concept check. I'm going to delete these. All right. So, and also note, in some earlier videos, I used the sixth edition. This is the seventh edition, but when I first started out, I use the sixth edition. Alright, let's jump right in, So Cucklis book chapters. The Hive Notes and Sections Playlists for each video of this chapter are listed below, so they're all here. There's vector functions, that's the full playlist, and now there's the Hive Notes, and then there's the Sections playlist of it, basically break it down into multiple short clips, and so on. There's the one over there. on right now is this Problems Plus. So this is the last part of this vector functions. Chapter, so I went over vector functions and space curves, derivatives and integrals of vector, arc length and curvature, motion and space, velocity, and acceleration, applied project Kepler's laws, then I did the review, the concept check, and true false squids, and now looking at the Problems Plus. And also these, note as always, these links are also on MES links, if you open this up. Alright, so here's the page, MASFM slash slanks, there's link to a whole bunch of stuff I did. There's the math tutorials view. Click here, you could drag, drop, and down. You could also click this to make everything dropped, and so on. So if I go over here, the math tutorials, go to the vector functions. and yeah, this opens all up, and I have the vectors in geometry of space, spherical harmonics, and all this and so on. So this is the one we're dealing with right now. Let's just go back in here. All right, so now let's take a look at the topics to cover. Well, there's going to be seven problems. All right, here, quickly just pause video and added labels for all these problems. So problem one is a projectile fired from origin, problem two, projectile fired down incline plane, and then problem three, ball rolls off table. and then problem four curvature from parametric equations, and then problem five total distance traveled by projectile, problem six cable wound around a spool, then problem seven vector equation on a plane All right, so now what I'm going to do now is I'm going to read all these problems, all seven first, just in case you want to try to solve them yourself, and then afterwards I'll solve, I'll go over the solution for each one. I'm going to read all this again. The timestamps are in the description below, so you can skip straight to the problem that you want, or you can just listen to my voice here, my pretty interesting voice. So problem one... projectile fired from origin, so a projectile is fired from the origin with angle of elevation alpha and initial speed v0, assuming that air resistance is negligible and that the only force acting on the projectile's gravity, we showed in my earlier video, and there's link to the hive notes there, that the position vector of the projectile is r of t equals to v.0, cosine, alpha, times by t, this is going to be the i vector component or the i standard basis component there or basically the x component and that's the horizontal is the vertical or the y so then the y component is v not sine alpha times t minus one half gt squared and then j vector We also showed that the maximum horizontal distance or range of the projectile is achieved when alpha is equal to 45 degrees, and in this case the range, or the horizontal distance, is capital R equals to v0 squared over G. Alright, and now part A of problem 1 states, at what angle should the projectile be fired to achieve maximum height, and what is the maximum height? And then B, fix the initial speed v0, and consider the parabola x squared plus 2 rY minus r squared equals 0, whose graph is shown in the figure below. Alright, so here's the figure, and I'm assuming this parabola, this one is this right here. And you can see that because when you plug in, well, x is 0, this goes to 0, when you plug in y equals to r, y equals to r, then you're going to get a 2R, actually no, instead of r, you do r divided by 2, and then you'll have r squared, and then this cancels equals 0. So then this height's going to be, well, this is going to be r squared. I mean, R over 2 is the height. All right, so that's that part. Anyways, and then, so here's the launching the parabola from the origin, you launch it this way, I mean, launching the projectile from this way, or you go backwards, and so on, but there's a parabola there. So it says, fix the initial velocity and consider this parabola's graph as though. And then we're asked, show that the projectile can hit any target inside or on the boundary, so anything here of the region bounded by the parable and the x-axis. and that it can't hit any target outside this region. So basically it is show that it cannot hit outside of this. That's pretty interesting. So if you have a fixed velocity there, fixed from the initial one, it's the initial, like that, V not. So it's basically, yeah, you can only fire within that dome there. So now Part C says, suppose that the gun is elevated at an angle of inclination alpha in order to aim at a target that is suspended at a height H directly over a point D units down range. and then it says the target is released at the instant the gun is fired. So you have this here, and you drop it right when this is fired. And then it says, show that the projectile always hits the target regardless of the value of v0, provided the projectile does not hit the ground before D. So yeah, this is a pretty interesting problem. So basically we've got to find D, and then we need to have an H here. So when you drop it, Regardless of this V-0 velocity that you launch it, there's going to be a distance D such that, whatever, this H is going to be related to D and V-not, and it's actually independent from V-0. So whatever this D is, as long as this doesn't hit the ground before this projectile goes all over to D. So it will always hit it. There's going to be a spot at height H and D related to each other such that... It doesn't matter what the velocity that you launch the projectile will always hit this target. It's pretty amazing. So again, this is why a lot of this motion in space vector calculus and so on could be, it can be an end is used in a lot of applications, especially military, which is pretty well. So problem two, projectile, fired down, inclined plane, not plan. So a projector is fired from the origin down an inclined plane that makes an angle theta with the horizontal, and then the angle of elevation of the gun. And initial speed of the projectile are alpha and v0, respectively. And it states, find the position vector of the projectile and the parametric equations of the path of the projectile as functions of T and ignore air resistance as usual. So part B says show that the angle of elevation alpha that will maximize the downhill range is the angle halfway between the plane and the vertical. and then as part C, suppose that the projectile is fired up an inclined plane whose angle of elevation is theta, show that in order to maximize the uphill range, the projectile should be fired in the direction halfway between the plane and the vertical. And then we're asked here, so in a paper presented in 1686, Edmund Haley, summarized the laws of gravity in projectile motion and applied them to gunnery. One problem he posed involved firing a projectile to hit a target, a distance, R up an inclined plane, show that the angle at which the projectile should be fired to hit the target, but use the least amount of energy is the same as the angle in Part C. Interesting. That's going to be a quite a big problem here. Show that the angle at which a projectile, yeah, we already did that. Use the fact that the energy needed to fire the projectile is proportional to the square of the initial speed. So minimizing the energy is equivalent to minimizing the initial speed. And so there's a figure right here, we're firing it down and inclined plane. There's the angle. Alpha with the horizontal and the plane. And then this is going to be the alpha as the launch angle. And there's the drop there. So problem three, ball rolls off table. So a ball rolls off a table with a speed of two feet per second, and the table's 3.5 feet high. And part A says, determine the point at which the table, at which the ball hits the floor, and find its speed at the instant of impact. So basically, find the point at which it hits it, and also the speed that you're going to be launching straight at it, V final. Alright, or V impact. And then part B states, find the angle theta between the path of the ball and the vertical line drawn through the point of impact, see figure below. Yeah, so that's this, find that angle right here. So as it falls and bounces, then it's going to bounce this way, so find that angle between the vertical and its bounce path. and there's the height 3.5 feet. So suppose the ball rebounds from the floor at the same angle with which it hits the floor but loses 20% of its speed due to energy absorbed by the ball on impact. Where does the ball strike the floor on the second bounce? So basically, where does it go over here? So at what distance is that? Alright, so now problem 4, curvature from parametric equations, so we're asked to find the curvature of the curve with parametric equations, x equals the integral from 0 to t of sine of 1.5 pi theta squared and then d theta, and then the y component is y equals the integral from 0 to t of cosine 1.5 pi theta squared and then d theta. Now problem 5 is on total distance traveled by projectile. So if a projectile is fired with angle of elevation alpha and initial speed V, then parametric equations for its trajectory are, as per my earlier video, x equals to v cosine alpha, and then t, and then the y component, y equals v sine alpha, t minus 1 1 1⁄2 gt square. So we know that the range, or horizontal distance traveled, is maximized when alpha is equal to 45 degrees. And now we're asked, what value of alpha maximizes the total distance traveled by the projector, not just horizontal, including the vertical? State your answer correct to the nearest degree. Alright, now let's take a look at problem 6. cable wound around a spool, so a cable has radius R and length L and is wound around a spool with radius capital R without overlapping. What is the shortest length along the spool that is covered by the cable? Interesting. And now problem 7, a vector equation on a plane. So it's asked, show that the vector equation below, this vector right here, lies in a plane and find the equation of the plane. So this is a vector equation, show that it's on the plane and find that equation. All right, and now that we read all of them, let's take a now look at the solution to each one, so we'll look at solution to problem 1. So, projectile fired from origin. So a projectile is fired from the origin with angle of elevation alpha and initial speed V-0. Assuming that irresistance is negligible and that the only force acting on the projectile is gravity, G, we showed in my earlier video that the position vector of the projectile is, So r of t equals v0, cosine, alpha, and then t, and then i vector, plus v0, sine alpha t minus one half gt squared, and then all times, and this is all times by the J standard basis vector. We also show that the maximum horizontal distance or range of the projectile is achieved when alpha is equal to 45 degrees, and in this case the ranges r equals to v0 squared divided by G. And again, I covered this in my earlier video. There is linked to the hive notes, sir. And then part A says, at what angle should the projector be fired to achieve maximum height, and what is the maximum height? And then we're asked, fix the initial speed v0, and consider the parable x squared plus 2 r, y, minus r squared equal 0. And that's this pink one right here. And again, as I stated earlier, if you made x equals 0, this vanishes, and then r if you make it equals to r over or over 2, this two is canceled, we get an r squared, r squared minus r is equal to zero. So in other words, this is going to be one half of r. Like that, that's the total height of this dome there. Alright, so if we have that, and then we're asked to show that the projectile can hit any target inside or on the boundary, so it can fire like this inside, or right on it, etc., but it can't hit anything outside of it, or on the boundary of the region bounded by the parabola and the x-axis, and that it can't hit any target outside, so it cannot hit past it. That's because this V-0, it's basically we're setting it equal to this V-0. I'm pretty sure, yeah, so it is. Yeah, that's, because we're going to be fixed initial speed, V-0, and so on. Alright, yeah, and that's basically you're firing initial speed and it's going to go down with gravity. So you cannot shoot past that from the origin. So, and then see, suppose that the gun is elevated at an angle of inclination alpha in order to aim at a target that is suspended at a height H directly over point D units downward. yet downrange, target is released at the instant the gun is fired, show that the projector always hits the target, regardless of the value V-not provided the projector has not hit the ground before D. So you have H, if you drop it, a distance D, so there's going to be a relation between these, that doesn't matter what speed that you're launching this out, it's going to hit this exact spot, assuming that this height H doesn't go through the ground, doesn't hit the ground before D, all right? So going further, so now let's take a look at the solution to part A. Alright, so let's scroll up to A again. So A, we're asked at what angle should the projector be fired to achieve the maximum height, and what is the maximum height, so we gotta find out what the angle is for the maximum height, and the maximum height itself. So I'm gonna copy this, so that is our position vector of the projectile, and we're gonna be launching it from the origin, it's gonna be like that. So we're going to be launching it. I'm going to copy and paste this. So this is, again, from my earlier video, I showed that this is the case of a projectile. So the projectile reaches maximum height when, and then basically, so that's the position vector. And then if you have the x, y, axis, just for completeness, x, y, so we're going to be launching it, and it's going to be like this. It's going to be launching it, it's going to land like here. So the maximum height, here is the maximum height, and that is at, this is going to be at the derivative Y prime is equal to 0. So basically, DY over DT is equal to 0. So this is flat. So that derivative is going to be zero there. And this is going to be, this is V0. Yeah, this is output of V0 like that. That's the speed there, initial speed, and there's the angle alpha. All right, so at y prime, again, that just equals to DY over DT, over DT equals to 0, and that equals to, well, we're going to be taking the derivative of the Y component, this is the Y component, that's the Y equals that, so that's not just going to be, well, D over DT of this whole segment there, V-0, sine alpha. t minus one half gt squared. Alright, so yeah, that's all we have to do, actually, I'm just going to put it like this. Alright, so if we take a derivative, this just equals, so we're going to have a 0 equals 2. Well, derivative of this part here, that's just, that's just, the derivative is just this right here, v, not sine theta, I mean, sine alpha, v, not sine alpha. and then minus, this is a two, so take the two down, so two times two over two, it's can cancel, and then minus the power, so GT, so that's without it. All right, so we have this, now we just solve for T, so we could just move this over to the other side, solve for T, we're going to get a, well, GT equals that, divide by T out, I'm divided by G, so we're going to have equals to, G is equal to V-0, sine alpha over G, so this is, T when at the maximum height. So this is the time of the projectile when it reaches the maximum height. All right, so now we have this. So what we're going to do is, well, we have the Y value is this, that's the Y component. So we have the maximum T value. We just plug that inside. So we have Y, so let's put a thus, thus, let's put it for completeness, So we have y equals to v0 sine alpha, v0, sine alpha, and then this is going to be T, so we're going to be throwing this inside there, minus G, uh, yeah, one-half gt squared. So this is going to be one-half g-t squared. So this is going to be one-half g-t squared. I'm going to throw that in there and in there. So, Yes, so this equals 2, so v0, sine alpha, fix this S, all right, and I just fix that there, so then now we're going to put the T inside, that's going to be, well, V0, sine alpha over G, and then minus one half, and now we're going to have a G, and then that whole setup there. So we're going to have a v0, sine, alpha, and then this is going to be over G, and then squared, like that. All right, now that we have this, so we could start doing some simplifications, or just expand it a lot. So we have v0 here, there's a V0, so that just becomes, well, v, not squared. Next one, sine, sine alpha, that's going to be, well, sine alpha, that's going to be, well, sine alpha squared. and then divided by G. Now the next part is going to be one half, and now this G, there's me a G squared, so then one of the Gs cancels, so we just out left of it divided by G. This one's V0 squared, sine squared, sine squared alpha. So we have v0 squared, sine squared alpha, all right, and then divided by G. And it's exactly the same thing on the other side, but this is a half fit, so a full divided by minus a half, just a half. So half, you could also do 2 over 2, but I'm just going to know that. So what we have is, yeah, what we have is, well, this minus half, it just equals the half. So what we're going to get is Y, this is going to be the maximum height, y equals 2, I'll put this all together, so v, uh, uh, not squared, yeah, so v, not squared, sine, uh, yeah, squared alpha, and then divided by, uh, 2G. So, put the G and put it, 2 like that, so that's what we have. All right here, I just fix up that box, and now this is the maximum height obtained when the projectile is fired with an angle of elevation alpha, and this maximum height is largest when alpha is equal to 90 degrees, and which makes sense as, yeah, as it means the launch angle is straight up vertically, yeah, basically we're just launching it straight in the air, so this is going to be a 90 degrees. So we're launching that projectile, or just the object. Yeah, we're just going to be launching whatever it is, bullet, ball, et cetera. So we're launching it in the air like that, and also, again, we know, recall that the trigonomic sign function. Just recall it looks like this, and this is going to be at the maximum height is going to be 1. This is going to be at sine theta, and this is at piver 2, radiance, here's the origin. Alright, and now this point right here, this is going to…

Comments

More videos on 3Speak

• Mending My Favourite Red Shoes with Micro Crochet by @theworldaroundme
• Problems Plus 6: Cable wound around a spool by @mes
• Dune: Part Three | Final Trailer Reaction! by @sergiomendes
• The Sounds in the Woods by @litguru
• Having fun at the UK Drum Show 2026 by @steevc
• MES Math Q/A 71: What is the Conservation of Angular Momentum? by @mes
• The Secret of Keeping Up by @elianaicgomes
• The light with you by @lesterluiz
• MES Math Q/A 72: What is a Ground Rod and how to find it in your house? by @mes
• Trailer for MES Livestream 141: John Hutchison shows Tom Sky levitation footage he missed by @mes
• Unboxing Sunstreaker Blokees Transformers Galaxy Universe 06 by @triverse
• 🎉Congratulations to the last week's ALIVE CHAT Monday activity ( 21-23 October) winners.🏆 by @rajpootg