Here we go. - Cool; I was gettin' worried there. - Me, too. Okay. They're comin' in now. I've got... 11. I'll wait until noon, because that's when I set this one to start. And, um... then I'll begin. This time, I didn't need password to log in. Good. All right, I've got Manon; I've got... Richard; I've got Causey. Ross. Yeah. I have Jordan... Don't have Jordan. [student] There you go. 15. [student] Oh my gosh; I was panicking for the last 30 minutes. - Oh, well—never panic, because— - [laughs] I promise, I—even if there's nobody here, I can record it... You know, just by myself. - Oh, sweet. Okay. [instructor] I'm kind of waiting on Jordan, because he was having trouble. Is that Luke, Luke, Luke...? Nope. There's Jordan. - [student] Finally. - Heh. Do I have Audrey? [student] So, what happened there? [instructor] Your guess is as good as mine. Okay. Um, so, it is 12:01, so I'm gonna go ahead and start. And hopefully you printed out, or, you know, copied the handouts that have been on Blackboard. Since we lost so much time today, I really think we'll probably just get through 4.1, and then try to save the 2.8 until next time, and maybe I can find a way to combine 2.8 with 4.2. We'll see. But 4.1, I'm glad you're here for me, because this one is a section on applications. It's called related rates. And it... can be a little bit challenging; but it's also really fun. So let me get my screen-sharing going for ya, here. Wow. Okay. That was a very long delay. I don't know if it's just that there's so many people on Zoom, and it's slow? I don't know. Hope this works. So, here's where we are. Today, I'm gonna definitely get through 4.1, and like I said, if I need to try to squeeze 2.8 into Wednesday, I'll do that. All right. And here's our handout for today. So this one is related rates. Try to work on my... focus. And this is the famous balloon problem. If I had a balloon, I would blow one up for you; but I don't have a balloon. And I didn't want to go to the store to get one. Hashtag: coronavirus. So this is the famous balloon problem which we shall just try to imagine. It says, "Suppose I can blow up a balloon at a rate of 3 cubic inches per second." So, [puffing noises] blowing it up; it's getting bigger. This is a unit of volume. So, (dv/dt) equals 3. It's the derivative of volume with respect to t, because it's a rate of change in volume. Both volume and radius are changing as you blow that thing up; the radius is increasing; well, everything is increasing. The radius; the diameter; the circumference; the surface area; the volume; everything. The rate at which the radius is changing, we're gonna call that (dr/dt). Remember, a derivative is a rate of change. Now, at first, the radius grows quickly. So imagine, you know, when you've got the balloon, it's about this long. And you put in that first couple of [puff-puff-puffff]. And then it grows fast. Like, all of a sudden, it's a round shape. But as the balloon gets larger, the radius grows more slowly. So imagine you got a big balloon here, and I put in a couple more puffs of air. [puff-puff] You hardly notice any change in the shape of the balloon. So, as the balloon gets larger, the radius grows more slowly, even though the rate at which the volume is changing remains constant. So, it's always 3 cubic inches per second. It's just that that's more noticeable when the balloon is this big; less noticeable when the balloon is this big. Volume and radius are related by the formula V = (4/3) π r³. That's just the formula for volume of a sphere. And A says, "At what rate is the radius increasing with respect to time, when the radius is 2 inches?" Okay. So here's my little handout here. "At what rate is the radius increasing with respect to time when the radius is 2 inches?' Okay. So what we need to find, is this. We need (dr/dt). Because this is the rate at which the radius is increasing. We need (dr/dt). Okay. So. I'm gonna take the formula that I have, which is V = (4/3) π r³; and we're gonna differentiate that with respect to t. Not r, but t. So, this is an implicit differentiation. I want it with respect to t. So I'm gonna find (dv/dt), and that will be the derivative with respect to t of the right-hand side, which is (4/3) π r³. So what I want you to notice here is that r is the wrong letter. t is the right letter; r is the wrong letter. So, with this implicit differentiation, when I differentiate that right-hand side, I've got to follow it with an r prime, 'cause it's the wrong letter. Now on the left-hand side, you know, we just leave that, (dv/dt), and here we go on the right-hand side differentiating. So we'll do 3 times (4/3) is 4. And that's π r². Now, the chain rule says, multiply by the derivative of the inside. And that inside is the r. So since it was the wrong letter, this is what we'd follow it with; r prime. Now here's the deal. So, instead of using r prime, which is a perfectly fine symbol; but instead of using that one— - Professor? - Uh-huh. [student] Um, how did you get the r²? The derivative function? Minus one? - [instructor] So, the 3 ti—mhmm. 3 times (4/3) is 4, π, and then r ⁿ - 1. So r². [student] Okay. Gotcha. [instructor] Instead of using this r prime, we're gonna use this notation for the derivative. The only reason is because with (dr/dt), it really makes it noticeable that we're talking about a rate of change of radius with respect to time here. And with r prime, that's not as obvious what the intent is. So we'll follow this with (dr/dt). So that's implicit differentiation involving a chain rule. Now, what we need is (dr/dt). Well, we've got an equation here; we're just going to isolate (dr/dt) in this equation. So if I isolate this equation... Let's see. How can I do this? I can... ummm... hmmm. I can first substitute in place of (dv/dt) the 3, because I was told that (dv/dt) equals 3. So now, 3 equals 4π r² (dr/dt). And I say we divide everything by the 4πr², and then we've got (dr/dt) all alone! So we'll say (dr/dt) = (3/4 π r²). Hmmmm. What was the r in this problem? What was the r. See right here? - [student] Two? - When the radius is 2 inches. So now we'll just substitute in the 2; we'll have a number. 3 over 4 times π times r². So, 2 squared is another 4. This point, I'm gonna go to my calculator; see what that is. And we'll clear. Okay. So I want 3 divided by... And I'm gonna put that denominator in parentheses, so the calculator understands I'm dividing by all of this stuff. So that would be a 16π down there. And I get... (dr/dt) is approximately 0.0597. I just chose to round that to four decimal places. I need a unit, though. So this was a rate of change of the radius with respect to time. So the unit for radius was... -Inches per second. [Prof] -Inches. And for time, it was seconds. So, I'll circle this by itself. So, (dr/dt) equals 0.0597 inches per second. The rate of change of the radius with respect to time. Okay, so that was the first example. And that's the way all of these are gonna work. You're gonna have a formula. Sometimes you're given the formula, and sometimes you have to figure the formula out; that's comin' up. And then once you get that formula, you're gonna be doing an, um— implicit differentiation with respect to time on all of these. With respect to time. And then, solving for an unknown. Okay, so let's look at part B. Still about the same problem. It s-says—"It s-ssays." It says, "Suppose I increase my effort when r equals 2 inches, and begin to blow air into the balloon at a faster rate. A rate of 4 cubic inches per second. Well, how fast is the radius changing now?" Okay. So in this case, the (dv/dt) is now equal to 4. So, I had (dv/dt) from above, so I can just write that down again. Instead of reinventing the wheel. And it was 4 times π times r² times (dr/dt). And now I'm going to substitute the 4 in place of the (dv/dt). So we've got that faster rate of change of volume. So 4 equals 4 times π. That radius is still 2, as it says; when r equals 2 inches. So that's a 2² (dr/dt), and we just need to solve for (dr/dt) again. So, to do that, I'll divide both sides by the 4π times 2²? 2² is 4. 4 times 4 is 16. So, there's that 16π again. So I'll bring out my calculator. And... turn it on. Ooh, I got a bad glare on that; sorry. Okay, so this one'll be 4 divided by. And then in parentheses, (16π). That would also reduce to ¼π. Dun't matter. At any rate, this (dr/dt) is approximately 0.0796. And again, that would be inches per second. So you notice, when you compare the two, that your radius here is changing at a faster rate than it did here. Obviously; because you were increasing your effort, and the rate of change in your volume was higher. Okay. So, um. Hold on one second, y'all. I've got to mute you for a second because my dog needs to go out. [exaggerated whisper] I'm so sorry. [Student] Hey, I just came back. My house just had a rolling blackout. [Instructor] A rolling blackout? [Student] Yeah, I've been gone for, like, five minutes. [Chuckles] [Instructor] Whoaaa. So, where do you live? [Student] Uh, right next to the Crow Bar. On South Congress. [Instructor] Oh, maaaan... Okay. That's all we need, are blackouts. [Student] Yep. [Instructor] How fun. Sorry. So, now, for part C it says, "At what rate is the volume increasing with respect to the radius, when the radius is 1 inch or 3 inches?" "At what rate is the volume increasing." Okay. So, I'm gonna write a little note here to be careful with this one. You've got to read it carefully. "At what rate is the volume." So, now, underline that. —"increasing with respect to the radius." So, what we want here, is (dv/dr). This is the rate of change of the volume with respect to the radius. We need (dv/dr). So, we're gonna start with the formula we were given again. And that was V = (4/3) π r³. That was our formula for volume of that sphere. We need (dv/dr). So, if you'll notice. When we get (dv/dr), uh... r is the right letter. So this one's not gonna require an implicit differentiation. This one's pretty straightforward. (dv/dr). 3 times (4/3) is 4. Times π. Times r², and there you have it. That's the rate of change of volume. Don't need to do— following it by a (dr/dt), because r was the correct letter in the first place. So, now we just need to evaluate this when r is one inch, and when r is 3 inches. So, let's see if I can get a little more room here. There we go. So, at r = 1, (dv/dr) is equal to 4π. Okay? If I'm looking for my units here, this is a rate of change of volume with respect to the radius. The units of volume were inches cubed. The units for the radius were inches. Now. I don't want you to reduce that to inches squared. [Chuckles] Don't do that. So, this is a rate of change of volume. So what it says is that when the radius is 1 inch, that your volume is changing at a rate of 4π cubic inches for every 1-inch change in radius. So, leave this be; it means something. It's describing how the volume is changing with respect to how the radius is changing. Does that make sense to y'all? [Student] Yeah. All right. Let's try r = 3. So, (dv/dr) in this case would be 4π times 3². 3² is 9. 9 times 4. This'll be 36π. And I'm gonna leave it like that. And my units, again, are inches cubed, per inch. Sounds better if I say "cubic inches per inch," I think. Okay. So, there's example one. And that was— [Student] Is there a place that we can get our... or find our graded tests? [Student] -Like, you have—okay. [Instructor] -Yes. Yeah. [Instructor] So, um, when you go to your gradebook, and go down to, like, the row that the test is on... There should be a place where you can see my feedback, and that's where I uploaded your graded test. Can anybody else jump in here; if you found it, can you explain that better than I just did? [Student #2] Just next to the grade, there's like, a little cloud thing in blue, which has the comment. [Student #1] -The little speech bubble. -Yeah. And you can find there. [Instructor] -Great. Thank you. [Student #1] -Yes, uh, thanks. [Instructor] Sure. All right; so now, related rates procedure. So we went through that first example pretty slowly. And so now I'm gonna show you; this is just the general way we're gonna handle all of these. So the first thing is, we're gonna draw a picture if we can. Uh, I didn't really need to draw a picture of the balloon problem. I could have drawn a sphere, I guess, if I wanted, but. For some of these, you need a diagram. You're gonna need a picture, and you'll need to label things. And that's the second point, is "label and assign variables." Okay. The third thing is, write down what you know, and what you need to know. So whatever the question's asking, that's what you need to know. And then what you know is usually gonna be a formula associated with the shape that you're drawing. Then you wanna find an equation or a formula that relates the variables. So, oftentimes, this is gonna be a formula for volume, or for area. It could be the Pythagorean theorem. Just depends on the picture that we end up drawing. And then we're gonna use implicit differentiation to differentiate with respect to time. And then the last thing is just substitute in your known values, and then solve for the unknown values. So, we're gonna follow that pattern on all of the rest of the problems. 'Kay, next up is the famous sliding-ladder problem. And I wish we were in a classroom, because in a classroom, I bring in my meter stick, and pretend it's a ladder, and then I prop it up against the wall, and I pull it out slowly from the bottom, and watch it slam down on the floor. So, you know, the ladder is sliding down the wall. And when you see it in class, I just think this makes a little more sense, but. Darn it! So this is the famous sliding-ladder problem. Says, "A 10-foot ladder rests against a wall." So I'm just imagining a ladder propped up against a wall. If the bottom of the ladder slides away from the wall at a rate of 1 foot per second— so that's steady, constant pulling the bottom of that ladder away from that wall— —"how fast is the top of the ladder sliding down the wall, when the bottom of the ladder is 6 feet from the wall?" So the first thing that we talk about is, when that ladder is propped up against the wall, and you're pulling the bottom of the ladder, pulling it out slowly, the top of that ladder is also falling down. But would it fall at the exact same rate at which you're pulling the bottom of the ladder away? I mean, it's all one ladder. So, this is my ladder. And this is my wall. And I'm pulling the bottom away. It seems like whatever rate I'm pulling it away, that the top should slide down at that same rate. But if you think about it... I mean, really think about it. If there really were a ladder there, and you had a string tied around the bottom, and you're pulling it out, it's gonna slide down the wall slowly at first, but what happens when it gets close to the floor? [Student] Speeds up. [Instructor] Yeah man, that thing is gonna smack the floor so hard, it's gonna damage the floor! Unless it's on a carpet. So, what really happens is, even though we're pulling the bottom out at a constant rate, the rate at which the top is sliding down is increasing. Craziest thing about physics. So let's try to draw this. Says, "Draw a picture if you can." That's the first bullet. So when I draw my picture, I'm just gonna draw a wall. And then, here's my ladder propped against the wall. This is the floor. Okay. So, that ladder. One thing I know about it is that it's 10 feet. So I can label that "10." And of course, you notice, I just drew a triangle. So, the hypotenuse of that triangle is definitely 10. Now, I can think of this as being in a coordinate system. And when I think of it that way, then the base of this triangle is x, and so what x really represents is the distance from the wall. And then y can be... Oh. The height of the wall where the ladder meets. So, x is the distance from the base, and then y is the height of the ladder propped against the wall. So, I've labeled it with what I know. And I've assigned variables to what I don't know. Um. There is something else I knew. It says, "If the bottom of the ladder slides away from the wall at a rate of 1 foot per second." So, here's the bottom of the ladder. It's being pulled away from the wall at a rate of 1 foot per second. That's one of our rates. And x is the thing that's changing there. So the distance from the base is changing. That's actually (dx/dt). So I'm gonna write down what I know; that (dx/dt) equals 1. So I really know two things about this: I know the length of the ladder. And I know (dx/dt) is 1. So I wrote down what I know and what I need to know— Oh no, I didn't. What do I need to know? What's it asking for? How fast is the top of the ladder sliding down the wall. Well, as the top of that ladder slides down— [Student] (dr/dt). [Instructor] True. It's y that's changing. So what I want to know, what I need to know, is (dy/dt). Which is why it's called "related rates." You're gonna have multiple rates in the same problem. (dx/dt) is given to us as 1. We wanna find (dy/dt). So now, the next thing says, "Find an equation or a formula that relates all the variables." So, back to our draw-ring. We have a 10, an x, and a y. What's a formula you know that relates these three numbers? [Student] Pythagoras theorem. [Instructor] Thank you, Pythagoras. Pythagoras makes our lives easy. "Py-tha-gor...as." So thank you, Pythagoras. And here's your theorem. It says that x² plus y² is equal to 10². Okay. So that's our equation. And that's the thing we need to differentiate. So once we have that equation, we use implicit differentiation to differentiate with respect to time. So, that means on the left-hand side, we want the derivative with respect to t, of x² plus y². On the right-hand side, the derivative with respect to t of 100. So now on the left, we're gonna split it up. We want the derivative of that sum. So I'm gonna write it as: the derivative with respect to t of x² plus the derivative with respect to t of y², equals. And then, what is the derivative with respect to t of 100? What is that? [Student #1] -Zero. [Student #2] -Zero. It's the constant, so we're gonna get a zero on the right-hand side. Now, on the ladder problems, when you know the length of the ladder, you'll have the constant on that side of the Pythagorean theorem, and that derivative is always going to be zero. Now, on the left, we need to differentiate. So getting the derivative with respect to t of x²? x is the wrong letter. So, we'll do our 2x, all right, but then we've got to follow it by... (dx/dt). And that's the chain rule. Since x is the wrong letter, 2 times x is the derivative of the outside. This is the derivative of the inside. Now for the y² term? Same thing. y is the wrong letter. So we'll do 2y, followed by (dy/dt) is equal to 0. Great. So now that we've differentiated, we're gonna sub in the things that we know. So, what do we know here? Um, let's see. It says, "10-foot ladder"... A rate of 1 foot per second; that was (dx/dt). And we also are stopping this; we're looking at this when the bottom of the ladder is 6 feet from the wall? Okay. So it's 6 feet from the wall right now. That's x. So I'll say 2 times 6 times (dx/dt), which is... what, now? [Students] -One. [Instructor] -One. Yeah. And then plus 2 times yyyy. [Student] You can find that using Pythagoras' theorem. [Instructor] Exactly right. So, 2 times y, and then times (dy/dt) equals 0. We don't know what y is, but we can find it. So I'm gonna go back over here, and rewrite my Pythagorean theorem, which is x² + y² = 10². I know what x is; x is 6. So this is— [Student] -Professor? [Instructor] -Yes. [Student] So the 2 times 6 times 1. Is the 1 a derivative of the x? [Ins.] Yes. That was the rate of change of x with respect to time. It was a 1. [Student] Okay. So, down here, 6² + y² = 100. That's y² = 100-36. y² is equal to... uh... Make that 64. And y must be 8. Negative-8 wouldn't make sense, so we're going with the positive square root of 8. So now, I can plug in that unknown. And this is somethin' that commonly happens. So, once you've got your equation, you do your implicit differentiation; you fill in the stuff you know. A lot of times, there's another unknown variable that you've gotta go find, but you will be given the information to find it, and it's usually from your formula. So you'll plug something in to find something else; then you can sub it all in, and finally, just be left with that one unknown, which is (dy/dt), and that's what we want. Well, 2 times 6 is 12. 12 plus 2 times 8 is 16; times (dy/dt) equals 0. Let's try to isolate (dy/dt), so I have 16 (dy/dt) = -12; and the last step is just dividing both sides by 16. (dy/dt) is equal to -12 over 16, and that is negative... -It's like, three-fourths? [Student] - Yep. So negative 0.75. And then our units for this, since this is a change in y with respect to t, is gonna be feet per second. The units of y were feet; the units of time were seconds, so in (dy/dt), units are feet per second. Man, I've almost run out of— [Student] Would you have a preference on fraction or decimal? [Instructor] Oh no, I don't. Nah. To me, on these kind of problems, though, the decimals... I guess I like 'em better because I can imagine that better. Like, I have an idea of -0.75 feet per second, but -3/4— Well, I guess it wouldn't matter. I don't care. Whatever makes you happy. Okay. Now, that was the famous ladder problem, and— [chuckles] because, in every calculus book since the history of calculus, there has been a ladder problem. And you will have more ladder problems in your homework. And you will most likely have a ladder problem on your next test. [Whispers] It's famous. Okay, the last question says, "How fast is the top moving down when the ladder is 9 feet from the wall." How about 9.9 feet. How about 9.99999999999 feet? So in other words: the ladder's only 10 feet. So, when you're pulling it out. When it's 9 feet— I mean, most of the ladder is down. It only has another foot to fall; so we're looking at the speed at which it's falling at that point. Okay. So, let's go back to when x equals 9. Because we need to figure out what y is at that point. 'Cause, you know, if I'm drawing a picture of it... It now looks like that, right? So it's almost all the way on the ground. So when x is 9, let's figure out what y is. So using our Pythagorean theorem, x² + y² = 10². That is, 9² is 81. Plus y² equals 100. So then y² is 100 minus 81, which would be 19, and y will be the square root of 19. So then, we'll go back to our "(dy/dt) equals." And our (dy/dt) was... Oh, man. Do I have to reinvent that wheel? Shoot. I do. So, (dy/dt) would equal. I'm gonna go back to this step so I can isolate (dy/dt) before I've substituted in numbers for x and for y. (dy/dt) would be... Will be -2x times (dx/dt), and then that would be divided by 2y. Think I got it. (dy/dt) would be -2x(dx/dt) when you subtract this from both sides, and then to isolate the (dy/dt), you're dividing both sides by 2y. So. It looks ugly, but this is what it looks like. Now we'll substitute in our new information. So, our new x is a 9. So this is -2 times 9. (dx/dt) is still 1. And then 2 times y would be 2 times the square root of 19. Now, I plugged all that into my calculator already, and that was approximately -2.06 feet per second. So it sped up. Remember when it was 6 feet away, the speed at which the top was falling was -0.75 feet per second. Now, it sped up to -2.06 feet per second. [Student] -Professor? [Instructor] -Yes, go ahead. [Student] The 9.9, did you round it up? [Student] The 10? The 10²? [Student] Know when it says x + y² = the 10². Is it from the— [Instructor] Yes. [Student] But it's a question, or you just rounded it up? [Instructor] So this is still going back to my Pythagorean theorem. [Student] Oh, okay. [Instructor] I still have a hypotenuse of 10 there; the base is 9. And we were looking for y. -Gotcha. -Yeah. Turned out to be... the square root of 19. That fits in there. So then I would do it again for 9.9, and then for 9.9999999... I don't have room, so I'm gonna talk you through it. So, when you get to 9.9. That ladder's almost all the way down. When you go through and calculate the rate of change of y with respect to t, when x is 9.9, your rate is then... Uh, -7 feet per second. When you go to 9.9999999, it's approaching infinity. It is negative, but so large, it's incredible. So, as it's slamming the floor, the rate at which it's slamming the floor? That rate is approaching infinity. Can't make this stuff up. It's really true. That's why it damages the floor. It's pretty darn fast. All right, and that is another famous sliding-ladder problem. We'll take that one away. And now I'm lookin' at number 10 from the exercises. This one's comin' up next. Probably better also check what time it is. 12:34? We're good. So exercise 10 says: "A particle moves along the curve; y = √(1+x³)." "As it reaches the 0.23, the y coordinate is increasing at a rate of 4 centimeters per second." That's (dy/dt). "How fast is the x coordinate of the point changing at that instant?" Okay. So here, the graph that we draw is the graph of the function. So the curve is y = √(1+x³). That's the graph we want to draw. So I'm gonna draw my coordinate system here. Like so. And I graphed this on a graphing calculator earlier to see what it looks like; and you don't have to be exactly right, but it looks something like that. And then this point, I'm gonna label this point right here at (2,3), because the particle is moving along, and at some point, it's gonna reach that point. Particle's moving along the curve. As it reaches the point (2,3), the y coordinate is increasing at a rate of 4 centimeters per second. So we know that (dy/dt) equals 4. The question is, how fast is the x coordinate of the point changing at that instant? So, what we want is (dx/dt). We know (dy/dt); we want (dx/dt). So if I look at, you know, my little bullets, and see where I'm at. I drew a picture. It says, "Label and assign variables." Well, I guess I kind of did. I've got the point labeled, and I wrote down what (dy/dt) is, and... I wrote down what I don't know, which is (dx/dt). So then I find an equation or formula that relates all of these variables. Well, that equation or formula is the y = √(1+x³). That's relating x and y. We wanna use implicit differentiation now to differentiate with respect to time. And then, we'll substitute in what we know; solve for what we don't know. So now I need to find the derivative with respect to t. So I want derivative with respect to t of the left-hand side. I want derivative with respect to t of the right-hand side, which I'm going to rewrite as (1+x³) to the ½ power. Just makes it easier for me to differentiate. So now on the left, it is just (dy/dt). And then on the right, derivative of that (1+x³) to the ½. So bring my ½ down in front. (1+x³) to the -½ power. Now multiply by the derivative of the inside. Okay, now. Your inside is this (1+x³). Derivative of (1+x³) is... 3x². But now, chain rule says, "Do it again"; it's a double chain. Now we need to multiply by the derivative of x with respect to t. Because x was the wrong letter. t's the right letter; x is the wrong letter, so I've gotta follow it with that (dx/dt). Now I've differentiated implicitly; now it's time to sub in what I know. So, I do know that (dy/dt) is 4. That's 4 equals ½ times 1 plus... What's x at this point? [Student] -2. [Instructor] -2. So that's a 2³, to the -½. And that's times 3 times a 2². And then that's times (dx/dt). (dx/dt) is the unknown; that's what I need to solve for. All right, so this is 4 equals. Ummm, 2³ is 8. 8 plus 1 is 9. 9 to the -½, so it's like 1 over the square root of 9, is... 3, I think. So this would be 1 over, 2 times... 8+1 is 9; square root of that is 3. So that's 1 over 6. And then 3 times 2²; that's 4 times 3; that's 12. (dx/dt). So, this is 4 equals. 2 goes into 12 six times; 6 over 3— that's just a 2 times (dx/dt). I think I'm ready to isolate my (dx/dt) by dividing both sides by 2. And (dx/dt) is 4 over 2, which is 2. And the rate is in centimeters per second. Okay. So sometimes, I guess, solving the equation after you substitute in your known values can get a little tricky, but you know; just take it one step at a time, and you'll get there. So let me know how that one went. [Student] Can you just go over what happened to, uh, 12? [Instructor] Yeah, sure. So, the ½, times the 12? Is 6. So I just canceled the 2 with the 12, leaving me a 6 on top; but 6 over 3 is 2. Good? [Student] Yeah. All right. So, you guys are so quiet. I don't—I don't like that about Zoom; it's different than being in a classroom; in a classroom, you know... We can see each other's eyeballs, and you can just ask a question; or sometimes I'll look at you, and I know you have a question, and I'll say "What's up." Um, jump in there; really. Stop me any time you wanna stop me. Don't be shy or embarrassed about it. Stop me, and ask your question. Because the most important thing is that you guys continue to learn. Exercise 4 says, "The length of a rectangle is increasing at a rate of 8 centimeters per second." Got a rectangle. "And its width is increasing at a rate of 3 centimeters per second." "When the length is 20, and the width is 10, how fast is the area of the rectangle increasing?" Okay. So the first thing we're gonna do? Draw a picture. So, I've got a rectangle here. Here we go. And I'm gonna label this thing. So it says, "The length of the rectangle is increasing at a rate of 8 centimeters per second; width is increasing at a rate of 3 centimeters per second." "When the length is 20, and the width is 10, how fast is the area of the rectangle increasing." So, like, right now, the area is 20 times 10, or 200, but. We're gonna be increasing the length and the width, and looking at how fast that area is changing. So I'm gonna write down the things that I know. I've given a lot in this problem. It says the length is increasing at a rate of 8 centimeters per second. So, that would be the derivative of l, with respect to time. That is 8. It says, the width is increasing at a rate of 3 centimeters, so. dw, the change in width, with respect to time. That one is 3. We know that we're kind of stopping this when l is 20, and when w is 10. So, there are four things that I know. What do I not know? What do I need. I need, or want to know... how fast the area— [Student] -(da/dt)? [Instructor] Yeah. How fast the area. Derivative of area with respect to time. I need the rate of change of the area with respect to time. So, if I'm looking for the rate of change of area, then I want to use the area formula here. Area of a rectangle? Length times width. So there's my formula relating all of my variables; it's time to differentiate implicitly. So now we'll get the derivative with respect to t of the left-hand side. And the derivative with respect to t of the right-hand side. Now on the left, there's your (da/dt). This is the very thing we're lookin' for. So then on the right, we need to get the derivative of length times width. So I said it: length times width. This is a...? [Student] -Product rule. [Instructor] -Product rule. So we want the first function, l. Times the derivative of the second function. Okay now, remember: t's the right letter. Everything else is the wrong letter. So when I do first times the derivative of the second, I don't know what the derivative of the second is, so I have to write (dw/dt). So, l times (dw/dt). And then plus the second, which is w, times the derivative of the first, which has to be (dl /dt). So now I'm gonna go back up here, where I was given these four pieces of information. I'm gonna substitute them in. (da/dt) = l, at this moment in time, is 20. (dw/dt) is 3. Plus w at this moment is 10. And (dl/dt) is 8. Okay, this one's gonna be easy. Don't have to isolate anything; just multiply and add. So, that's gonna be 60 plus 80, and 80 plus 60 would be 140. Now, units. What are the units for the area? [Student] Centimeters squared per second? [Instructor] Mhm. So, units for area are centimeters squared. The units for time are second. So it says that, at this point in time, when our rectangle is this big, and it's increasing? That the rate of change in the area is 140 square centimeters for every second. Okay. Almost done. That was an easy one. Maybe we should've done that one first. Okay. The last one on this handout, I believe. Exercise 32. Exercise 32. Oh, but this is a good one. So, exercise 32 says, "Two sides of a triangle have lengths 12 meters and 15 meters." Two sides of a triangle. It didn't say a right triangle. Just said "a triangle." "The angle between them is increasing at a rate of 2 degrees per minute." "How fast is the length of the third side increasing when the angle between the sides of fixed length is 60?" [Exaggerated shriek] If it were a right triangle, this would be so much easier to draw! But it didn't say that; and it's not; and it's changing; so, man! Let me just go for it. So I'm gonna draw a triangle. Maybe something like—I'm gonna make this pretty big. [Chuckle] Something like that. And your triangle doesn't have to look exactly like mine, but. I'll be danged if that doesn't look like a right triangle. That looks like a right angle right there. I just couldn't help myself. It's not. Not a right triangle. So I'm gonna label my sides. I'm gonna call that one 12, and this one 15, because it looks longer. And then there's an angle between them, and that angle between them we'll call θ. So, if that's θ, and here are the two sides. What's happening is, that is opening up. So as that opens up, we're looking to see how that third side is changing. It's obviously growing; it's getting longer. We're looking for that. "How fast is the length of the third side increasing when the angle between the sides of fixed length is 60 degrees." So guess let's start writing down the things that we know here. So, we know that two sides of the triangle are 12 and 15... OK, got that. The angle between them is increasing at a rate. Ah. This is a rate that we know. And it's the rate of change of that angle with respect to time. So, we know (dθ/dt). (dθ/dt). And that is a rate of 2 degrees per minute. So (dθ/dt) is 2. "How fast is the length of the third side increasing when the angle between the sides of fixed length is 60? So, it's telling us that we're kind of stopping this, looking at when that angle is 60 degrees right then, how fast is the third side changing. Well, we need to give a name to that third side. Hmm, I don't—what do you wanna call that third side? Anybody? Any variable? [Student] -x. [Instructor] -Why not? So we'll call that third side x. Works for me. Now we need a formula. We need a formula that relates what's going on here. So, look at your picture. Your knowns; your unknowns. Does a formula come to mind— and it cannot be Pythagoras, because this is not a right triangle. [Student] This is the double-angle thing? I mean... It's sine over hypotenuse equals sine over hypotenuse? [Instructor] Not that one. [Student #2] Is this a sine-angle-sine problem? Or a side-angle-side problem? [Student #3] -Is it the law of sines? [Instructor] -Yes. Yes, it's SAS. So, think trigonometry. [Student #3] It's not the law of sines or anything, is it? [Instructor] Keep thinkin'. You're close. [Student] Law of cosine? [Instructor] Yeah, that might help if I write that in there. So when you know two sides and the included angle, that's a law of cosines. And we know two sides. And we know the included angle at this moment is 60°. So definitely a law of cosines. Yayyy, I love the law of cosines! That part of trig was so fun; solving for the triangles using the law of sines and the law of cosines. I loved doing those problems. Remember the vector problems? They were great. Okay, now, what does the law of cosines say? Well, the law of cosines says this: That your side opposite, which we're calling x. We're gonna square it. x² is equal to. And it's the sum of the squares of the other two sides; so it starts out kind of looking like the Pythagorean theorem. But then it's minus 2 times a times b times the cosine of θ. This is the law of cosines. So that's our formula relating everything. Ummmm, what do we next? Implicit differentiation. So, we want derivative with respect to t. Of the left-hand side, which is x². And the derivative with respect to t of the right-hand side, which is (a² + b² − 2abcosθ). Well, I say before we get this derivative, maybe we substitute in what we know, with the sides 12 and 15? Ummm, I can do that. Or not; I don't have to. I can live with it like this. Y'all, give me a preference. Do you want me to substitute in those numbers now, or get the derivative first? If I get the derivative first, you know, these will just be zeros, because there are only constants. Weigh in with your preference here. [Student] Put the numbers? [Instructor] -Put the numbers in? [Student] -Yes. So, derivative with respect to t of x² equals the derivative with respect to t; and we'll put those numbers in. So, the a? I guess I'll just call a the base; 15. That would be 15². Plus the other side squared; so that's 12². Minus 2 times 15 times 12. Times the cosine of θ. And then we can clean that up a bit. This is the derivative with respect to t of x². Notice I'm not taking the derivative yet; I'm just cleaning this up a bit. Equals derivative with respect to t of. If I do 15² + 12². Go into my calculator here. 15² plus 12². Okay. That is 369. So that would be 369 minus. Now, the 2 times 15 times 12? That is 360. Sitting in front of the cosθ. Okay. Now, let's differentiate. Let's do it now. So then on the left-hand side, remember that x is the wrong letter. So when I get the derivative of x², it's 2x, but follow it by...? (dx/dt). On the right-hand side. The derivative of 369 is just 0, so we won't worry about that. So now let's look at the derivative of -360cosθ. Well, that constant in front just hangs out. What's the derivative of cosine? [Student] Negative-sine. [Instructor] So since it's negative-sine, then we can do... That. So 360sinθ. Now, θ is the wrong variable, so what do we follow this by? (dθ/dt). And that's the chain rule. So if you have a cosθ, derivative is -sinθ (dθ/dt). That's the derivative of the inside. Okay, great. So now we're ready to substitute in things that we know; and we're solving for... What are we solving for? I didn't write down what we needed to know. We need... And it says, "How fast is the length of the third side increasing?" We need (dx/dt), the rate of change of x with respect to t. Okay, got it. So I'm solving for (dx/dt). Well, then on the left-hand side, I'll have 2 times x. Ummm. How are we gonna find x here? How we gonna find x. I'm bringing my picture right down in front of your face, there. If I'm looking for this side... [Student] Is it a [inaudible] equation? [Instructor] We're gonna plug it into the law of cosines to find out what this third side would be when the two sides are 12 and 15, and, at this moment, that angle is 60°. So we're going to go back to the law of cosines just to determine this unknown. Remember, we had to do this before on one of the ladder problems. Okay. So then, using a law of cosines, it would say... I'll try to do this over here on the side. It would say that x² is equal to a² + b². So that's 15² + 12² again. Minus 15 times 12 times the 2; times the cosine of 60°. So our x² equals. That 15² +12² ? That was the 369. And then 15 times 12 times 2, that was the -360cos60°. So x² is 369 minus 360 times. And the cosine of 60° is one that we know. [Student] -One-half? [Instructor] -Is what? [Student] -One-half, I think? [Instructor] -One over two. One-half is right. So this is ½. x² is 369 minus... I guess that'd be 180? And then 369 minus 180 is— [goofy voice] I unno. [Instructor] -I got— [Student] -189. [Instructor] 189. So x would be the square root of that. Which is not real pretty; it's 13.7-ish. So I'm just gonna leave it at 13.7. More decimal places would be better; but I kinda messed myself up by not giving myself very much room to write any number in here at all, sooo, I'm gonna have to just round it off. So 13.7. And that equals the 360 times the sine of θ. Oh, but the sine of θ is the sine of... 60°, times (dθ/dt), which was 2. Hold on. I can fix this. My Calc 2 student told me on Wednesday, "So why don't you just use a pencil?" "Then if you mess up, it's no big deal!" Well. I've got a... Wite-Out tape here. So let's fix all that. Like, really? You're not gonna work? [Cries] Why is my life so hard?! All right. So I'll just rewrite it. 2 times 13.7, times (dx/dt). Equals 360 times the sine of 60°; times (dθ/dt), which was 2. There we go. Now, 2 times 13.7—aw, heck. You know what I'm gonna do? Say (dx/dt) is equal to. 2 times 360 would be... 720. Sine of 60°. I know that one, too. That would be... [Student] -Square root of 3 over 2. [Instructor] -Square root of 3 over 2. And then, let's divide that by 2 times 13.7. All right, I'm going to my calculator to figure this one out. 720. Times the square root of 3. Divided by 2. And then that divided by 2 times 13.7. Hey y'all; if I didn't fat-finger this, I got approximately 22.7. And now I need a unit for that. This was a rate of change of x with respect to time. And x was measured in meters, and time was measured in minutes. So, 22.7 meters per minute. Okay. [Student] Umm. [Instructor] Yes? [Student] -Oh, nothing; I just said "Wow." [Instructor] -Oh, okay. Wow! So, yeah. This was a pretty challenging section; but also doable. So if you look at those bullets, draw the picture; label; write down what you know and what you don't know; what you need to know. Find a formula that relates everything. If you try to go through that step-by-step, I think you'll be just fine. And I tried to pick problems— most of them— are like the ones that we did... in class today? On Zoom today? And so, hopefully you'll have an example for almost everything in the homework. But definitely stop by during office hours; um... I've got that all figured out now, and I'm in Blackboard from 10 to 11; so, you know, before your class for an hour, stop in. Or this afternoon from 3:45 to 4:45. I'm on Blackboard again, and Collaborate. So, there's a link in your Blackboard. Just go to it; and I'll be there, and I can help you with homework problems, so. Especially some of you who were used to coming by when we were still at Hays. Come on by! I wanna still be able to help you, even though it's not... quite as effective this way? It's still better than nothing. Umm. And then the tutoring labs, the learning labs, have gone online using Brainfuse. They were supposed to send out an email about that. I never got one. I'm hoping that the students did. Somebody let me know if you've got anything about that? [Student #1] -I didn't. [Instructor] -Ahh. [Student #2] I think I remember seeing something about [inaudible]... [Student #3] So, I've been talking to a tutor. Um. And we've been meeting on Zoom. But the way that she sees it is that it's almost like a ticketing system, kinda like how Highland works now; where you send in a ticket for a singular question, and then they can reach out to you and help? [Instructor] Do you do it from the website? [Student #3] Yeah, I believe so. There was a post on it on the front page; I don't know if it's still there. [Instructor] Oh, okay. So, maybe just go to austincc.edu. And if there's not anything on the front page, maybe do a search for "learning lab," and then hopefully their page will come up with information. I did not recieve anything about it. I just—I heard from someone who works there that they were gonna do Brainfuse. So yeah, the ticketing system, that would be okay, I guess; and just kinda wait until it's your turn. [Student #3] What problem set are we working on? [Instructor] Sorry? [Student #3] What problem set are we going to be working on, for next class? [Instructor] Oh, um. So this is Section... uhh, what is this? 4.1. So you'll have homework from 4.1. And then on next class, we're gonna try to do... 2.8, which will go fast. And then, 4.2. We'll try to do two sections. We'll see. [Student] So the homework is 4.1, right? And 4.2? [Instructor] -Yeah. [Student] -Okay. [Student #3] Wait, so um... We're doing the homework for 4.1 and 4.2 for next class? -We're not doing the— -[Instructor] No-no-no-no-no, no-no. So, all you need to be working on right now is 4.1. [Student] Okay. So we'll just do the normal homework. So we're not doing problem sets... between Mondays -and Wednesdays anymore? -Oh, oh; I see what you're saying. So I'm not giving you a problem set this week, because you just had a test. So I don't really have anything to problem-set you over. -I appreciate that. -[laughs] You're welcome. But we will next Monday. Next Monday, you'll get a problem set. Um, I was gonna say that in my Calc 2 class, some of those students are also meeting on Zoom, to work on homework together. So, they had study groups going, and they're just keeping those going on Zoom. So, I'm gonna throw that out there. If any of you guys had study groups. You know, continue to do that. [Student] Um, before we go. Is this meeting going to be posted in Recorded Meetings? [Instructor] Yes, it is. It just takes a while. So, once I'm finished, it has to convert it, or something? I don't know. And that can take hours. So, hopefully... hopefully by tonight I'll have it posted? But in the morning, as a last resort. [Student] Okay. Heard. Okay. Well, so, we are early. And I'm gonna let you go; so if you wanna go, just go ahead and exit the meeting. I'm gonna just stay here for a minute, in case anybody wants to talk or ask me a question. And if you're leaving, bye— [Student] -I actually have a question. [Instructor] -Sure. You can hang out. So, I actually was struggling with the particle moves along a curve equation. [Instructor] Okay. I would just like you to break down a little bit more what you did, um, a couple steps through it. [Instructor] Sure. Let me turn my screen-sharing back on. It's really taking forever. [Student] Uh, you said this is already recorded already, right? [Instructor] Mhmm. [Student] How do we go to view it? Do we just go on Blackboard, then... [Instructor] Yeah; so. Once it's ready to post, then you'll see a link. I think the link's already there, right? That says... Zoom Recordings, or Recorded Meetings. I forgot what I called it, but. "Recorded" is in it. So you'll just click there, and then you'll see them. [Student] -All right. Thank you. [Instructor] -Sure. Okay. So here's my particle problem, again. So the deal was, um. Little dude is moving. Little particle is moving. And the curve, the y = √(1+x³)? That is the formula that relates your variables, x and y. And then, let's see. The y coordinate was increasing at a rate of 4; so as it's moving, the rate of change of the y coordinate is 4. And what we wanted here was the rate of change of dx coordinates. So we want (dx/dt). Okay, so my formula was y = √(1+x³)? And one thing that's nice about these problems is you don't have to find the formula, or think about it, or figure out what it is, because it's just handed to you. It's whatever the equation is. It's kinda nice, really. So we're always differentiating with respect to t; so I wrote (d/dt) of both sides. But what I did over here was, I just rewrote it in the form of a rational exponent, because it makes it easier for me to differentiate. So my left is (dy/dt). And... here's differentiating on the right-hand side. ½ down in front. Rewrite 1 +x³. Decrease by 1, so -½. This is the derivative of the inside. The inside is the (1 +x³). But the derivative of (1 +x³) is 3x²(dx/dt). Any time it's the wrong letter, gotta follow it by that (dx/dt). And then I just substituted all of the stuff in. (dy/dt) was 4. ½ times 1 plus. x was 2, because that's the point we're kind of lookin' at here. Here's 3 times 2²; and then (dx/dt). And then I did a bunch of stuff in my head. Maybe that's where... [Student] That's the part I had a problem with. Yeah. [Instructor] Sorry about that. I could see that I didn't have much room left; and so, that's why I did that. So the 2³ plus 1 is 9. But it's 9 to the -½. That's 1 over the square root of 9. So that's ⅓. And that's why I wrote this 3— oh, it paused on me— That's why I wrote this 3 down here. So my ½ is one over two. This, when I bring it downstairs, is 3. And then, 2² is 4. Times 3; there's the 12. And then (dx/dt). Did that help? [Student] -Uh, just give me a moment. [Instructor] -Sure. So you're multiplying ½ by -⅑ ? So, it's more like this. I'm gonna show y'all the steps out here to the side. The 1 + 2³, that's a 9. But it's a 9 to the -½. Well, that's the same as ⅑ to the positive ½. And then times 3 times 4. So that's ½ times ⅓; 'cause 9 to the ½ is square root of 9, and that's 3. And that's times 12. So, this is 12 over 6, which is 2. And that's where that comes from. -Thank you. -Sure. -I really needed to see that. -Good. Glad to help. Anybody else still in the room, you have a question; go ahead. [Student] So, is this considered as a multivariable calculus, or... This is just single-variable? [Instructor] Uh, yeah; actually, that's a good question. Uh, no. It's not considered multi-variable calculus. It's not. [Student, softly] Okay. There you'll get into, you know... xyz, and all this cool stuff called partial differentiation, and you'll have double and triple integrals. It's good. So, no; this is not that. -Can't wait to learn that. -Yeah, thank you for today; I'm leaving. Oh, you're welcome. Bye-bye. -Manon, you said you can't wait? -Yeah, I can't wait to learn that. Yeah, it's beautiful stuff. You'll love it. I was looking at one of the hardest challenging problems in the mathematics right now, which is Riemann's data function, -that they can't prove it. -They did? Oh wow. Yeah; they can't, um. There's no proof of it. We know the answer, like... The numbers, but. We can't prove it, basically, right now. Yeah. So, if you can't prove it, then you don't know it. We can graph it, and look at the values, and we know where it's approaching. The answer, but. Yeah, there's a $1,000,000 prize on it, if someone solves it. -Oh ho-ho, nice. -Yeah. Anybody else? [Student] All right, I'm gonna leave. Thank you for the lecture. Okay. Bye, Manon! -Yeah. Stay safe. -You, too. All right, everybody; I'll end the meeting, and um... I'll see ya Wednesday. Bye!