Hi, good. Sorry. It's been so long since we've had a lesson. I've been here and there and everywhere the last few weeks. But I thought today we could dive in would be the tricky geometric series question that we started last time. And I thought that would tie together all the the work we did on geometric series and see how much of it sunk in, whether there's any bits we need to go over. So I'm not expecting you to remember it. So let me just get that up. I think I remember one of the two formulas. Great. So let me just get that question up for you. So we're told the first three terms of a geometric series are k plus ten k and k minus six. We're told this is geometric. The first question is find for value. And just to say again, I'm not fuif. You haven't completely memorized the formula at this point. I want you to be aware of what the formula are and then I'm very happy to quote them for you if you find that you need them. But actually, I don't think you need very much in the formula department for this purpower. So we had a little to think about it last time, but we didn't really get it to it. So if I give you permission on the board, see if I can get can we find the value of k. I have the permission now. Okay, have a go. Is it correct though? Taking the difference? Well, you're right to be questioning that. Remind me what geometric means? What's the definition of a geometric series plus a constant? So we have arithmetic mehave geometric. Can you remember what the differis between? How would you characterise those two different types of series? One has a constant difference. Excellent. So the arithmetic has a constant difference. Can you remember what the kind of equivalent is for a geometric? It's not a constant difference that there is a constant something. Can you remember what that is? Multiple Yeah sort. Can you explain what you mean by that? Like a constant power of something. So for arithmetic, we're adding the same thing each time to get from one term to the next. There's something kind of similar for geometric. Can you remember we're multiplying the same thing. Excellent. So we know that this series, we're multiplying by the same time to get from one term to the next. That's all we know, really. We know that the three terms are given by these expressions, and we know that we're multiplying by the same time, the same amount each time to go from one to the next. So can we do this? Well, let's have a think about that. That first thing you've written. So you're saying the first term is twice the second term. That's not necessarily visit it. What if we make a diagram? So we have these three terms and we know we're times ging by something. Each time. Yeah, Yeah. So maybe you could translate that into some algebra. How can we capture the fact that these three terms to get from one to the next were times invby something? Could you maybe set up some equations? Okay, so nearly so I like that first thing you've written, maybe tried doing without I splitting it up. Try not to do everything at once. Try writing two separate equations. Think about the first pair and then the second. So think you're getting muddled because you're trying to do too much at once. Wait, but the a should go here. Well, depends what you want to be your a I've put a question mark in our diagram. You could give that a name, which we could call that a. But depends whether you're also thinking about a being the first term. Sometimes a represents the first term. You thinking maybe a constancy. So maybe a different letter might be good. Yeah exit. So that first equation you've written is representing the fact that to go from here to here, we're times thing by C C, lots of k 20X K excellence. And can you set up a similar one for the second pair? So it looks like you've introduced another constant. Can you preempt what I'm going to say there? So should be the same constant excellent, because we're told that it's geometric. Lovely. So what you've done there is you've nicely set up system of equations that capture the fact that this is a geometric series. So looking at your system of equations, I think that's a good situation to be in. Can you see why I'm happy with that system of equations? Yeah, you can. You can make three equations from that. Sort of unmore thinking we have two unknowns, we have two equations. That means we can solve it. As long as you have the same number of unknowns since and equations, then you're happy. So it's a bit of a fidy system of equations. But see if you can go ahead and solve that because what we want is definfind k. It doesn't seem right. Just think about how you've Yeah so you've copied that the first one wrong. So you've put A C, so that's what is it. But good, good radar, spot on. Something thin's come wrong. Shouldn't this? Can we just substitute that? Oh, no, sorry, this. Yeyep if we do it correctly. So you're thinking substitute? Yeah. Okay. So ck equals that whole expression plus ten. Absolutely. But we don't need to not forget the minus six. But we've forgotten the minus six then, have you? So we'll need the minus six up here. So you've got the right idea, H, to be a little more careful. The substitution. Nice. So then can we find k. Excellent. Very nice. So then the follow up question, you probably just need to keep our value of k. Next question, can you find the sum to infinity? Of this. Write me with the equation again. So quite a nice easy one to remember. This is our sum to infinity. So you just need to remember what the a and the R stand ds for in this equation. So tell me about what you're writing here. R would be the. Multiple absolutely. The common ratio, which we called c here. Nice. Yeah that comes out right. Possibly a little arithmetic slip there, but basically I'm happy about this. Here is the right Sumy beef with this is you've got pulled into doing fancy notation that we don't need to do. So what you've written here, this whole sentence isn't quite right because what you've said is the sum to infinity of the formula of the sum to infinity series there. Yeah. So the sum to infinity wanted to abandon this. Yeah as long as you understand that. Yeah, excellent. Good stuff. So is that kind of coming back to you? All the geometric series stuff we did last time? Yeah, excellent. I think let's move on and make a start on calculus. Let me just get an answer up. Okay. So it seems to me you've already done some bits and bobs on differentiation. So I think the most important thing we can do is really deepen your knowledge of how differentiation works. So to start off, could you tell me what you understand by differentiation? What is your current understanding of what that is? That's finding the gradient of a point a on a not line, but on a function by measuring two values that are infinitely close together. Excellent. That is a very good description. So you've got a key point that if we have a curve, some function of x. The gradient at each point Yeah is given by our derivative. And just always through a bit of notation, we have both of these of points here. When we say the gradient at a point, what we actually mean by that is the gradient of the tgents, because gradient kind of only makes sense to speak of a straight line. So we're taking the tgenent at each point, and you've absolutely got the idea that what we're doing, so we're saying, well, if this is my x, then I'll go a little bit further along x plus a tiny bit, and I'll take the gradient of that straight line. And you're quite right that the smaller we make that gap, the more and more accurate that becomes. And the limits as that gap tends to nothing. That's our derivative. Excellent. So what. We're just doing our differentiation day to day. We kind of just learn a couple of our most useful functions that we use all the time. We learn what their derivatives are, and we use a couple of useful properties that means we can combine them. So for example. I think you've come across this. What's the derivative of x? The power N N to the n minus one. Excellent. And this works for Yeah all powers of n. The first useful property we have of the derivative is that it respects addition. And that's probably something you haven't even thought of as a property that we need to establish. But if we have some function, that's two functions added together, then a derivative of that, we can just do the individual bits together. So what that means is any combination of functions that we know how to handle, if we had x to minus one plus three x to the two, we know how to differentiate. That it respects addition, we just do them separately, also respects subtraction. Another useful property, phthat, again, you've probably used without even really thinking about it, is that if we have some function, we know how to differentiate. So some f of x. And we know it's derivative. If we stick any old scalar multiout, the front that just stays there. So we know x squared differentiate 22x. If there's any number at the front, you just times it by, the times it by and then apply it in minus one. Absolutely. But the really important thing to clock is that this multiple has to be a scalar. So it can't be a function in its own right. So we know how to do with three x squared, but something like sine X X squared, we can't just pop the sx back in. So this property only works for scalers. So something that doesn't depend on x. When we get to a second year differentiation, we'll look at how do we handle products. So when we have a non scalar multiple at the front, the product rule, Yeah, absolutely nice. So that's kind of all of your sort of day to day differentiation for the first year content. However, something we've added in recently, which I'm really glad about, is from first principles. So I think you're going na have no problem with this at all because you already told me what the derivative means. We have the gradient of a really tiny. What's the worand? A straight line across. So is this a formula that you've seen before? Seen the definition yeso. This is the definition of a derivative. And there's more interesting examples as we go further along. But all we need for now is to think about our most basic derivatives. We've already seen that. We know that x squared goes to two x. Could you prove that for me? Can you do that from first principles? Now not quite think about how you've plugged that in. Have another go at that. Think about how you plug it in. Squared plus H squared? No. If our function is x squared, then what's the function of x plus H? Try and think about it just one step at a time rather than knowing what you're aiming for. No Kono. If a function of x is x squared, then the bit that we're not quite getting is this what's function of x plus H? Yeah. But I'd like to see that you understand that it's this or squared that's the really important bit. Yeah. Yeah, okay at it. Carry on. Wait x here. Nice. See if you can evaluate that limit. But we use rait also ruling this. Lopertelrule, did you say, Yeah, it's much simpler than that derivative. Much, much simpler. That would be like trying to do the hovering with a bulldozer. We don't need to do that. Try something simpler. Try something neat and elegant and doesn't use any big machinery. You don't need de by age. Well, we need to rewrite this hole inside bit, right? Because just staring at that, we don't immediately see what would the limit be. So is there a way to think of simplifying the top? Yeah, you expand that out and then we get some cancelling. That's nice. Well, let's do it one step of time. I liked what you had before. Now I would say the simplest response to that is to say, Oh Yeah, my x squares cancel. Yeah. So have a look what that looks like then what we've got left on the top. Can divide by a Yeah and then we can just sort out that fraction, can't we? Yeah but don't forget we're still in the limit. Yeah Yeah we just don't have enough space to write the limits. Like absolutely. So just simplifying it down a bit. We get it to a point where we isolate the H and then it's super obvious what the limit is gonna to do. So then we could do the x squared case, but it's exactly the same. Once we've done binomial, we can do it for general x to the n. You haven't come across binomial yet. If you binomial, I don't remember keywords, but could you show me how well I think it would make sense to do that once we as a separate topic otherwise with sort of combining a bunch of things, but basically we can prove the xn derivative once we've done binomial. So we'll come back to that. Excellent. I think that's everything I wanted to talk about for now. What I'd really like to do is look at some questions so see if you can apply this in practice. And there might be bits involved in these questions that we haven't explicitly discussed yet. So if we need to take a step back, that's fine. I'm just seeing it as a jumping off point. Okay, let's try say what I'm going to send you the sheet that will be quicker. Okay, here's the worksheet again. Yeah, so there might be bits that you haven't quite come across yet. If that's the case, just try thinking really clearly and slowly one step at time and see how far you can get. I think let's start with question two. Have a look at questions to you for me. Question two. Yeah. Oh, sorry, that wrong question. 2 cm under the question, sorry. There two things there. Yeah, that's the first thing. Technically what you've written there is that f of x is equal to one minus three over two x to minus a half. Now I suspect that's not what you mean, so keep your workings nice and clear. The first line is not the same as the second line, is it? You've done a process. Yeah, a little thing I know. I think I'm being fussy, but it only invites getting things mixed up. If we're not clear in our. So I agree that fractional powers are kind of nasty, but have a little think about exactly how you've executed that. Not quite. Just do one step at a time. What's x to the minus a half and you put that in a kniof form. Goes x over one. That. So just do it one step at a time. X to the minus a half. Can you rewrite that? My advice would be to ignore the three of the two for a second, just do the x one of a half. How would you rewrite that without the nasty power? Just x the minus, minus half. Absolutely. That's one over root x, and then we just times it by three over two. So you were trying to bring the three over two into the power, but that we need brackets for that. Okay, so tell me about what you've written there. No, no, I've I don't know what I've written there. I tried to plug the coordinate. It's in. Yeah so we do need to do some plugging in, but we have to do it really understanding what each thing stands for. So I'm right to want to plug something in. Can you formulate a plan? What is it you're trying to calculate first? Can we have like a sort of a sort of general plan for this question? We need the gradient at the position. Yeah, we're looking for the gradient at p actually the gradients of the tgents of P. I guess just formulating really clearly what we're doing sometimes is all we need. But this doesn't work now. What do you mean it doesn't work to find the point where y, where x is equal to zero? Yeah. So you found the gradient. The other ingredient we need for any straight line is the see the y interset. Absolutely. So now we're just into a technique we've done before. We don't know what it is. So you can give it a name. Let's just call it c for now. And we're told that this passes through the point 41. So this is just a coordinate geometry technique that we've done a few times. And to find the area of the triangle of B Q Yeah. So so that's Yeah I mean, for us to show that it pasounces through the origin, that's kind of obvious. Yeah. So the next thing is, do you know what the normal is? Do you know what that means? Yeah, the line, that's what you call it perpendicular to the another line specifically. It's a to a specific line almost. What's the relationship between the tangent and the normal genome? They're 90 degrees. Yeah so the normal is the line perpendicular to the tangent at that point. No, so then this would just be negative for x right? Not quite. So we know that the gradients minus four. So now we just need to find the c the y interept. And we can use the same technique as before here. Alright, so now you've got the equation of your normal. Nice. And now it's just a bit of kind of problem solving. We put it all together, find this area of this triangle. Plus the y axis at point q. So I need to find its White intercept. Yeah exactly. Yes, 17. So than one and 17. Wait, not wait, I think 17, right? But what's for four and 17? The normal to her sex. Okay, so then okay only that then. Then x should I yes sorry x would be zero at the y intercept. Excellent Yeah. Nice. So there are three vertices of our triangle. Yeah so. Could we vtoriize them and then use and then find their magnitude to I think, well, you could do something much simpler. Why this? Quite a convenient triangle, but this is just 17. That's it, right? So divided. Yeah so you're doing 17 dividing by four times four it's only divided by two right? Yeah so doing half base time type Yeah nice. Yeah, excellent. So 34, nice. Nicely done. So lots of quite simple things, but lots of them going on at once. Excellent. Next up. Let's do five. Look, get five for me. Any ideas how you might start? Question five. No, quite. I wouldn't have given you a question that needed something we haven't looked at yet. This is going to be a real exercise in finding the simple but neat thing to do because you can differentiate that using just the things we've talked about today. So it requires a bit of subtlety to spot how can we use today's content. To differentiate that equation. What's the only type of function we've looked at today? What's the only type of derivative we've covered? Yeah, next to the end. So I promise you, you can do this just using this. Or just be really careful with your brackets, then you've got basically the right idea. All right, Yeah. Yeah, one over x squared is x the minus two. Absolutely great. So that's probably the first tricky thing. Sort Ted here I can get. Just be really careful with your exponent laws. You're getting some bracket issues coming in here. So the minus one is only applying to the x squared. So the twelve is totally separate. There's nothing we need to do to the twelve. Yeah, absolutely. Because we're using that rule that says. Yeah, but going what am I going to say about your notation? Why am I not going to be happy with that? Still quite right. Yeah so it might seem a fussy point, but you've said two things are equal that aren't equal. So that that will cause us problems. We do that nice, excellent. So I need to put the negative three there. Let's think about it one step at a time, try and take all the guesswork out of it and just go one step at a time for a straight line. We need the gradient and we need the c, the y intercept. What we've been doing so far is finding the gradient first. So one step at a time, you need to find the gradient of the tangent at this point at a. So how are we going to do that just that little job? How are we going to find the gradient of the tangent at a. Yeah, absolutely. We're going to take the x coordinate of the point we're interested in and put it into our gradient function. Next. Excellent. Nice. More specific. Yeah, that would be a nice web b rising. Yeah, good stuff. So that's our gradient sorted. What's next? Wait, I need to check if we need it though. Again, try one step at a time. Can you write the equation with the c value? Excellent. So now we've broken it down. Now we just need to find the c. Excellent. Let's finish it up. Excellent. So something I'd like you to think about is you're capable of understanding every single step you do. There should never be anything mysterious or Oh, and then by magic, I take that number, I put it there, everything should make sense. You're definitely capable of understanding exactly why you're doing every single step. So really try and get into the nitty gritty. Every single step should have a really precise reason, no kind of magical vagueness. Excellent. And then the second part of this question, this is a question where you might not have come across. Yeah verify. What exactly are we being asked to do if if we have a verifier question? Plug the values in. Yeah, it's kind of a show that with a calculation. So if this were an exam webe thinking, how can I really clearly show my working? Because just arriving that the answer isn't very impressive because they gaexcellent show me how you do that. I think I'm doing something wrong. A couple of ways you could approach this. We could either kind of give us a harder question than it really is and just find a coordinate where the tangent are a intersecthe curve again, and then hopefully itbe minus 1:12, job done. How would we do that? There is another way as well. But let's start with that. Let's just think of it as fine. The coordinates where the tangent to a intersects the curve again, and hopefully itbe -112. How would we do it that way? Ist minus Trix plus nine. Set it to be equal to. See. Excellent. Then can we solve that equation? Yeah. So right about now you might realize why they've given us this question in the form of a verify. Yeah, because what kind of equation is this? Hey. Regarto, how you call it, it's a cubic and we don't have a go to method for solving cubic. So now I'd like you to without the whole verify thing is the fact that we know what the answer is, we're just double checking. What could we do now? Could just plug it in now Yeah, plug it in, confirm that that is the solution to that equation or I don't like what you've done in that last step. What me you that. It's just there setting it to the line that intersects with both of the. Both of the points. Yeah but that's so the fact that we've set it up here that's going to give us all points that are both on the tangent and on the curve. You've rearranged it down to this and then moving it all to one side will get zero. We can't just bung in another y Yeah and then we confirm Med that that hopefully gives us zero. Yeah, it does. Excellent. Very nice. Be happy with that one. Yeah, excellent. Let's scroll down, see if we find something a bit different. Let's just try and do one tricky one with the rest of our time. Why not let's skip to the last one thatwill be nice and challenging. Have a look at question twelve. Okay. Oh. No. My differentiation still correct. So was that last statement meant to be a differentiation? Yeah, so that's not quite right. Again, let's let's try not to make anything up. Believe me, I wouldn't give you any questions that need anything other than what we've covered today. So can you spot why that's not quite right? I think that would be instructive. So bearing in mind that we've said the only things we've looked at today are this and the fact that adding and subtracting just behaves as weexpect and also weacan times by a scalar that behaves as weexpect, there are only three properties we've covered so far. So can you tell me what. Excellent. Yeah just to pause on that point, it's really important what you did there is to treat k minus x as a scalar, but it's not because it has an x in there. So a really good idea then is to expand the brackets. Absolutely. So then Yep. So you've expanded the bracket, right? So the first time you've done really nicely. So you've realized that the most convenient form will be that, and then you've differentiated that perfectly. Let's just think about this term here. So you've expanded that, right? X root x. Tell me about this step. Why is that the same as Oh, wait, no, just have another go at that bit. Excellent. And then that's in a nice form that we know how to differentiate. Should have a think about that. Not quite. Excellent. So the nack really is to find the simplest thing that works. That's always the way with matths. It's not better matths to use a fancier technique. It's always best to use the simplest, most elegant thing. And I think the way we can do that is by building up the ideas just one step at a time. I know you've kind of come across all these fancier ideas, but I think itgoing be really good and stretching for you to try and always do with the simplest machinery possible. Excellent. I'll let you keep going. So talk me through what you're doing here, and I think you'll spot what you need to speak. Wait, Yeah, do I put the results in the wrong order? Nice. Just got a little slip somewhere. This is right. Just let me think about this stuff. Yeah, Yeah. Yeah. Yeah, nice. Excellent, good stuff. And I think we've just about got time to do puppy. I can scroll up if youlike. Yes, please miss. You spot where you might be doing extra work that you don't need to be doing. That's another a good general skill. So it's still point p Yep. So we're still thinking of p so that we actually already know. So just see we need to find. Sorry, could you repeat what you we just need to find c as you say? And bear in mind, we're doing the normal. So what do we need to tweak? You're nice. So. Yeah. And really, we should just rearrange it into exactly the form they want, but that's not a big move. Excellent. We're just about out of time. Excellent stuff today. Do you have any preference what we look at next? Do you have any 16 plus work? If you need to look at next time? Would you like to so that's basically first gecalculus wrapped up. So would you like to move on to second gocalculus? Or would you like to look at some more first gcontent maybe more to the second year? Great. You should do that next. Excellent stuff. See you guys yebye. Thank you, miss.