The Road to Exams [Week 17, Parametric differentiation and integration]

Added: 2026-05-05

[00:00:01]Okay, week 17 of January to June, the road to exams and we are looking at some stuff to do with parametric calculus. This is a tough question. If you do want extra guidance with parametric calculus, I think I covered it in my pure B aiming for a star session that took place in the Easter holidays. Those were my paid for sessions, but you can get the recordings for those and you will look at lots more questions like this and lots of other topics, too. If that is something you're interested in, of course, it is linked in the description, but you've probably heard me talk about that loads and loads already. So maybe we should just get on with the question.

[00:00:32]A curve C has parametric equations. X= sin^ 2 T, Y = 2 tan T and T is between 0 and pi /2. It's referring to this curve that we've got here. The point P with parameter t =<unk> / 4 lies on C. So this bit here, I'm just going to jot this down. This is where T is equal to<unk> over4.

[00:00:55]The line L is the normal to C at P. The amount of times I do a question and I do it with a tangent. It's the normal. So obviously we're going to do a negative reciprocal when we get the gradient for this. Show using calculus that an equation for this line is 8 y + 2x = 17.

[00:01:13]I'm going to just take it a little bit easier to begin with. I think I'm just going to figure out what is this coordinate of p cuz I know I'm going to need that later on. Okay, so let's deal that to begin with. When t is equal to<unk> / 4, x, which is sin t all squared, would be the sin of<unk> / 4 squared. I don't need my calculator to do this cuz I know that sine of 45 or sine of p<unk> / 4 is 1 over <unk>2. So, it's going to be a half. But if you do check on your calculator, you do of course end up with that x is a half and y is 2 tan t. 2 tan t. I'm not going to use my calculator for this because I know that the tan of pi over 4 is the tan of 45 which is just one. So y is equal to 2. Meaning that the coordinate for p is a half and two. And I sometimes find adding these things onto the diagram can help. This is a half and the ycoordinate is 2. Now of course we're trying to find a normal. So we need to have some stuff to do with dy by dx. We need to find out what the gradient is for this. So continuing with this, I am going to find out what dy by dt and dx by dt are. I'm going to tell you why actually because we know that dy by dx for parametric is dy by dt divided by dx by dt. If you don't know that, that's just something you need to kind of get memorized. But you can think if you do this as like a fraction division, that one flips and the dt's cancel.

[00:02:39]So let's figure out what dxdt is. I'm going to be differentiating this thing up here. I will pull the power down. I will reduce the power by one. And I will multiply by the derivative of sin t, which is cos t. Okay, I don't want that mult that line underneath. And then dy by dt is the derivative of 2 tan t. I would like you to have memorized what the derivative of tan is. If you haven't, I think it's in the formula booklet, but tan differentiates to sequ.

[00:03:11]So it is going to be 2 se 2 t meaning that dy by dx is the 2 se 2 t divided by the 2 sin t cos t. Now the only simplifying I'm going to do is just get rid of these twos so that it's just the se 2 t over sin t cos t. You could probably simplify this because obviously se is 1 over cos^2 blah blah blah blah blah. I'm probably just going to work with doing the substitution here. So I'm going to find dy by dx when t is equal to<unk> over 4.

[00:03:47]So I'm going to figure out what sec should I do sec to begin with? I'm going to do cos and sign. So I know that co of p<unk> / 4 is 1 / <unk>2. If you do it on your calculator, you'll get <unk>2 over2. But for some reason, I just always memorize it as 1 over<unk>2. I know that s of p<unk> /4 is also 1 over <unk>2. So if I want to find what se is, it will be the reciprocal of this one.

[00:04:11]Again, you could just put it in your calculator, but I kind of like doing things without sometimes. So sec of pi over 4 is the<unk> of 2, meaning that se squared of p<unk> over4 is just two. So I've now got these things. I've got the se squ part that can go here. I've got the cos pi over4 part that can go here.

[00:04:30]And I've got the sin pi over4 part that can go here. Of course, you could type it in your calculator, but I sometimes find make mistakes doing that. So, sin t is 1 / <unk>2 cos t is 1 / <unk>2. So, it's 2 / 1 / 2, which is 4. Check it on the calculator if you want to, but we are doing the normal. So, the normal will be the negative reciprocal of this.

[00:04:58]So, the normal gradient is -14. Of course, we're going to need an extra page here. So to find the equation of the normal, we will use our y - y1 = m brackets x - x1 formula for this gradient with the position p, which is a half and 2.

[00:05:23]y - 2 = -4 x - 12. And then the thing we're aiming for is all in terms of integers. There's no fractions. So I'm going to aim to get rid of all of the fractions. I'll probably do that by multiplying by the 4 to start with. So that's my 4 y - 8 = - x - a half. So 4 y - 8 is - x + 12. I've got another fraction, so I'm going to double everything. So 8 y - 16 = -2x + 1. And then I think I've got what they wanted. We have an x y, sorry, an 8 y + 2x = 17. An 8 y. Good. Plus the 2x. I'll put the 16 on the other side. And we get the 17, which was the thing that they were looking for. Okay. So now I know something. I now know the equation of this line. Even though they told me it was 8 y + 2x = 17. We know what the next part is going to be. We can see a shaded area. Of course, it's just going to be find the exact area of S that we've got here. Okay, I see it as two shapes. One of the shapes is going to be pretty easy or easier than the other one is, of course, we have a triangle. Okay, so for this triangle, we just need to find out the base of the triangle. Okay, the height of the triangle is two. That was why it was good. I labeled that on the diagram earlier. I now need to just find the base of this triangle. Well, I know that this is a half. I can probably find out this part here which is where it crosses the x- axis. So it crosses the x axis when y is equal to zero. So 2x is 17.

[00:07:10]I've just said that y is zero. So 2x is equal to 17. So x is 17 / 2. So this value is 17 /2. So I can now figure out what the base of the triangle is. Of course, we've got the height over here, which is two. We're going to figure out the base.

[00:07:29]So, for part B, the base of the triangle is the 17 over 2 minus the half, right? It's got to be.

[00:07:40]That's 17 over two. That's a half. I'm going to subtract them and that will tell me how wide it is. So, that's 16 over two or that's eight.

[00:07:49]The height of the triangle we just know is two.

[00:07:54]So the area of that triangle is the base time the height / 2 which is 8. Again annotating things in your diagram can be really helpful. I'm going to just put an eight and say that area there is 8. Now my last part is to figure out what this section is. This kind of like curvy section that's here.

[00:08:17]And we just want to figure out what the limits are going to be because it's going to be um parametric integration.

[00:08:21]So our limits are going to be to do with t. At this point we know that t is equal to p<unk> /4. We can see that the range is between 0 and p<unk> / 2. I think it's pretty obvious that t is going to be zero here because when t is zero you get x is the s of 0 which is zero. And y is 2 tan of 0 which is zero. In other words you get the origin. So this little bit the origin is when t is equal to zero.

[00:08:50]Oh my gosh. So the green the algebraic part that we're now going to do the integration on we are going to be trying to find for parametric integration we are trying to do y with respect to x but for parametric we do y dx dt dt and the t limits are going to be between 0 and pi over 4. This is the part that makes it parametric. We normally do y dx but because our limits are d uh dt or t limits we need to change it as y dxdt.

[00:09:19]These dts are sort of like canceling out with each other. So we're going to be doing between 0 and p<unk> / 4 of y * dx dt. Well, y is 2 tan t and dx dt is 2 sin t cos t. So y is 2 tan t and dx dt is 2 sin t cos t dt. And this is good because it's now all in terms of t. is being integrated with respect to t. The limits are between 0 and pi over4 which are t limits and we had to use the t limits because everything is in terms of t. So that's why I'm doing it from t to 0 to t to p<unk> /4 to find out that green area that we've got down here.

[00:10:02]This is the green area that we are trying to work out. Okay.

[00:10:07]So simplifying some of this we're going to have between 0 and pi over 4. The two and the two gives me the four. Tan t is sin t over cos t and that's getting multiplied by a sin t cos t dt.

[00:10:26]So these causes are going to cancel and what we're actually integrating is between 0 and p<unk> / 4 of 4 sin^ 2 t dt. And this is the part that they've asked this a number of times. If you're a further math student, you probably know very quickly how to integrate sin^ square t. This cannot be integrated to like a sin cubed. This is not something we can do the reverse chain rule for. To be able to integrate something of this form, we have to use the double angle formula for cos 2 t rearranged.

[00:10:58]So cos 2 t is 1 - 2 sin^ 2 t. Some people might have this just memorized off the top of my head off the top of their head. But if you can't memorize what sin^ square is equal to, we're going to take the double angle formula for co and rearrange it. It's the only way you can integrate sin^ 2 to when it's by itself like this. So the 2 sin^ 2 t is 1 - cos 2 t. And we want 4 sin^ 2 t. So 4 sin^ 2 t is just going to be this thing doubled which is 2 - 2 cos t.

[00:11:37]If you haven't seen this yet, you will hopefully with enough practice start to see that when you're trying to integrate sin^ square or cos^ 2, you have to use a double angle formula for cos. So that is 2 - 2 cos 2t.

[00:11:50]We haven't even done the integration yet. We're going to do the integration finally. So it is going to be a 2t.

[00:11:57]It's going to be sin 2t. I think just that I think it's just going to be minus sin 2t because when you differentiate that you do get - 2 cos 2t and the limits are 0 and p<unk> / 4. Wow, they've really made us work hard for this question. So I'm going to sub in<unk> / 4. So that's 2 *<unk> / 4 - the s of 2 *<unk> / 4 - 2 * 0 - the s of zero. Well, luckily that's zero and that's zero. So we can just completely ignore them. 2<unk> / 4 is<unk> / 2. And then the s of um oh my god, have I got something wrong with this? No, I haven't got something wrong with this because this is the s of 2<unk> over4. That's the s of p<unk> / 2 which I think is 1. But at this stage I'm checking absolutely everything to make sure I'm not making any mistakes.

[00:12:49]Sine of p<unk> /2 is 1. Thank goodness.

[00:12:52]So this is p<unk> / 2 minus 1. Now that was the green bit in our diagram. That was the green bit. The blue bit we've already got labeled was eight. So the blue bit that we had labeled was eight.

[00:13:10]So this means the area I think the name of the shape was S. I think they called it the area of S is p<unk> / 2 minus one plus the 8 which is pi / 2 + 7. I'm imagining this is the hardest question we've done so far in this series.

[00:13:31]This was fine because it was a show that we do have the pi over 2 + 7 for that answer. So, I've kind of been doing these videos as like a you know the maths, let's just go through some exam questions together. But if you need to go into more detail of why these things work, do dive into my chapter summaries.

[00:13:47]Do dive into my my playlists going through these things. And of course, like I've mentioned many many times, do go to my like Emmy Freay Star session recordings if you want to get some of that extra support on some of these things cuz these are very, very tough questions. These are really, really hard. This one I imagine was kind of towards the end of an exam paper. It's international, so we can't really do that as a comparison. Anyway, next week we're doing a hard mechanics topic. We are doing ladders resting against walls, the moments kind of topic, which is week 18 of 24. So, I hope you're sticking with me for this series, and I will see you next