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    Hello,

    I was just wondering if someone could help me with part c of this questions...

    The curvec, has equation y = 3x - x^2. It passes through the origin and the point B on the x axis.

    a) Find, in the form ax + by + c = 0, an equation of the normal to c at the point A(1,2) DONE

    x + y - 3

    b) Show that this normal [asses through the point b. DONE

    The shaded region R is bounded by the curve and the line AB

    c) Find by integration the area of R.

    I am really sorry that I can't include the diagram because I am not sure what edexel paper this is from.

    Anyway I have tried to do it by taking the linear equation away from the curve equation -
    3x - x^2 - (-x+3)
    and then integrating that resulting equation... Which I got to be -

    2x^2 - 3x - x^3/3

    The answer was wrong - when I looked at the mark scheme it said to take away the area of the triangle from the area of the curve equation after integrating.

    I do understand that method - but I was just wondering why the method I tried didint seem to work - Ibecause it usually does work - is there something I have missed that you can see that I cant.....

    Again - I wish I could put the diagram on!
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    (Original post by christinajane)
    ...
    If I've understood you well, you want to find the area bounded between y = 3-x and y = 3x - x^2, which is as below:



    As you can see, the curve is greater than the line. So you want to do curve - line, which is:

    \displaystyle \int (3x - x^2) - (3-x) \, \mathrm{d}x = \int (3x - x^2 - 3 + x) \, \mathrm{d}x which seems to be different to what you have?

    Whoops, I've mis-read your post, when I integrate that, I get:

    \displaystyle \int (-x^2 + 4x -3) \, \mathrm{d}x = -\frac{x^3}{3} + \frac{4x^2}{2} - 3x + c whereas your x^2 term has a coefficient of 2 and mine a coefficient of 1.

    If I put limits from x=1 to x=3, I get: 4/3
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    (Original post by Zacken)
    If I've understood you well, you want to find the area bounded between y = 3-x and y = 3x - x^2, which is as below:



    As you can see, the curve is greater than the line. So you want to do curve - line, which is:

    \displaystyle \int (3x - x^2) - (3-x) \, \mathrm{d}x = \int (3x - x^2 - 3 + x) \, \mathrm{d}x which seems to be different to what you have?

    Whoops, I've mis-read your post, when I integrate that, I get:

    \displaystyle \int (-x^2 + 2x -3) \, \mathrm{d}x = -\frac{x^3}{3} + \frac{2x^2}{2} - 3x + c whereas your x^2 term has a coefficient of 2 and mine a coefficient of 1.
    But dont you simplify that to become 4x -x^2 - 3 and then integrate that?
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    (Original post by christinajane)
    But dont you simplify that to become 4x -x^2 - 3 and then integrate that?
    Right... yes, you do. I'll fix that, what's your answer and what's the markschemes answer?
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    (Original post by Zacken)
    Right... yes, you do. I'll fix that, what's your answer and what's the markschemes answer?
    The mark scheme says:

    R = area under curve - area of triangle

    Area of triangle = 2

    Area under curve = 10/3

    R = 10/3 - 2 = 4/3

    I am seem to get zero! Which obvioulsy isnt correct!
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    (Original post by christinajane)
    The mark scheme says:

    R = area under curve - area of triangle

    Area of triangle = 2

    Area under curve = 10/3

    R = 10/3 - 2 = 4/3
    I get 4/3 as well, it looks like you've simply made a slip in your arithmetic somewhere.

    \displaystyle \bigg[-\frac{x^3}{3} + 2x^2 - 3x\bigg]_1^3 = (-9 + 18 - 9) - (-\frac{1}{3} + 2 - 3)

    \displaystyle = 0 - (-1/3 -1) = -(-4/3) = 4/3
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    (Original post by Zacken)
    I get 4/3 as well, it looks like you've simply made a slip in your arithmetic somewhere.

    \displaystyle \bigg[-\frac{x^3}{3} + 2x^2 - 3x\bigg]_1^3 = (-9 + 18 - 9) - (-\frac{1}{3} + 2 - 3)

    \displaystyle = 0 - (-1/3 -1) = -(-4/3) = 4/3

    Yeah - i bet its those bloody minus signs again! Let me go and check - I have probably forgotten to take something away from something else - its usually something simple with me!

    Thanks for help zacken - will give it another go - at least I was kinda on right track!
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    (Original post by christinajane)
    Yeah - i bet its those bloody minus signs again! Let me go and check - I have probably forgotten to take something away from something else - its usually something simple with me!

    Thanks for help zacken - will give it another go - at least I was kinda on right track!
    I'd have only docked one mark for you for this kind of error, if that's any consolation, your method is entirely perfect.

    Let me know if you find your mistake!
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    (Original post by Zacken)
    I'd have only docked one mark for you for this kind of error, if that's any consolation, your method is entirely perfect.

    Let me know if you find your mistake!
    Can you mark my exams! Ha.

    Yup I found my mistake - I am afraid to say its rather embarrassing - I set the second limit to zero instead of 1.

    I think I was just thinking of where the curve crosses the x axis instead of where the line crosses the curve!! :-/

    Got there in the end and got the correct answer

    At least I got the hard part correct ......

    Thank you again for your help now and all the other times you've helped me!
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    (Original post by christinajane)
    Thank you again for your help now and all the other times you've helped me!
    Well done for getting it almost correct. And no problem.
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    Just a quickish question I wanted to ask you Zacken I have been working on differentiation and integration and I was wondering if you could just explain what the findings and numbers actually mean.

    One in particular I was just doing and whilst I got the question right I just wasn't sure how to interpret that into English if you will...Question:On a car journey, the average speed of a car is v m/s-1. For v greater or equal to 5, the cost per km, C pence, of the journey is modelled by

    C = 160/V + V^2/100

    Using this model,

    a) Show, by calculus, that there is a value of v for which C has a stationary value, and find this value of v.So I have differentiated it and got V = 20 What does this actually mean in terms of C??

    b) Justify that this value of v gives a minimum value.I got the second derivative and proved that it was in fact a minimum value of C - what is this showing me exactly...???

    c) Find the minimum value of C and hence find the minimum cost of a 250 km car journey.

    So - I put my minimum value of v which is 20 back into the original equation and got 12 - what does this show me? I did have to look at the answers at this point because I wasn't sure what to do with my 12 - turns out I had to multiply it by 250 to get 3000 so the cost was £30

    So like I said I know how to do the questions, for the most part, but just not sure what all those numbers actually mean in a real life situation. I think they should do more questions like this instead of just using graphs all the time because you can't really relate to graphs.

    Anyway, if you have the time I would appreciate it any help and teaching you can offer!
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    (Original post by christinajane)
    Just a quickish question I wanted to ask you Zacken I have been working on differentiation and integration and I was wondering if you could just explain what the findings and numbers actually mean.
    I'm afraid that I'm going to have to reply to this question with numerous appeals to graphs.

    Basically, what the question is saying that, for a given average speed V, we'll be modelling the cost of the journey per kilometre (in pence) by the function

    C = \frac{160}{v} + \frac{v^2}{100}.

    That is, if I tell you that today, on my way to school, I travelled at an average speed of 20 kilometres per hour, then the cost of my trip in pence per kilometre is C = \frac{160}{20} + \frac{20^2}{100}.

    a) now the whole point of this sort of question is to find the value of (in this case V) the average speed such that the cost per kilometre is minimised (or maximised, depending on the question) - that is, I want you to find what average speed I need to be travelling at on my way to school for it to cost me the least (in pence per kilometre).

    So, if we look at the graph of the function:



    All this is is the graph of my cost per kilometre in pence for a given average speed V.

    Now as you can see from the graph, initially, for low average speeds, the cost is really high, then as you get higher and higher speeds, the cost decreases nicely. But once you pass a certain point, it starts to go up again, oh no!

    BRIEF INTERLUDE:

    We want to find the point at which the cost is the absolute minimum, which is precisely what differentiating does - the minimum point occurs when the tangent to the graph at that point has a gradient of 0, which I'll illustrate with a simpler example below:



    The tangent to the graph at some random point (the red line) has some random gradient, but as we can see by the green line, the tangent to the curve at the minimum point, the gradient of that line is zero, it's a straight horizontal line. That is precisely why we differentiate and set equal to 0, because dy/dx gives us the gradient of the curve at any point and we want the point where they gradient is 0, i,e: dy/dx = 0.

    END BRIEF INTERLUDE.

    So, yes - back to what we were saying about costs in terms of average speed. We have C = some function of v, we differentiate to get dC/dv, now we're setting this equal to 0 because we want the gradient of the tangent at some point (which we want to find) is 0.

    You got V=20, this is the value of V at which the minimum point occurs, that is - when my average speed to school is 20 kilometres per hour, then the cost of my journey is at the very lowest possible. However, this is only locating my "x coordinate" it is telling me what my best/most optimised speed is, the speed that gets me the lowest cost, but it's not telling me what the lowest cost actually is; we'll deal with that in c)

    b) Of course, I've been talking about it being a minimum point all along, but that was slightly sneaky of me, because when the gradient is 0, it might possibly be the case that I'm actually finding a maximum instead of a minimum (Because the gradient is 0 at a maximum too) Hence, if I differentiate again and get \frac{dC^2}{dv^2} then I plug in V=20 into this expression, if it's a negative number, that means the point v=20 is a maximum point and if I get a positive number then the point v=20 is a minimum point.

    c) So, yeah - we know that v=20 is the best speed for us to drive at to get a lowest cost, but what actually is this cost? :woo:

    Well - we know what C is for any given V, so let's just plug V=20 into the formula for C to find what cost this gives us, and lo and behold - this tells us that if I want to go to school, the best speed for me to drive at is v=20 and the cost associated with that speed is 12 cents per kilometres. No other speed will get me a cost lower than that.

    But bleh, we know that the cost is 12 cents per kilometre and we're travelling 250 km, so the total cost is 250 kilometres * 12 cents per kilometre = 3000 cents = 30 quid for the total journey.

    That is, for every kilometre we drive, we sped 12 cents. If we drive two kilometres (guess you got tired of me!) then it's 24 cents - if we go for a long romantic 250 kilometre drive (ooh) then it's 250 * 12 = 3000 cents).

    (I apologise, I've used the word cents instead of pence in writing this up...)

    Also - you might be questioning how I know that the cost is in centre per kilometre, well - this is given to us at the start of the question (the cost per km C in cents).

    Please point out anything you didn't understand.
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    Wow Zacken - you are actually amazing!

    Sounds so simple when its explained properly - seriously thank you!

    I think it helps when you understand why you actually do these things rather than just how you do these things. Which is something I am trying to work on as I think it helps you remember things better!

    The graphs certainly helped with that

    I really really hope you get to be whatever you want to be, if you're not already! And the hand is ok.

    I should have paid you to teach me maths rather than the college I go to at the moment.

    It all makes sense now.

    I think I can overlook the cents instead of pence ;-) hehe
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    (Original post by christinajane)
    Wow Zacken - you are actually amazing!

    Sounds so simple when its explained properly - seriously thank you!

    I think it helps when you understand why you actually do these things rather than just how you do these things. Which is something I am trying to work on as I think it helps you remember things better!

    The graphs certainly helped with that

    I really really hope you get to be whatever you want to be, if you're not already! And the hand is ok.

    I should have paid you to teach me maths rather than the college I go to at the moment.

    It all makes sense now.

    I think I can overlook the cents instead of pence ;-) hehe
    Awesome, glad it makes sense! And yes, it certainly does help if you understand why you do things rather than how you do things, and it's admirable that you seek out the former instead of being content with the latter. :yep:

    Thank you! I hope I get to where I want to be too, we'll see if that happens after these exams. :woo: What about you? Have you decided on any courses/unis yet?
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    Hello Zacken,

    Well I have already done university etc.

    I am learning maths just for fun and out of interest! Ha. I was always bad at maths in school and strangely found a love for it when I did a classical mechanics course at a night school. Plus I am bored with my job and would like to see if this will lead anywhere.

    I haven't done maths properly for about 10 years, so as you can imagine struggling a little bit with the A level I am doing - hence why I ask a lot of silly questions on here!

    Just a quick question.

    Sum of arithmetic sequences equation. When I am asked to find the value of n I am not sure how best to manipulate the equation.

    So for example I am working on this at the moment:

    sn = n/2( 42 +5n -5)

    sn = n/2(5n +37)

    What should I do with that /2 at the bottom of n?

    I always always get caught out by that and I just want to know definitively how best to handle it!!

    If I multiply the brackets by two to get rid of the fraction do I then have to multiply the RHS by two?

    See I just loose marks because I dont have the basics with manipulating equations.... :-(
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    (Original post by christinajane)

    If I multiply the brackets by two to get rid of the fraction do I then have to multiply the RHS by two?

    See I just loose marks because I dont have the basics with manipulating equations.... :-(
    If you mean that you have some number like:

    500 = n/2(5n+37), then yes - you should multiply both sides by 2 to get 1000 = n(5n + 37) which you can the expand out and collect everything on one side to get a quadratic in n.
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    http://www.thestudentroom.co.uk/show...9#post64641649
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    Hey! Hijacking others thread is not okay. Your thread will be replied to in due course. Thanks.
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    (Original post by Zacken)
    Hey! Hijacking others thread is not okay. Your thread will be replied to in due course. Thanks.
    The title 'Integration' was relevant?
 
 
 
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