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Mega A Level Maths Thread - Mark V Watch

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    (Original post by joodaa)
    Nice Nice hmm It wont let me attatch a photo of the questions. Well basically I'm not sure how to approach questions where you're given the parametric equations and you have to find tangents to the curve without them giving you a specific point on the curve to use.

    1.A hyperbola has parametric equations x=secθ, y=tanθ Prove dy/dx is cosecθ and find the equations of the two tangents to the curve which are parallel to the y axis.I was able to prove that it made cosecθ but not sure how to do the last part.

    2.An asteroid has parametric equations x=4cos^3θ,y=4sin^3θ. Prove dy/dx is -tanθ and find the equations of the four tangents to the asteroid which are equally inclined at 45 degrees to both axes.Again I was able to prove the differentiation but dont know how to find the tangents.Thanks in advance ladd
    In the first question, because the tangent is parlallel to the y-axis then the gradient is infinite. That is, the denominator of dy/dx is 0. So since \frac{dy}{dx} = \frac{1}{\sin \theta} then the tangents occur when \sin \theta = 0. You find the values of theta for which that is true and then plug it into x = \sec \theta to find the tangent in the form x=k for some constant k.
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    (Original post by Zacken)
    In the first question, because the tangent is parlallel to the y-axis then the gradient is infinite. That is, the denominator of dy/dx is 0. So since \frac{dy}{dx} = \frac{1}{\sin \theta} then the tangents occur when \sin \theta = 0. You find the values of theta for which that is true and then plug it into x = \sec \theta to find the tangent in the form x=k for some constant k.
    Ahhh I didn't think to look at the gradient in that way. Yes that makes sense now
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    (Original post by joodaa)
    Ahhh I didn't think to look at the gradient in that way. Yes that makes sense now
    Can you think of something similar to do with the second question? In this case - it's not parallel to the y-axis, but it's inclined at 45 degrees to any axis. Can you think of what that entails for \frac{dy}{dx} = -\tan \theta = ?
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    (Original post by Zacken)
    Can you think of something similar to do with the second question? In this case - it's not parallel to the y-axis, but it's inclined at 45 degrees to any axis. Can you think of what that entails for \frac{dy}{dx} = -\tan \theta = ?
    Hmm do you do -tan45? Or maybe-\tan\theta =45?
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    (Original post by joodaa)
    Hmm do you do -tan45? Or maybe-\tan\theta =45?
    The latter, I think
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    (Original post by RoseGatz)
    Logistically that makes sense. How many people are in your biggest class at the moment? I was thinking about this the other day, if you're taking 5 A2 classes how is your timetable structured? Does your day start early and finish later than over here?

    Those are great uni's. Good luck! Let me know when you get your offers through.

    Thanks. I think my main one is obvious, it's a shame my parents aren't too keen on it, plus you won't have heard of the other uni's I've applied to anyway. They're fairly local to me.

    People just think that those who do FM are trying to get away from writing and whilst this is true for some, it really isn't the same for others. I was planning on taking GCSE German but alas my high school didn't offer it so I was stuck with French. To be honest I prefer the listening and reading side of learning a language, I'm not so fond of speaking or writing element.

    Zacken Really?:confused: Nobody really mentions it over here... not that I've noticed anyway. I'm guessing private/grammar schools encourage it more than most state schools, that's why I've never noticed.
    In my maths classes we have about 10, and physics about 15 (German 5 lel). Well in our school we have 5 periods and 2 breaks. Each period is an hour long. Starts around 7 and ends 1:30

    Thank you very much - I hope so

    I wasn't sure What's your main?

    I agree - Actually I'm opposite - best at speaking :3 To be honest the writing aspect at A2 annoys me. It can be very tedious!
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    (Original post by Zacken)
    The latter, I think
    Heyy man so I tried it but I keep hitting a standstill :/ The book comes with a diagram to help but it's just making me more confused
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    Need help on an FP2 question! :erm:

    Question 7 on the link:
    http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF
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    (Original post by RoseGatz)
    Need help on an FP2 question! :erm:

    Question 7 on the link:
    http://filestore.aqa.org.uk/subjects...W-QP-JAN09.PDF
    Sure thing. Which part do you need help with, what's your specific trouble?
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    (Original post by Zacken)
    Sure thing. Which part do you need help with, what's your specific trouble?
    Part (a) in particular, no idea where the -1 and x (the one which is outside the square root) come from.
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    (Original post by RoseGatz)
    Part (a) in particular, no idea where the -1 and x (the one which is outside the square root) come from.
    Ah, okay - well you're using the chain rule here:

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }}{\mathrm{d}x}\left(\cosh^{-1} \frac{1}{x}\right) = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{x}\right) \times \frac{1}{\sqrt{\frac{1}{x^2} -1}} \end{equation*}
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    (Original post by Zacken)
    Ah, okay - well you're using the chain rule here:

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }}{\mathrm{d}x}\left(\cosh^{-1} \frac{1}{x}\right) = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{x}\right) \times \frac{1}{\sqrt{\frac{1}{x^2} -1}} \end{equation*}
    Chain rule, why?
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    (Original post by RoseGatz)
    Chain rule, why?
    Because you're differentiating a function of the form f(g(x)), in this case: f(x) = \cosh^{-1} x and g(x) = \frac{1}{x}.
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    (Original post by Zacken)
    Because you're differentiating a function of the form f(g(x)), in this case: f(x) = \cosh^{-1} x and g(x) = \frac{1}{x}.
    Oh yeah :facepalm:Of course, differentiating a function means you have to times by the derivative. Okay so once I've differentiated 1/x, I get -1/x^2 ... where do I go now? (Feels like I'm asking a stupid question, just so tired atm)
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    (Original post by RoseGatz)
    Oh yeah :facepalm:Of course, differentiating a function means you have to times by the derivative. Okay so once I've differentiated 1/x, I get -1/x^2 ... where do I go now? (Feels like I'm asking a stupid question, just so tired atm)
    Well - for example, if you were differentiating \sin x^2, you would do: \frac{d}{dx} (x^2) \times \cos x^2 - that is, you'd multiply the derivative of x^2 by the derivative of sin x except replacing x with x^2 in the derivative of \sin x. If you get what I mean?

    So in this case, you know that \frac{d}{dx} \cosh^{-1} x = \frac{1}{\sqrt{x^2 - 1}}, then because you're differentiating \cosh^{-1} \frac{1}{x} you'll replace all instances of x with \frac{1}{x} and multiply that by the derivative of \frac{1}{x}.
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    (Original post by Zacken)
    Well - for example, if you were differentiating \sin x^2, you would do: \frac{d}{dx} (x^2) \times \cos x^2 - that is, you'd multiply the derivative of x^2 by the derivative of sin x except replacing x with x^2 in the derivative of \sin x. If you get what I mean?

    So in this case, you know that \frac{d}{dx} \cosh^{-1} x = \frac{1}{\sqrt{x^2 - 1}}, then because you're differentiating \cosh^{-1} \frac{1}{x} you'll replace all instances of x with \frac{1}{x} and multiply that by the derivative of \frac{1}{x}.
    Yes I get all that, that's not the bit I'm struggling with. It's the obvious bit, so to speak, that I just can't get. Once I've differentiated it, how do I get it into the form it wants?
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    (Original post by RoseGatz)
    Yes I get all that, that's not the bit I'm struggling with. It's the obvious bit, so to speak, that I just can't get. Once I've differentiated it, how do I get it into the form it wants?
    Ah, okay:

    \displaystyle

 \begin{equation*}-1 \times \frac{1}{x^2} \times \frac{1}{\sqrt{\frac{1}{x^2} - 1}} = -1 \times \frac{1}{x^2 \sqrt{\dfrac{1-x^2}{x^2}}} = -\frac{1}{x^2 \dfrac{\sqrt{x^2-1}}{\sqrt{x^2}}} = \cdots\end{equation*}
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    (Original post by Zacken)
    Ah, okay:

    \displaystyle

 \begin{equation*}-1 \times \frac{1}{x^2} \times \frac{1}{\sqrt{\frac{1}{x^2} - 1}} = -1 \times \frac{1}{x^2 \sqrt{\dfrac{1-x^2}{x^2}}} = -\frac{1}{x^2 \dfrac{\sqrt{x^2-1}}{\sqrt{x^2}}} = \cdots\end{equation*}
    Oh! Now it just seems so obvious. My mind just kept drawing blanks, I blame it on the tiredness haha. Thanks for the help!
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    (Original post by RoseGatz)
    Oh! Now it just seems so obvious. My mind just kept drawing blanks, I blame it on the tiredness haha. Thanks for the help!
    Haha, yeah I get you with the tiredness bit, it's currently 2 a.m here and I can't for the life of me multiply these stupid confidence intervals in S3 correctly, argh! :lol:

    No problemo!
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    Got a question I'm a bit stuck on, care to help Zacken or anyone else for that matter? It's more pure stuff :flutter:

    The function \displaystyle \begin{equation*}f\end{equation*  } is defined by

    \displaystyle \begin{equation*}f(x)= \frac{1}{x^3(1+Inx)}\end{equatio  n*} where \displaystyle \begin{equation*}x>0\end{equatio  n*}

    Show that \displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*} as \displaystyle \begin{equation*}x \rightarrow{0}\end{equation*}

    I've managed to show that \displaystyle \begin{equation*}f(x) \rightarrow\infty\end{equation*} instead of showing \displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*}, got no idea where the minus is coming from.
 
 
 
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