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    I need help with q7... I tried to change the limits in terms of x but I was getting the same number!!! How am I supposed to change them
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    (Original post by pondsteps)
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    I need help with q7... I tried to change the limits in terms of x but I was getting the same number!!! How am I supposed to change them
    Correct, you get the same number so it is like cancelling itself out.. however, the graph in the first part is supposed to give you an idea as to how you can integrate it

    *At least I think so, I haven't attempted it
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    (Original post by pondsteps)
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    I need help with q7... I tried to change the limits in terms of x but I was getting the same number!!! How am I supposed to change them
    I haven't sketched the curve out but I'm assuming the limits are different and you're doing something wrong.

    To find the new limits:
    Sub x limit into equation of t^2 - 1 and same for other x limit and find new limits for t.

    Use the chain rule to make it all in terms of t and integrate.

    Remember dx/dt * dt is dx.


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    Can someone pls explain to me part 8bi and ii.. Why is it y>0 and y<0 .. I don't understand the whole question :/
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    (Original post by pondsteps)
    Can someone pls explain to me part 8bi and ii.. Why is it y>0 and y<0 .. I don't understand the whole question :/
    When t is 0, y is what value?

    With this in mind, if t is greater than 0, what range of values can y take?

    Use the parametric equations rather than trying to figure it out from the graph
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    (Original post by SeanFM)
    When t is 0, y is what value?

    With this in mind, if t is greater than 0, what range of values can y take?

    Use the parametric equations rather than trying to figure it out from the graph
    Ohh right.. But why are we using the y and noy the x
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    I just trued it for the x and for iii I got x<-2 but that isn't possible because that's not even part of the limit!! Why did I do wrong ? The y part worked fine but idk why the x isn't working SeanFM
    (I mean the limit of the x from the graph not the t)
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    (Original post by pondsteps)
    I just trued it for the x and for iii I got x<-2 but that isn't possible because that's not even part of the limit!! Why did I do wrong ? The y part worked fine but idk why the x isn't working SeanFM
    (I mean the limit of the x from the graph not the t)
    Easy mistake to make, but it is not x<-2. It is actually x>-2. Can you see why?
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    (Original post by SeanFM)
    Easy mistake to make, but it is not x<-2. It is actually x>-2. Can you see why?
    Does the sign flip when we square?
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    (Original post by pondsteps)
    Does the sign flip when we square?
    If I had the graph y = x^2 - 2 for all values of x, what is the stationary point? And is it a minimum or maximum?

    If I limit it to just x<0, is there any difference in the stationary and minimum or maximum point?

    Now what does this have to do with your t<0 situation?
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    (Original post by SeanFM)
    If I had the graph y = x^2 - 2 for all values of x, what is the stationary point? And is it a minimum or maximum?

    If I limit it to just x<0, is there any difference in the stationary and minimum or maximum point?

    Now what does this have to do with your t<0 situation?
    Ugh I'm so sorry I understood what ur saying but I don't know how it relates to our case! I feel quite dumb
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    (Original post by pondsteps)
    Ugh I'm so sorry I understood what ur saying but I don't know how it relates to our case! I feel quite dumb
    It is identical, just doesn't involve parametrisation.

    Our situation is that we have t<0 and x= t^2 -2.

    This is just like saying y = x^2 - 2 where our x = the y in our example and our t = the x in our example.

    Or if you want to look at it again, what different values can x take when t<0? This is in x = t^2 - 2.

    We know that at t=0 (which is not part of the domain) x = -2.

    Keeping this in mind, if we choose random and easy values t<0 such as t = -1, t= -3, x = -1 and 7 respectively. So we can see that since x^2 will always be a positive number and since x can never be -2 exactly because t<0 doesn't cover t=0, so we can conclude that x > -2.

    You can also draw a graph of y = x^2 - 2 where x < 0, it is identical to our situation.
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    (Original post by SeanFM)
    It is identical, just doesn't involve parametrisation.

    Our situation is that we have t<0 and x= t^2 -2.

    This is just like saying y = x^2 - 2 where our x = the y in our example and our t = the x in our example.

    Or if you want to look at it again, what different values can x take when t<0? This is in x = t^2 - 2.

    We know that at t=0 (which is not part of the domain) x = -2.

    Keeping this in mind, if we choose random and easy values t<0 such as t = -1, t= -3, x = -1 and 7 respectively. So we can see that since x^2 will always be a positive number and since x can never be -2 exactly because t<0 doesn't cover t=0, so we can conclude that x > -2.

    You can also draw a graph of y = x^2 - 2 where x < 0, it is identical to our situation.
    So we just switch the sign if it doesn't satisfy the limits
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    (Original post by pondsteps)
    So we just switch the sign if it doesn't satisfy the limits
    No... there is a reason why the sign is the way round it is.

    Try rereading what I said, ask questions if anything is unclear, plot the graph of x = t^2 - 2 for t<0.

    If you are stuck after those things then let me know.
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    (Original post by SeanFM)
    No... there is a reason why the sign is the way round it is.

    Try rereading what I said, ask questions if anything is unclear, plot the graph of x = t^2 - 2 for t<0.

    If you are stuck after those things then let me know.
    I didn't understand this part!!
    Keeping this in mind, if we choose random and easy values t<0 such as t = -1, t= -3, x = -1 and 7 respectively. So we can see that since x^2 will always be a positive number and since x can never be -2 exactly because t<0 doesn't cover t=0, so we can conclude that x > -2.
    I just don't get why the sign is flipped ughhhhh
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    (Original post by pondsteps)
    Can someone pls explain to me part 8bi and ii.. Why is it y>0 and y<0 .. I don't understand the whole question :/
    When t=0 you have (-2,0) as a point


    When t&gt;0 you have x=t^2-2&gt;-2 \Rightarrow x&gt;-2

    and y=2t&gt;t&gt;0 \Rightarrow y&gt;0


    Then for t&lt;0 what do we have??


    What's confusing?
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    (Original post by SeanFM)
    If I had the graph y = x^2 - 2 for all values of x, what is the stationary point? And is it a minimum or maximum?

    If I limit it to just x<0, is there any difference in the stationary and minimum or maximum point?

    Now what does this have to do with your t<0 situation?
    Okay I understood this part.. There is no change..And?
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    (Original post by pondsteps)
    Okay I understood this part.. There is no change..And?
    Firstly, did you understand why when t > 0, y > 0 and when t = 0, why y = 0?
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    I am hoping you understand my solution here because I have really simplified what others are saying above using different concepts, hopefully a simpler approach:

    The parametric equation is x = t^2 - 2...
    If you draw that graph out, it is x = t^2 and then translated 2 downwards. Think of it was y = x^2 but with the variables replaced and y = f(x) so in this case x = f(t). If you do x = f(t) - 2 you will get x = t^2 - 2 which is a translation of f(t) or t^2 by 2 units downwards (down the y/vertical-axis).

    What you've got to realise is that x = t^2 graph is symmetrical because b^2 = 4ac , it has two equal roots of 0.
    Therefore, when you translate it 2 units downwards - it is still symmetrical. So, if when t > 0, we have acknowledged that t^2 - 2 > 0, then when t < 0, t^2 - 2 will still be > 0 because the graph is symmetrical - it is a reflection in the line t = 0 or the x-axis. I hope that makes things clearer, it is a symmetrical graph and therefore x > -2 when -∞ < t < ∞, t ≠ 0. This is the same as t > 0 or t < 0 lol.
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    (Original post by Chittesh14)
    Firstly, did you understand why when t > 0, y > 0 and when t = 0, why y = 0?
    Yeah you just substitute in the zero and put the sign right
 
 
 
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