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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 10022017 14:50

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 10022017 18:37
(Original post by pondsteps)
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
*At least I think so, I haven't attempted it 
Chittesh14
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 10022017 23:13
(Original post by pondsteps)
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
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.
Posted from TSR MobileLast edited by Chittesh14; 10022017 at 23:15. 
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 11022017 07:18
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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 11022017 10:42
(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 :/
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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 11022017 10:45
(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 
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 11022017 11:29
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)Last edited by alevels2k17; 11022017 at 11:33. 
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 11022017 11:41
(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) 
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 11022017 12:14
(Original post by SeanFM)
Easy mistake to make, but it is not x<2. It is actually x>2. Can you see why? 
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 11022017 12:30
(Original post by pondsteps)
Does the sign flip when we square?
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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 11022017 13:06
(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? 
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 11022017 13:15
(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
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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 11022017 15:01
(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. 
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 11022017 16:01
(Original post by pondsteps)
So we just switch the sign if it doesn't satisfy the limits
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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 11022017 16:32
(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.
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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 11022017 16:44
(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 you have
and
Then for what do we have??
What's confusing?Last edited by RDKGames; 11022017 at 16:46. 
alevels2k17
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 11022017 17:19
(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? 
Chittesh14
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 11022017 19:29
(Original post by pondsteps)
Okay I understood this part.. There is no change..And? 
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 11022017 19:49
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/verticalaxis).
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 xaxis. 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.Last edited by Chittesh14; 11022017 at 19:55. Reason: mistake 
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 11022017 20:05
(Original post by Chittesh14)
Firstly, did you understand why when t > 0, y > 0 and when t = 0, why y = 0?
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