ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Over9000 on May 22, 2009, 10:25:10 pm
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I am having trouble with question 2d.
Can someone show me a proof?
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It must be an inverted parabola. The part smoothly joined to A is defined for x < -3. Clearly it does not pass through D (14, -126).
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Yeh but the inverted parabola doesnt satisfy smooth join, it makes an elbow, I asked my teacher and he said smooth join means must continue.
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The parabola is y = (-192/17)x^2 + (2112/17)x + 5922/17, x < -3.
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Yeh, I got that, the tricky part is, it says must prove cant have a smooth join, so how do I prove that the parabola doesnt smoothly join (i.e how to I prove it elbows?)
The proof must be rigourous also otherwise they might not allocate full marks
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You misunderstood the question!
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You misunderstood the question!
What do you mean? Please explain
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It is rigorous enough. Further explanation is unnecessary.
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So saying parabola equation is The parabola is y = (-192/17)x^2 + (2112/17)x + 5922/17, x < -3.
is enough for an assignment question?
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The question asked you to show that there is no parabola that meets both requirements, smoothly joined to A AND passing through D.
You can find a parabola that is NOT smoothly joined to A but passes through D.
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The question asked you to show that there is no parabola that meets both requirements, smoothly joined to A AND passing through D.
You can find a parabola that is NOT smoothly joined to A but passes through D.
I know but how do you explain algebraically that it isnt a smooth join, if you just put the equation as your answer how would teachers know that that implies it isnt a smooth join. You can just sketch a graph of both and show that the inverted parabola elbows but how do u prove it with algebra instead of just saying by inspection it isnt a smooth join?
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In fact it can be an inverted parabola or a 'normal' parabola.
The part smoothly joined to A is defined for x < -3. Clearly it does not pass through D (14, -126).
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Say what I said.
It must be an inverted parabola.
The part smoothly joined to A is defined for x < -3. Clearly it does not pass through D (14, -126).
Im not quite sure what you mean for x<-3, isnt the graph restricted to x>=3?
Also the parabola you found, which is the same to mine, passes through D
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x>=-3 is for the cubic, not for the parabola.
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I think that a smooth join means the first and second derivatives should be the same.
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I think he means that attatching a parabola to A is just extending the domain of the function to x less than -3, hence the parabola has domain (-infintiy,-3). When TT discussed this with me I assumed that you can just have a parabola with domain R, in that case both conditions(smoothness at A and passing through D) can be clearly satisfied). However in this case our new function (the union of the cubic and parabola) is not really a function since it is one-to-many for x>-3 and hence the reason why parabola is restricted to x<-3. (which trivially does not pass through D). (however the question never mentioned that the parabola and cubic are to form a piece-wise(hybrid) function) It's the wonderful ambiguity of vce questions me thinks :-\
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I think that a smooth join means the first and second derivatives should be the same.
I know it means first derivative, I just dont know how to prove that the parabola elbows with the cubic to show that its not a smooth join
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I think he means that attatching a parabola to A is just extending the domain of the function to x less than -3, hence the parabola has domain (-infintiy,-3). When TT discussed this with me I assumed that you can just have a parabola with domain R, in that case both conditions(smoothness at A and passing through D) can be clearly satisfied). However in this case our new function (the union of the cubic and parabola) is not really a function since it is one-to-many for x>-3 and hence the reason why parabola is restricted to x<-3. (which trivially does not pass through D). (however the question never mentioned that the parabola and cubic are to form a piece-wise(hybrid) function) It's the wonderful ambiguity of vce questions me thinks :-\
Yeh, what you have to understand is that basically, theyre trying to form a racetrack right, so I assume that u can really have a parabola going to - infinity as that wouldnt work, so theres on parabola that fits everything which is the inverted parabola, but how do you prove it isnt smoothly joined (i.e prove it elbows to the cubic), thats where im stuck.
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Another example, parabola y = 15x^2 + 282x, x < -3, smoothly joins A, but it does not pass through D.
Because the gradient of the cubic at A is positive, all parabolas that smoothly join to the cubic at A are defined for x < -3.
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I think he means that attatching a parabola to A is just extending the domain of the function to x less than -3, hence the parabola has domain (-infintiy,-3). When TT discussed this with me I assumed that you can just have a parabola with domain R, in that case both conditions(smoothness at A and passing through D) can be clearly satisfied). However in this case our new function (the union of the cubic and parabola) is not really a function since it is one-to-many for x>-3 and hence the reason why parabola is restricted to x<-3. (which trivially does not pass through D). (however the question never mentioned that the parabola and cubic are to form a piece-wise(hybrid) function) It's the wonderful ambiguity of vce questions me thinks :-\
Now Bigtick is a male.
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ok lol didn't read the racetrack bit, just the problematic part d.
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This is probably too late now to be of any use to you - sorry!
You seem to have worked out the equation to the parabola y = (-192/17)x^2 + (2112/17)x + 5922/17 and to do that you must have made the gradients of the cubic and the parabola equal to each other AND have used the points (-3, -126) and (14, 126).
Therefore you have found a parabola that satisfies the conditions that it smoothly join at A and passes through points A and D.
I think you have been given a trick (or poorly worded) question. From Question e (which talks about completing the track circuit from D back to A) , it appears as though the required parabola may also have been intended to complete the track from D to A. If that is the case, then they would want a positive parabola which would have a turning point at x = 5.5. However, that would have a negative gradient at x=-3 which would produce the "elbow" since the cubic has a positive gradient at that point.
You could argue that the parabola y = (-192/17)x^2 + (2112/17)x + 5922/17 fits the specification except it would be one helluva mess with half a dozen runners trying to turn at the resulting cusp at point A.
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Thanks plato.