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#1 2007-01-23 10:19:24
- luca-deltodesco
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implicit curve, stationary points
this came up in my exam today for Core 4, im wondering if i did it correctly:
A curve is defined implicitly by the equation
Show that the curve has two stationary points, and find their coordinates:
dy/dx:
since were looking for stationary points, dy/dx = 0
any value of x and y satisfying this will give dy/dx = 0, substitute y = -2x
quadratic equation, so there are two roots, two stationary points
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#2 2007-01-23 10:53:57
- mathsyperson
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Re: implicit curve, stationary points
That's how I'd have done it as well. I'd say you got full marks on that question, let's hope the examiner agrees. 
How did you think you did with the rest of it?
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#3 2007-01-23 10:55:45
- luca-deltodesco
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Re: implicit curve, stationary points
apart from one part of one question, well. (this particular single part of a particular question i didnt even manage to take an educated guess at, it just completely stumped me, i cant remember what it was now)
exam was deffinately alot harder than core 3 exam was, i flew through that one (did core 3 on thursday)
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#4 2007-01-26 17:05:36
- George,Y
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Re: implicit curve, stationary points
"Neither x nor y is 0." must be an additional condition
Last edited by George,Y (2007-01-26 17:06:06)
X'(y-Xβ)=0
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#5 2007-01-26 18:58:43
- luca-deltodesco
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Re: implicit curve, stationary points
"Neither x nor y is 0." must be an additional condition
why?
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#6 2007-01-26 22:45:35
- George,Y
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Re: implicit curve, stationary points
Because under the condition x=-2y, x and y could be both 0, then dy/dx could be any number while the equation still stands.
X'(y-Xβ)=0
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#7 2007-01-27 01:17:13
- mathsyperson
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Re: implicit curve, stationary points
Oh, I see what you're saying.
luca just substituted dy/dx=0 once he'd done the differentiation, but technically he should have rearranged it to get dy/dx as the subject.
If he had, he would have got
.So then by equating that with 0, he would have got the same conclusion that 2x+y = 0, but also the additional one that x+2y ≠0.
In this case though, it doesn't matter, because the solutions that he found satisfied the missing condition anyway.
Why did the vector cross the road?
It wanted to be normal.
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