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watchmath
 5 years ago
Limit and circle
The problem is attached
watchmath
 5 years ago
Limit and circle The problem is attached

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0\(C_1\) is a fixed circle given by \((x1)^2+y^2=1\) and \(C_2\) is a shriking circle with radius \(r\). \(P\) is the point of \((0,r)\), \(Q\) is the upper point of intersection of the two circles,a nd \(R\) is the point of intersection of the line \(PQ\) and the \(x\)axis. WHat happens to \(R\) as \(C_2\) shrinks, that is, as \(r\to 0^+\)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it moves toward the y axis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0@ishan Do you mean \(\lim_{r\to 0^+}R=0\)? That is not correct!! Try again :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the Y(op)>0 and PR>0R

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0eventually the shrinking circle will have radius 0 and it will just be a point on the origin and P=Q drawing a line from P to Q would then give a horizontal line on the x axis

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0P=Q because C1 and C2 intersect at the origin

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0You need to work out this more rigorously. Your intuition can deceive you here.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0the slope of PQR is r/R (using P=(0,r) and R=(R,0)) so as r goes to zero the slope of PQR are getting closer to zero

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0But we are interested in finding R (as r>0 ). How is the fact that you mention helping us?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0R is the xintercept of PQ so i'm not seeing how the line PQ isn't approaching the xaxis or y=0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0meaning R would be every point on the xaxis since PQ is that line I don't know its late

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i go to sleep and dream about it tonight k?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0oops this morning its 3:34 am

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0well if the slope is approaching 0 and the line isn't moving toward the xaxis, then that would mean the line PQ is getting closer to not having a xintercept (R) since PQ is a constant not equal to 0

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0It is 4:42 here :). I think dumbcow will answer this, right? :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not 100% but i was able to get R as a function of r

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, then now it is just an exercise of computing limit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[R =\frac{r^{2}}{2\sqrt{4r^{2}}}\] use L'hopitals rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first i set the 2 circle equations equal to each other to find Q in terms of r then determined slope with P and Q used equation of a line to find R the xintercept

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0thats right Q is the intersection of the two circles :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btw thanks for the diagram could not have done this without that, im a visual person :)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok i guess i see how it can be 4 lol

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0what about what i mention? why is it wrong? i mean i know its wrong because it doesn't give the right answer, but what makes it wrong to do

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0P=(0,r) R=(R,0) the slope of PQ is r/(R) this shows as r goes to zero tha PQ is getting closer to being a horizontal line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes slope is getting flatter which pushes the xintercept farther out so there is just a limit to how far out it can go so you are not wrong on that point

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0but it never totally goes flat okay i get it

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0meaning there is a xintercept

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right P approaches the origin and PQ approaches the xaxis but dont ever get there :)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0because we never get to the origin lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your welcome watchmath did that answer all your questions

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath give me another i want to prove i can think like a pro

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0don't do it now wait til tomorrow let me sleep if you give it to me now then i will never sleep

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok the fixed circle EQ is (x1)^2 +y^2=1...the other circle is at the center with (h,k)=(0,0) therefore the EQ is x^2 + y^2 =r^2 are you geting it watch?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now in this two EQ, you can compute the intersection PQ and get its X and Y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now once you get the x and y, you can get its hypotenuse which is the PQ..then from there you can set the limit as r>0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now i dont know if Q is fixed there at that point, or changed position with the other circle, as r>0
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