watchmath
  • watchmath
Limit and circle The problem is attached
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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watchmath
  • watchmath
\(C_1\) is a fixed circle given by \((x-1)^2+y^2=1\) 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 \(r\to 0^+\)?
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anonymous
  • anonymous
it moves toward the y axis
anonymous
  • anonymous
if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

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anonymous
  • anonymous
PLS HELP
watchmath
  • watchmath
@ishan Do you mean \(\lim_{r\to 0^+}R=0\)? That is not correct!! Try again :D
anonymous
  • anonymous
the Y(op)-->0 and PR-->0R
anonymous
  • anonymous
if a^3 = b^3 + c^3 + d^3 , then least value of a is ?
myininaya
  • myininaya
eventually 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
  • myininaya
P=Q because C1 and C2 intersect at the origin
watchmath
  • watchmath
You need to work out this more rigorously. Your intuition can deceive you here.
myininaya
  • myininaya
the 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
  • watchmath
But we are interested in finding R (as r-->0 ). How is the fact that you mention helping us?
myininaya
  • myininaya
R is the x-intercept of PQ so i'm not seeing how the line PQ isn't approaching the x-axis or y=0
myininaya
  • myininaya
meaning R would be every point on the x-axis since PQ is that line I don't know its late
myininaya
  • myininaya
i go to sleep and dream about it tonight k?
myininaya
  • myininaya
oops this morning its 3:34 am
myininaya
  • myininaya
well if the slope is approaching 0 and the line isn't moving toward the x-axis, then that would mean the line PQ is getting closer to not having a x-intercept (R) since PQ is a constant not equal to 0
watchmath
  • watchmath
It is 4:42 here :). I think dumbcow will answer this, right? :D
dumbcow
  • dumbcow
limit as r->0 is 4
dumbcow
  • dumbcow
not 100% but i was able to get R as a function of r
watchmath
  • watchmath
Yes, then now it is just an exercise of computing limit.
dumbcow
  • dumbcow
\[R =\frac{r^{2}}{2-\sqrt{4-r^{2}}}\] use L'hopitals rule
dumbcow
  • dumbcow
first 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 x-intercept
myininaya
  • myininaya
thats right Q is the intersection of the two circles :(
dumbcow
  • dumbcow
btw thanks for the diagram could not have done this without that, im a visual person :)
myininaya
  • myininaya
ok i guess i see how it can be 4 lol
myininaya
  • myininaya
what 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
  • myininaya
P=(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
dumbcow
  • dumbcow
yes slope is getting flatter which pushes the x-intercept farther out so there is just a limit to how far out it can go so you are not wrong on that point
myininaya
  • myininaya
but it never totally goes flat okay i get it
myininaya
  • myininaya
meaning there is a x-intercept
dumbcow
  • dumbcow
right P approaches the origin and PQ approaches the x-axis but dont ever get there :)
myininaya
  • myininaya
because we never get to the origin lol
myininaya
  • myininaya
thanks cow
dumbcow
  • dumbcow
your welcome watchmath did that answer all your questions
myininaya
  • myininaya
watchmath give me another i want to prove i can think like a pro
myininaya
  • myininaya
don't do it now wait til tomorrow let me sleep if you give it to me now then i will never sleep
anonymous
  • anonymous
ok the fixed circle EQ is (x-1)^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
  • anonymous
now in this two EQ, you can compute the intersection PQ and get its X and Y
anonymous
  • anonymous
now 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
  • anonymous
now 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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