Question

find the values of a,b and c that make the function \[f(r) = ar^{2} + br + c; \] \[r<0 , 2\sin^{2}(r) + 3\cos^2(r)\] \[r \ge 0\] twice differentiable at r=0.

Answer 1

ok ready?

Answer 2

^.^i already pasted it to a notebook xD let me delete the old post

Answer 3

ok yes i'm ready lets go

Answer 4

no damn!

Answer 5

o.O ?

Answer 6

let me say it in english

Answer 7

for this to be continuous, the limit from the left must be equal to the limit from the right at 0

Answer 8

the limit from the right at 0 is 3
the limit from the left at 0 is C
this means C = 3

Answer 9

hmm ahh yea i see!

Answer 10

now i will try to write it in math
\[\lim_{r\rightarrow 0^-}f(r)=\lim_{r\rightarrow 0^-} ar^2+br+c=c\]
\[\lim_{r\rightarrow 0^+}f(r)=\lim_{r\rightarrow 0^+}2\sin^2(r)+3\cos^2(r)=2\sin^2(0)+3\cos^3(0)=3\]

Answer 11

so we know C = 3

Answer 12

AHH WoW, now i see man

Answer 13

now to b
\[f'_-(r)=2a+b\]
\[f'_+(r)=2\cos(r)\sin(r)-6\cos(r)\sin(r)\]
so
\[f'_-(0)=b\] and
\[f'_+(0)=0\] so
\[b=0\]

Answer 14

and finally
\[f''_-(r)=2a\]
\[f''_+(r)=-2\cos(2r)\] so
\[f''_-(0)=2a\] and
\[f''_+(0)=-2\] thus
\[2a=-2\iff a=-1\]

Answer 15

ahh yea

Answer 16

really good job :D

Answer 17

and your function is
\[f_-(r)=-r^2+3\]

Answer 18

yeah well it comes with no guarantee, but i believe it is correct

Answer 19

sorry it took so long to write fingers are slow today

Answer 20

np man :D

Answer 21

and thanks much

Answer 22

btw this line
\[f'_+(r)=2\cos(r)\sin(r)-6\cos(r)\sin(r)\] was a mistake, it should have been
\[f'_+(r)=4\cos(r)\sin(r)-6\cos(r)\sin(r)\]
but you still get 0 at 0 so no harm done

Answer 23

and the second derivative is correct. i just typed it wrong here

Answer 24

:D oka :) Really gj

Answer 25

cool ......lol

Answer 26

:D can you have a look at the other question which is new :)