Answer:
f(0,0)=ln19
Step-by-step explanation:
is given as continuous function, so there exist
and it is equal to f(0,0).
Put x=rcosA annd y=rsinA

we know that
, so we have that


So f(0,0)=ln19.
to calculate the slope m , use the ' gradient formula '
m = 
with (
,
) = (0, - 1 ) and (
,
) = (1, 5 )
m =
= 6
0