y = 9ln(x)
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2
curvature k = |y''| / (1 + (y')^2)^(3/2)
<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2)
= (9/x^2) / (1 + 81/x^2)^(3/2)
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)]
= 9x(x^2 + 81)^(-3/2).
To maximize the curvature, </span>
we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2)
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2]
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2)
Setting k' = 0 yields x = ±9/√2.
Since k' < 0 for x < -9/√2 and k' > 0 for x >
-9/√2 (and less than 9/√2),
we have a minimum at x = -9/√2.
Since k' > 0 for x < 9/√2 (and greater than 9/√2) and
k' < 0 for x > 9/√2,
we have a maximum at x = 9/√2. </span>
x=9/√2=6.36
<span>y=9 ln(x)=9ln(6.36)=16.66</span>
the
answer is
(x,y)=(6.36,16.66)
X by its self is always 1x
1x+8=-15 you have to get rid of 8 by subtracting on both sides
1x=-23 now get rid of 1 by dividing on both sides
x=-23
Answer:
for the 1st one the answer is x= 3 y=1 and the second one is x= 5 and y = 3 and 1/3
Step-by-step explanation: so basically for the first one -5 times x which is 3 is -15 right? and then 5 times y which is 1 is 5 right so -15 plus 5 = -10
second one is x = 5 and y = 3 and 1/3 because 5*x which is 5 is = to 15 right? and then 3 times y which is equal to 3 and 1/3 is equal to 10 which when you do 15-10 in pretty sure it equals 5 correct me if im wrong
Hope this helps xD ;)
Wats the question?????????
