Question

Find the length of the curve y=(x^2/2) - (lnx/2) ; 2 less than or equal to x which is less than or equal to 4.

Answer 1

parameterize the curve as

(t, (t^2/2 - ((ln t)/2))

arc length is
\[\int\limits_{t_{1}}^{t_{2}} ds\]

Answer 2

where

\[ds = \sqrt { (dx/dt)^2 + (dy/dt)^2} dt\]

Answer 3

u understand>??

Answer 4

hmmm i think so, i'm trying to follow along

Answer 5

so whts dx/dt and dy/dt?

Answer 6

derivatives of "x^2" and "ln(x)"?

Answer 7

ill mae it easy

x=t
y=t^2 / 2 - (lnx)/2

now whts dx/dt and dy/dt?

Answer 8

sorry, but i still don't get it :/

Answer 9

DO U KNOW PARAMETERIZATION?

Answer 10

ok lets try something else

\[ds = \sqrt{ 1 + (dy/dx)^2 } dx\]

Answer 11

now can u tell me dy/dx??

Answer 12

u ther?

Answer 13

no i don't know paramertization, sorry

Answer 14

ok tell me dy/dx now

Answer 15

where r u?

Answer 16

i'm right here. i'm just looking up parameterization

Answer 17

dont

Answer 18

u dont need to

Answer 19

just tell me the value of dy/dx?

Answer 20

okay umm would it be t- 1/x?

Answer 21

y = (x^2 - ln x)/2
dy/dx = x - 1/2x

Answer 22

how did you get x? oh wait no, it's be
cause t=x nvm

Answer 23

leave t out now

Answer 24

okay

Answer 25

so now i told u

ds = sqrt [ 1 + (dy/dx)^2] dx

Answer 26

now give me ds..whats ds? in terms of x?

Answer 27

okay hang on

Answer 28

would ds = (x^2 - 2x +2)/(4x^2)

?

Answer 29

yes most prob

Answer 30

so now evaluate this wrt integral dx from x1 to x2

Answer 31

square root of this

Answer 32

its not right ...ds

Answer 33

\[ds = \sqrt { x^2 + 1/4x^2}dx\]

Answer 34

oooohhhh okay i'm writing this down and taking notes