Question

Find an expression for the arc length of the upper half of a circle of radius 25 on the interval [0,t] where (0 11 years ago

Answer 1

@satellite73 yo can u try this

Answer 2

finally !!!

Answer 3

i wrote 25 pi/2 but it was wrong then i wrote t*pi/2 that was wrong too

Answer 4

we have to use arclength formula just incase

Answer 5

i forget
is it someting like
\[\int_a^b\sqrt{1+(f'(x))^2}\]

Answer 6

ya

Answer 7

ya thats correct

Answer 8

25 is a huge radius

Answer 9

in this case the circle with radius 25 has the equation
\[f(x)=\sqrt{25^2-t^2}\]

Answer 10

yep

Answer 11

it is rather large, but not as large as i am old

Answer 12

hey i dont judge a question by its size @myininaya

Answer 13

=P

Answer 14

take the derivative, get
\[\frac{t}{\sqrt{25^2-t^2}}\]

Answer 15

and you could actually compare this answer by doing it the easy algebraic way

Answer 16

you know to check the answer

Answer 17

that is what i started to do actually but it required too much thinking

Answer 18

^

Answer 19

square it, add one, and take the square root of the result
halas

Answer 20

C=2pi*r
C=2pi*25=50pi
0 to 25 means we are looking at a quarter of the whole circle
C/4=50pi/4
50pi/4=25pi/2
hmmm...I got 25pi/2 like you did

Answer 21

is it 0<t<25
or -25<t<25?

Answer 22

0 to 25

Answer 23

oh it says find the expression
and doesn't say to evaluate

Answer 24

yeah

Answer 25

lol and its confusing because interval is from 0 to t and 0<t<25...so...

Answer 26

I think we she do the integral for 0 to t

Answer 27

I think that inequality is just saying where t has to be less than 25

Answer 28

basically we have to find such a t that it the arc length was from 0 to t (t<25) that we can get the answer

Answer 29

\[\int\limits_{0}^{t} \sqrt{1+(\frac{-t}{25^2-t^2})^2 } dx\]
and satellite already found the derivative of f

Answer 30

oops those t's are x's

Answer 31

except the limit t

Answer 32

the question just asks for the expression

Answer 33

\[\int\limits\limits_{0}^{t} \sqrt{1+(\frac{-x}{25^2-x^2})^2 } dx \]

Answer 34

still a typo there dear

Answer 35

whatever I missed the square root

Answer 36

\[\int\limits\limits_{0}^{t} \sqrt{1+(\frac{-x}{\sqrt{25^2-x^2}})^2 } dx\]

Answer 37

\[\int\limits\limits\limits_{0}^{t} \sqrt{1+(\frac{-x}{\sqrt{25^2-x^2}})^2 } dx \]

Answer 38

too many latex symbols to type

Answer 39

or \[\sqrt{1+\frac{x^2}{25^2-x^2}}\]

Answer 40

alright you guys i got it right

Answer 41

i had it right before just that i was evaluating from 0 to 25 when i was just spposed to sub in for t...sigh didnt read...