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come on no has helped you with this yet?

no

should i solve this or go to sleep.....

solve before going, please

Oops! forgot to sleep after going

okay, first things first, lets find the first derivaitve of the function your where given

can you find it?

9sec^(2)x

okay good

now can you please evaluate:
f(pi/4)=9sec^2(pi/4)

square root of 2

is it right?

basically to make it more accesible we can write this as:
f(pi/4)=9[sec(pi/4)]^2

further more we can write it as: f(pi/4)=9[1/cos(pi/4)]^2

then evalutaing we get:
f(pi/4)=9[1/sqrt(2)/(2)]^2

what square root 2/2???

further: f(pi/4)=9[sqrt(2)]^2
which the gives us: f(pi/4)=9(2), which is 18
Thus f(pi/4)=18

oh, thought your doing sec

i rewrote secx as 1/cosx

how come?

just to make is easier for me, but if realise that sec(pi/4) is sqrt(2) then more power to you

Now we have found the slope (m) of our line, it is 18

y-y=m(x-x)

yes, but we will use this form, y=mx+b

i am getting sleepy lets finish this problem

ok

what would be the b in the slope formula

now lets substitute everyting in:
y=18x-(36-18pi)/(4)

are we good?

36??

i combined 9 and 18pi/4

yes?

still don't see how you can 36??

|dw:1316152625201:dw|

you could've just simplify it to 9pi/2

okay, then smarty pants do it your way, regardless we came to a correct answer

sorry, so the finally would look like this 18x+9(1-pi/2)

sure

distribute the negative sign, first

huh?

what negative sign?

|dw:1316153351852:dw|

got it ?

but was there a 9-18pi/4 at the beginning

yes there was

can you not see it:
|dw:1316153548845:dw|

I meant why was 9-18pi/4 at the begginig

okay, so can we lay this problem to rest?

I guess, but still don't get why they have 9-18pi/4 plug into b instead of just 18pi/4 by itself

|dw:1316153804630:dw|

oh, ok now we can lay this problem to rest, thanks so much for being patience

its good you ask questions, i know i can confusing sometimes :)

Thanks for giving me a metal

yeah I just want to understand the problem instead of just getting the answer

because I might see it on the test, but thanks again

anytime