Question

F'(t)=2cost+sec^2t -pi/2 10 years ago

Answer 1

@zepdrix

Answer 2

\[\large\rm f'(t)=2\cos t+\sec^2t,\qquad\qquad f\left(\frac{\pi}{3}\right)=4\]So what are you stuck on? :o

Answer 3

I think I did the f(t) right

Answer 4

F(t)=2sint+tan(t)+c?

Answer 5

Is that right? :O

Answer 6

k good

Answer 7

So we have a "general solution".\[\large\rm f(t)=2\sin t+\tan t+c\]They gave us `initial data` which we can use to figure out the unknown constant.
This will give us a "particular solution"

Answer 8

:O

Answer 9

So does it have to equal 4?

Answer 10

Hmmmm no?

Answer 11

it has to equal 4 when \(t=\frac{\pi}{3}\)

Answer 12

So the t will be pi/3 and the whole thing equals 4?

Answer 13

yes, then you can solve for \(C\)

Answer 14

Okay so

Answer 15

Square root of 3+1=4

Answer 16

\[\large\rm f(t)=2\sin t+\tan t+c\\
\large\rm f(\frac{\pi}{3})=2\sin(\frac{\pi}{3})+\tan (\frac{\pi}{3})+c=4\]

Answer 17

not quite, but close
\[\tan(\frac{\pi}{3})\neq 1\]

Answer 18

It doesn't equal 1? :O

Answer 19

Oh yeah ughhhh

Answer 20

Square root of 3

Answer 21

So square root 3 + square root of 3 = 2square roots of 3

Answer 22

So .539

Answer 23

Can you help me with one more?