Question

Find f.
f '(t) = 6 cos (t) + sec^2 (t), −π/2 < t < π/2, f(π/3) = 3

Answer 1

this requires integration

Answer 2

or if you don't know that word
antiderivatives!

Answer 3

what is the antiderivative of
cos(t) and sec^2(t)?

Answer 4

sin(t) + C and tan(t) + C

Answer 5

\[f'(t)=6 \cos(t)+\sec^2(t) \\ f(t)=6 \sin(t)+\tan(t)+C\]
then we can agree on this right?:

Answer 6

now just apply f(pi/3)=3 to find C

Answer 7

I'm not sure how to do that

Answer 8

f(pi/3)=3
meants replace f(t) with 3
and replace t with pi/3

Answer 9

\[3=6 \sin(\frac{\pi}{3})+\tan(\frac{\pi}{3})+C\]
then solve for the constant C

Answer 10

I got C = 0.40 (rounded to 2 decimal places)

Answer 11

when you solved for C?

Answer 12

how did you get that

Answer 13

yes, exact answer is -3Root3 / 2

Answer 14

[(-3Root3) / 2 ]+3 I meant

Answer 15

something looks a little fishy about that
you know sin(pi/3)=sqrt(3)/2
and cos(pi/3)=1/2
so tan(pi/3)=sqrt(3)
right?

Answer 16

\[3=6 \sin(\frac{\pi}{3})+\tan(\frac{\pi}{3}) +C \\ 3=6 \frac{\sqrt{3}}{2}+\sqrt{3}+C \\ 3=\frac{6}{2} \sqrt{3}+\sqrt{3}+C \\ 3=3 \sqrt{3}+\sqrt{3}+C\]
how many sqrt(3)'s do you see on the right hand side

Answer 17

2

Answer 18

I see 4.

Answer 19

ah 3 + 1

Answer 20

\[3 \sqrt{3}+\sqrt{3} \\ \sqrt{3}+\sqrt{3}+\sqrt{3}+\sqrt{3} \\ 4 \sqrt{3}\]
yep yep :)

Answer 21

\[3=4 \sqrt{3}+C\]

Answer 22

now try solving for C

Answer 23

-4sqrt3 + 3

Answer 24

tons cute :)

Answer 25

Now they were looking for your f that satisfied that given condition
so just replace out C earlier with this number you found

Answer 26

our*

Answer 27

\[f(t)=6\sin(t)+\tan(t)+C\]
that is replace this C with -4 sqrt(3)+3
and you are done

Answer 28

so f(x) = 6sin(t) + tan(t) + -4sqrt(3) + 3