Solving for f(t) in the equation f'(t)=8cos(t)+sec^2(t), given f(pi/3)=4

[Rasine]

Find f if f '(t)=8cost+sec^2t and f(pi/3)=4 .

so f(t)=8sint+tant+c

4=8sin(pi/3)+tan(pi/3)+c

c=-3^1/2

so f(t)=8sint+tant-3^1/2 is that right

 

[neutrino]

You need to re-evaluate the values of sin and tan at pi/3.

 

[Rasine]

thank you very much

 

[Rasine]

sin(pi/3) is sqroot3/2 and tan(pi/3) is sqroot3 and 8(sqroot3/2)+sqroot3 is
4sqroot3+sqroot3=5sqroot3

is that right?

 

[neutrino]

Yes.

But for the sake of clarity, use something like sqroot(3)/2 instead of sqroot3/2 when referring to $$\frac{\sqrt{3}}{2}$$.

 

[Rasine]

ok. so now i have (5sqroot(3)/2)+c=4

then c=4-5sqroot(3)/2 so that would be -sqroot3

is that right

 

[neutrino]

then c=4-5sqroot(3)/2

Why the 2 in the denominator?

so that would be -sqroot3

I don't understand. You've already solved for c, so...

f(t) = 8sin(t) + tan(t) + 4-5sqroot(3)