Question

f '(t)=t + (1/t^3), t>0, f(1)=6 Find f

I got f (t) = (t^2/2) + (1/2t^2) + c

.. stuck after that..

Answer 1

oh! i put the one in where t's are, and make it equal to 6..?

Answer 2

i think you integrated \(\frac{1}{t^3}\) wrongly

Answer 3

i did?

Answer 4

oh it should be -2

Answer 5

yeah looks like so
\[\frac{1}{t^3} => t^{-3} => \int t^{-3}dt\]

Answer 6

it shoulkd be \(\frac{1}{2x^2}\)

Answer 7

there's a minus infront of it lol

Answer 8

yes so f (t) = (t^2/2) - (1/2t^2) + c

Answer 9

yes, sub in 1 then set it = 6 then calculate for c then you're done

Answer 10

like t= 1

Answer 11

c= 5! thanks for the catch!

Answer 12

:)