A cup of coffee at 90 degrees celsius is put into a 30 degree celsius room when t =0 . The coffee's temperature, f (t ) , is changing at a rate given by f '(t )=-8(0 .8) t degrees celsius per minute, where t is in minutes.

Question

Estimate the coffee's temperature when t =10 :

Answer 1

If you mean f'(t) = -8 * 0.8^t, then
f(t) = -8/ln(0.8) * 0.8^t + C
since f(0) = 90,
90 = -8/ln(0.8) + C
C = 90 + 8/ln(0.8)

So,

f(t) = -8/ln(0.8) * 0.8^t + 90 + 8/ln(0,8)
= 90 + 8/ln(0.8) (1-0.8^t)
= 90 - 35.85 (1-0.8^t)

So, plug in t=10