the temperature T °C inside a house at a time t hours after 6.00 am is given by T=18-4cos(pi t/12) for 0\leqslant t\leqslant 24
the temperature \theta °C outside a house at time t hours after midnight is given by \theta =11-6cos(pi t/12) for 0\leqslant t\leqslant 24
if the difference between the inside and outside temperatures can be represented by D=T-\theta , sketch the graph of D. determine when the difference between inside and the outside temperature is less then 6°C.
May 30th 2011, 04:01 AM
Yes- that you should show some effort yourself first! All you are asked to do is subtract two functions. One thing to be carfule about- the variable "t" in the two equations represent two different things. In the formula for "T", t is "hours after 6.00 am" while in the formula for " ", t is "hours after midnight".
If we call the "t" in the formula for T t' so we can distinguish then t'= t- 6 because, of course, 6 am is 6 hours after midnight.
May 30th 2011, 04:10 AM
so for function T & theta, you substitute 6 into the equation. which will give you 18 and 11. 6pm and 11am