Application of Differentiation

用戶3618

2009-01-01 4:48 am

The temperature, T℃, of a heated liquid, t minutes after it is placed in a freezer can be modelled by the equation
T = 1800/(t^2 + 2t + 25) - 10, for t > 0 or t = 0.

(a) Find the initial temperature of this liquid.
(b) Show that the temperature T is strictly decreasing with the time.
(c) Find the rate at which the temperature of this liquid is changing 10 minutes after it placed in the freezer.
(d) What is the final temperature of this liquid?

回答 (1)

myisland8132

2009-01-01 5:03 am

✔ 最佳答案

(a)
sub t=0
T=1800/25-10=62
The initial temperature of this liquid is 62 ℃
(b)
T = 1800/(t^2 + 2t + 25) - 10
dT/dt
=-1800(t^2 + 2t + 25)^(-2)(2t+2)
=-3600(t^2 + 2t + 25)^(-2)(t+1)
<0
So T is strictly decreasing with the time
(c)
When t=10
dT/dt
=-3600(10^2 + 20 + 25)^(-2)(10+1)
=-1.8835
The rate at which the temperature of this liquid is changing 10 minutes after it placed in the freezer is -1.8835℃/min
(d)
Let t tends to infinity, 1800/(t^2 + 2t + 25) tends to 0 and so T=10
The final temperature of this liquid is 10℃

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