Integration application:(

(a) T = ∫ A 5^(0.5t) dt

= A (2/ln5) 5^(0.5t) + CSub. t = 0, T = 20 and t = 2, T = 40, we have2A/ln5 + C = 20 and 10A/ln5 + C = 40Solve it, A = 2.5ln5 and C = 15So, T = 15 + 5^(1 + 0.5t) (b) Sub. T = 9015 + 5^(1 + 0.5t) = 905^(1 + 0.5t) = 75(1 + 0.5t)ln5 = ln 75t = 3.3652 min

(a)
dT/dt = A*5^(0.5t)
T = ∫ A*5^(0.5t) dt
= A*[ln5/0.5 * 5^(0.5t)] + C, where C is a constant.

When t = 0, T = 20.
A*[ln5/0.5 * 5^(0.5*0)] + C = 20
A*(ln5/0.5) + C = 20
A*2ln5 + C = 20 ... (1)

When t = 2, T = 40.
A*[ln5/0.5 * 5^(0.5*2)] + C = 40
A*(ln5/0.5 * 5) + C = 40
A*10ln5 + C = 40 ... (2)

(2) - (1): A(10ln5 - 2ln5) = 40 - 20
A (8ln5) = 20
A = 20 / 8ln5

From (1), C = 20 - A*2ln5
= 20 - (20/8ln5) * 2ln5
= 20 - 5
= 15

So, T = 20/8ln5 * [ln5/0.5 * 5^(0.5t)] + 15
= 20/8 * [1/0.5 * 5^(0.5t)] + 15
= 5*5^(0.5t) + 15
= 5^(1+0.5t) + 15
= 15 + 5^(1+0.5t)



Alternatively,

dT/dt = A*5^(0.5t)
T [40 20] = ∫ [2 0] A*5^(0.5t) dt
= [A * 1/0.5 * ln5 * 5^(0.5t)] [2 0]
= 2ln5 A* [5^(0.5t)] [2 0]
40 - 20 = 2ln5 A * (5^1 - 5^0)
20 = 2ln5 A * (5-1)
= 2ln5 A * 4
= 8ln5 A
A = 20 / 8ln5


(b) When T = 90,
T = 90 = 15 + 5^(1+0.5t)
75 = 5^(1+0.5t)
ln75 = (1+0.5t) ln5
1 + 0.5t = ln75/ln5
= 2.682606195
0.5t = 1.682606195
t = 3.365212389

It takes about 3.37 minutes for the temperature to reach 90

5^ (0.5t) = [e^(ln 5)]^(0.5t) = e^(ln 5 x 0.5t)
so dT/dt = A x e^(ln 5 x 0.5t)
dT = A x e^(ln 5 x 0.5t) dt
Integrating both side, we get
T = A S e^(ln 5 x 0.5t)dt = A/(ln 5 x 0.5) [e^(ln 5 x 0.5t)] + C.........(A)
When t = 0, T = 20
20 = A/(ln 5 x 0.5) + C..............(1)
When t = 2, T = 40
40 = A/(ln 5 x 0.5)[e^(ln 5)] + C .......(2)
(2) - (1)
20 = A/(ln 5 x 0.5)[e^(ln 5) - 1] = A/(ln 5 x 0.5)[5 - 1] = 4A/(ln 5 x 0.5)
5 = A/(ln 5 x 0.5)
A = 5(ln 5 x 0.5) = 2.5 ln 5
Sub into (1)
20 = (2.5 ln 5)/(0.5 ln 5) + C
20 = 5 + C
C = 15
Sub into (A),
T = (2.5 ln 5)/(0.5 ln 5) [5^(0.5t)] + 15
T = 5 x 5^(0.5t) + 15
T = 5^(1 + 0.5t) + 15.
When T = 90
90 = 5^(1 + 0.5t) + 15
75 = 5 x 5^(0.5t)
15 = 5^(0.5t)
ln 15 = 0.5t ln 5
t = (2 ln 15)/ln 5