Step-by-Step Solution
Topic: Application of Differential Equations
(I) ORDER AND DEGREE
The given differential equation is dV/dt = kS.
This involves the first derivative of Volume with respect to time.
- • Order: 1 (Highest derivative is first order)
- • Degree: 1 (Power of the highest derivative is 1)
(II) GENERAL SOLUTION
Given differential equation: dr/dt = (2/3)k
Integrate both sides with respect to t:
∫ dr = ∫ (2/3)k dt
Apply Initial Condition: r(0) = 5 mm.
Substitute t = 0 and r = 5:
5 = (2/3)k(0) + C ⇒ C = 5
(III)(A) FIND k AND t (Case 1)
Given: r = 3 mm when t = 1 hour.
Substitute into the equation from (II):
3 = (2/3)k(1) + 5
3 – 5 = 2k/3 ⇒ -2 = 2k/3
-6 = 2k ⇒ k = -3
Specific Equation: r = (2/3)(-3)t + 5 ⇒ r = -2t + 5
Find t when r = 0:
0 = -2t + 5 ⇒ 2t = 5
(III)(B) FIND k AND t (Case 2)
Given: r = 1 mm when t = 1 hour.
Substitute into the equation from (II):
1 = (2/3)k(1) + 5
1 – 5 = 2k/3 ⇒ -4 = 2k/3
-12 = 2k ⇒ k = -6
Specific Equation: r = (2/3)(-6)t + 5 ⇒ r = -4t + 5
Find t when r = 0:
0 = -4t + 5 ⇒ 4t = 5