ANSWER on Question #84877 – Math – Differential Equations
QUESTION
Under certain condition cane sugar in water is converted into dextrose at a rate which is proportional to the amount unconverted at any time. If of 75 g m 75\,gm 75 g m t = 0 , 8 g m t = 0,8\,gm t = 0 , 8 g m
SOLUTION
Let
C − is converted sugar cane → { d C d t is conversion rate ( 75 − C ) is unconverted sugar cane C - \text{is converted sugar cane} \rightarrow \left\{ \begin{array}{c} \frac{dC}{dt} \text{ is conversion rate} \\ (75 - C) \text{ is unconverted sugar cane} \end{array} \right. C − is converted sugar cane → { d t d C is conversion rate ( 75 − C ) is unconverted sugar cane
Then, by the condition of the problem
d C d t ∼ ( 75 − C ) → d C d t = k ( 75 − C ) , where k is the coefficient of proportionality \frac{dC}{dt} \sim (75 - C) \rightarrow \frac{dC}{dt} = k(75 - C), \quad \text{where } k \text{ is the coefficient of proportionality} d t d C ∼ ( 75 − C ) → d t d C = k ( 75 − C ) , where k is the coefficient of proportionality
Then,
d C d t = k ( 75 − C ) ∣ × ( d t 75 − C ) → d C 75 − C = k d t → ∫ d C 75 − C = ∫ k d t → − ln ∣ 75 − C ∣ = k t − ln ∣ A ∣ × ( − 1 ) → ln ∣ 75 − C ∣ = − k t + ln ∣ A ∣ → e ln ∣ 75 − C ∣ = e − k t + ln ∣ A ∣ → 75 − C = A ⋅ e − k t → C ( t ) = 75 − A ⋅ e − k t \begin{array}{l}
\frac{dC}{dt} = k(75 - C) \Bigg| \times \left(\frac{dt}{75 - C}\right) \rightarrow \frac{dC}{75 - C} = kdt \rightarrow \int \frac{dC}{75 - C} = \int kdt \rightarrow \\
- \ln |75 - C| = kt - \ln |A| \times (-1) \rightarrow \ln |75 - C| = -kt + \ln |A| \rightarrow \\
e^{\ln |75 - C|} = e^{-kt + \ln |A|} \rightarrow 75 - C = A \cdot e^{-kt} \rightarrow C(t) = 75 - A \cdot e^{-kt}
\end{array} d t d C = k ( 75 − C ) ∣ ∣ × ( 75 − C d t ) → 75 − C d C = k d t → ∫ 75 − C d C = ∫ k d t → − ln ∣75 − C ∣ = k t − ln ∣ A ∣ × ( − 1 ) → ln ∣75 − C ∣ = − k t + ln ∣ A ∣ → e l n ∣75 − C ∣ = e − k t + l n ∣ A ∣ → 75 − C = A ⋅ e − k t → C ( t ) = 75 − A ⋅ e − k t
By the condition of the problem:
C ( 0 ) = 0 = 75 − A ⋅ e − k ⋅ 0 → 75 − A = 0 → A = 75 C ( t ) = 75 − 75 ⋅ e − k t → C ( t ) = 75 ⋅ ( 1 − e − k t ) \begin{array}{l}
C(0) = 0 = 75 - A \cdot e^{-k \cdot 0} \rightarrow 75 - A = 0 \rightarrow A = 75 \\
C(t) = 75 - 75 \cdot e^{-kt} \rightarrow C(t) = 75 \cdot (1 - e^{-kt})
\end{array} C ( 0 ) = 0 = 75 − A ⋅ e − k ⋅ 0 → 75 − A = 0 → A = 75 C ( t ) = 75 − 75 ⋅ e − k t → C ( t ) = 75 ⋅ ( 1 − e − k t ) C ( 30 ) = 8 = 75 ⋅ ( 1 − e − k ⋅ 30 ) → 1 − e − k ⋅ 30 = 8 75 → e − 30 k = 1 − 8 75 → e − 30 k = 75 − 8 75 → ln ∣ e − 30 k ∣ = ln ∣ 67 75 ∣ → − 30 k = ln ∣ 67 75 ∣ → k = − 1 30 ⋅ ln ∣ 67 75 ∣ → k = ln ∣ ( 67 75 ) − 1 / 30 ∣ = ln ∣ ( 75 67 ) 1 / 30 ∣ → k = ln ∣ ( 75 67 ) 1 / 30 ∣ \begin{array}{l}
C(30) = 8 = 75 \cdot (1 - e^{-k \cdot 30}) \rightarrow 1 - e^{-k \cdot 30} = \frac{8}{75} \rightarrow e^{-30k} = 1 - \frac{8}{75} \rightarrow \\
e^{-30k} = \frac{75 - 8}{75} \rightarrow \ln |e^{-30k}| = \ln \left| \frac{67}{75} \right| \rightarrow -30k = \ln \left| \frac{67}{75} \right| \rightarrow k = -\frac{1}{30} \cdot \ln \left| \frac{67}{75} \right| \rightarrow \\
k = \ln \left| \left(\frac{67}{75}\right)^{-1/30} \right| = \ln \left| \left(\frac{75}{67}\right)^{1/30} \right| \rightarrow k = \ln \left| \left(\frac{75}{67}\right)^{1/30} \right|
\end{array} C ( 30 ) = 8 = 75 ⋅ ( 1 − e − k ⋅ 30 ) → 1 − e − k ⋅ 30 = 75 8 → e − 30 k = 1 − 75 8 → e − 30 k = 75 75 − 8 → ln ∣ e − 30 k ∣ = ln ∣ ∣ 75 67 ∣ ∣ → − 30 k = ln ∣ ∣ 75 67 ∣ ∣ → k = − 30 1 ⋅ ln ∣ ∣ 75 67 ∣ ∣ → k = ln ∣ ∣ ( 75 67 ) − 1/30 ∣ ∣ = ln ∣ ∣ ( 67 75 ) 1/30 ∣ ∣ → k = ln ∣ ∣ ( 67 75 ) 1/30 ∣ ∣
Then,
C ( t ) = 75 ⋅ ( 1 − e − t ⋅ ln ∣ ( 75 67 ) 1 30 ∣ ) = 75 ⋅ ( 1 − e ln ∣ ( 75 67 ) − t 30 ∣ ) = 75 ⋅ ( 1 − ( 75 67 ) − t 30 ) → C(t) = 75 \cdot \left(1 - e^{-t \cdot \ln \left| \left(\frac{75}{67}\right)^{\frac{1}{30}} \right|}\right) = 75 \cdot \left(1 - e^{\ln \left| \left(\frac{75}{67}\right)^{\frac{-t}{30}} \right|}\right) = 75 \cdot \left(1 - \left(\frac{75}{67}\right)^{\frac{-t}{30}}\right) \to C ( t ) = 75 ⋅ ⎝ ⎛ 1 − e − t ⋅ l n ∣ ∣ ( 67 75 ) 30 1 ∣ ∣ ⎠ ⎞ = 75 ⋅ ⎝ ⎛ 1 − e l n ∣ ∣ ( 67 75 ) 30 − t ∣ ∣ ⎠ ⎞ = 75 ⋅ ( 1 − ( 67 75 ) 30 − t ) →
Conclusion,
C ( t ) = 75 ⋅ ( 1 − ( 67 75 ) t 30 ) C(t) = 75 \cdot \left(1 - \left(\frac{67}{75}\right)^{\frac{t}{30}}\right) C ( t ) = 75 ⋅ ( 1 − ( 75 67 ) 30 t )
The amount converted in one and half hours is
C ( 90 ) = 75 ⋅ ( 1 − ( 67 75 ) 90 30 ) = 75 ⋅ ( 1 − ( 67 75 ) 3 ) = 121112 5625 ≈ 21.53 ( g m ) C(90) = 75 \cdot \left(1 - \left(\frac{67}{75}\right)^{\frac{90}{30}}\right) = 75 \cdot \left(1 - \left(\frac{67}{75}\right)^3\right) = \frac{121112}{5625} \approx 21.53\,(gm) C ( 90 ) = 75 ⋅ ( 1 − ( 75 67 ) 30 90 ) = 75 ⋅ ( 1 − ( 75 67 ) 3 ) = 5625 121112 ≈ 21.53 ( g m )
**ANSWER:**
C ( t ) = 75 ⋅ ( 1 − ( 67 75 ) t 30 ) C(t) = 75 \cdot \left(1 - \left(\frac{67}{75}\right)^{\frac{t}{30}}\right) C ( t ) = 75 ⋅ ( 1 − ( 75 67 ) 30 t ) C ( 90 ) = 121112 5625 ≈ 21.53 ( g m ) C(90) = \frac{121112}{5625} \approx 21.53\,(gm) C ( 90 ) = 5625 121112 ≈ 21.53 ( g m )
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