Q. 3 - Calculus

Question

Theorems: Fill in the blanks to complete each of the following theorem statements.

If a quantity $Q\left(t\right)$ changes over time at a rate proportional to its value, then that quantity is modeled by a differential equation of the form $\frac{dQ}{dt}=\_\_\_\_\_$, with solution $Q\left(t\right)=\_\_\_\_\_$.

Step-by-Step Solution

Answer

If a quantity $Q\left(t\right)$ changes over time at a rate proportional to its value, then that quantity is modeled by a differential equation of the form $\frac{dQ}{dt}=kQ$, with solution $Q\left(t\right)=Q_0{e}^{kt}$.

Step 1. Given information

If a quantity $Q\left(t\right)$ changes over time at a rate proportional to its value, then that quantity is modeled by a differential equation of the form $\frac{dQ}{dt}=\_\_\_\_\_$, with solution $Q\left(t\right)=\_\_\_\_\_$.

Step 2. Filling the blanks

If a quantity $Q\left(t\right)$ changes over time at a rate proportional to its value, then that quantity is modeled by a differential equation of the form $\frac{dQ}{dt}=kQ$, with solution $Q\left(t\right)=Q_0{e}^{kt}$.

The differential equation is: $\frac{dQ}{dt}=kQ$

By separation of variables and integrating, we get,

$$\begin{gathered} \int \frac{dQ}{Q} =\int k dt \\ \Rightarrow \ln \left|Q\right| = kt + C \\ \Rightarrow \left|Q\right| = e^{kt+C} \\ \Rightarrow Q = A{e}^{kt} \end{gathered}$$

Since, $Q\left(0\right)=Q_0$ and $Q_0=A{e}^{k\left(0\right)}$ thus $A=Q_0$

Therefore, $Q\left(t\right)=Q_0{e}^{kt}$