Solve the problem.
-In a chemical reaction, the rate at which the amount of a reactant changes with time is proportional to the amount present, such that $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$, when $\mathrm { t }$ is measured in hours. If there are $78 \mathrm {~g}$ of reactant present when $t$ $= 0$, how many grams will be left after 4 hours? Give your answer to the nearest tenth of a gram.
A) 4.7 g
B) 2.4 g
C) 0.1 g
D) 7.1 g

Question

Quizplus · Accepted Answer

The given differential equation $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$ is a first-order linear differential equation, which can be solved to find the amount of reactant $y$ at any time $t$. The solution to this equation is $y = y_0 e^{-0.7t}$, where $y_0$ is the initial amount of the reactant. Given $y_0 = 78$ g and $t = 4$ hours, the amount of reactant left after 4 hours is $y = 78e^{-0.7 \times 4}$. Calculating this gives approximately $4.7$ g.

Solve the Problem dydt=−0.7y\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }dtdy​=−0.7y

Solve the Problem $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$