Question

A chemical substance has a decay rate of 6.9% per day. The rate of change of an amount N of the chemical is given by the equation dN/dt = - 0.069 N where t is the number of days since decay began. Complete parts a) through c). a) Let N_0 represent the amount of the chemical substance at t = 0. Find the exponential function that models the situation. N(t) = N_0 b) Suppose that 800 g of the chemical substance is present at t = 0. How much will remain after 2 days? After 2 days, g will remain. After 2 days, g will remain. (Round to the nearest whole number as needed.) c) After how many days will half of the 800 g of the chemical substance remain? After days, half of the chemical substance will remain. (Round to one decimal place as needed.)

Answer 1

Answer:

a)  $N=N_0e^{-0.069t}$

b)  $N=696.9$ grams

c)  $t=10$ days

Step-by-step explanation:

a)

We are going to use separation of variables to solve.

Get all your t's to one side and your N's to opposing side.

$\frac{dN}{dt}=-0.069N$

Multiply both sides by $dt$:

$dN=-0.069N dt$

Divided both sides by $N$:

$\frac{dN}{N}=-0.069 dt$

Integrate both sides:

$\ln|N|=-0.069t+C$

The equivalent exponential form is:

$e^{-0.069t+C}=N$

Using law of exponents you can write this as:

$e^{-0.069t}e^C=N$

$e^C$ is just a positive constant that I'm going to replace with K:

$e^{-0.069t}K=N$

Applying the symmetric property of equality:

$N=e^{-0.069t}K$

Applying the commutative property of multiplication:

$N=Ke^{-0.069t}$

K actually represents the initial amount of chemical substance since when plugging in 0 for t you get K for N, like so:

$N=Ke^{-0.069 \cdot 0}$

$N=Ke^{0}$

$N=K(1)$

$N=K$

We are given at time 0 the amount of chemical substance,N, is K. They want us to represent this value with $N_0$ instead. So the exponential equation is:

$N=N_0e^{-0.069t}$

b)

We are given $N_0=800$ at $t=0$.

We are asked to find how much of the chemical substance, N, remains after 2 days.  So we replace t with 2 in $N=800e^{-0.069t}$:

$N=800e^{-0.069 \cdot 2}$

Put into calculator:

$N=696.9$ (this was rounded to the nearest tenths)

c)  

The last part is asking for how many days will it take a initial 800 grams to go down to half of 800 grams.

We need to see the following equation:

$\frac{1}{2}(800)=800e^{-0.069t}$

$400=800e^{-0.069t}$

Divide both sides by 800:

$\frac{400}{800}=e^{-0.069t}$

Reduce the fraction:

$\frac{1}{2}=e^{-0.069t}$

Convert to logarithmic form:

$\ln(\frac{1}{2})=-0.069t$

Divide both sides by -0.069:

$\frac{\ln(\frac{1}{2})}{-0.069}=t$

Input into calculator:

$10.0=t$

$t=10.0$

$t=10$