Question

JUIVE Suppose that the amount in grams of a radioactive substance present at time t (in years) is given by A(t) = 800e 0.86t. Find the rate of change of the quantity present at the time when t = 5. 9.3 grams per year 0 -72.7 grams per year -9.3 grams per year 0 72.7 grams per year

Answer 1

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data

A(t) = 800e^(-0.86t)

Step 02:

Rate of change

t1 = 0

A(t) = 800e^(-0.86t)

A(t) = 800e^(-0.86*0)

A(0) = 800

t2 = 5

A(t) = 800e^(-0.86t)

A(t) = 800e^(-0.86*5)

A (t) = 800e^(-4.3)

A(5) = 10.85

Step 03:

`(\Delta y)/(\Delta x)=\frac{A(5)\text{ - A(0)}}{5-0}`
`(\Delta y)/(\Delta x)=(10.85-800)/(5-0)=(-789.15)/(5)=-157.83`