Question

Chlorofluorocarbons (CFCs) were used as propellants in spray cans until their buildup in the atmosphere started destroying the ozone, which protects us from ultraviolet rays. Since the 1987 Montreal Protocol (an agreement to curb CFCs), the CFCs in the atmosphere above the US have been reduced from a high of 1915 parts per trillion (ppt) in 2000 to 1640 ppt in 2014.23 The reduction has been approximately linear. Let C(t) be the concentration of CFCs in ppt in year t.
(a) Find C(2000) and C(2014).
(b) Estimate C′(2000) and C′(2014).
(c) Assuming C(t) is linear, find a formula for C(t).
(d) When is C(t) expected to reach 1500 ppt, the level before CFCs were introduced?
(e) If you were told that in the future, C(t) would not be exactly linear, and that C″(t) > 0, would your answer to part (d) be too early or too late?

Answer 1

Answer:

a) C(2000)=1915

C(2014)=1915

b) $C'(2000)=-\frac{275}{14}$

$C'(2014)=-\frac{275}{14}$

c) $C(t)=-\frac{275}{14}t+\frac{288405}{7}$

d) t=2021

e) too early

Step-by-step explanation:

a)

Since C(t) is the concentration of CFCs in ppt in year t, all we need to do to solve this part is determine what the concentration of CFCs is in years 2000 and 2014. Luckily for us, the problem already gives us those values, so:

C(2000)=1915    concentration in year 2000

C(2014)=1915      concentration in year 2014

b)

By definition, the derivative of a function at a given point is interpreted as the slope of the tangent line to the point of interest, so in order to find this answer, we need to find the slope of the line. Since the problem specifies that the behavior is linear, this means that the slope will always be the same no matter the year, so we get:

$m=C'(t)=\frac{C'(t_{2})-C'(t_{1})}{t_2-t_1}$

so:

$C'(t)=\frac{1640-1915}{2014-2000}=-\frac{275}{14}\approx -19.64$

therefore:

$C'(2000)=-\frac{275}{14}$

$C'(2014)=-\frac{275}{14}$

c)

Since the behavior is linear, we can calculate it with the point-slope form of the line which is:

$y-y_{1}=m(x-x_{1})$

in this case:

$C(t)-C(t_1)=C'(t)(t-t_{1})$

so we get:

$C(t)-1915=-\frac{275}{14}(t-2000)$

and we can now solve for C(t) so we get:

$C(t)=-\frac{275}{14}t+\frac{275000}{t}+1915$

for a final answer of:

$C(t)=-\frac{275}{14}t+\frac{288405}{7}$

d)

So next we solve the equation for C(t)=1500 so we get:

$1500=-\frac{275}{14}t+\frac{288405}{7}$

$1500-\frac{288405}{7}=-\frac{275}{14}t$

$-\frac{277905}{7}=-\frac{275}{14}t$

$t=2021 \frac{7}{55}$

so we will reach that concentration level at the year 2021 approximately.

e)

Since the second derivative of the concentration function is greater than zero, this means that the original function might be a function in the form: .

This means that the decrease of the concentration levels is slower than that of a linear equation. So the projected date will be too early than the real date.