Let P(t) be the population of a country, in millions, t years after 1990, with P(7) = 3.32 and P(11) = 3.67. (a) Find a formula for P(t) assuming that it is linear. P(t) = (b) Find a formula for P(t) assuming that it is exponential. P(t) =

Respuesta :

a)

Linear function has a form,

[tex]y=mx+b[/tex]

Where

m is the slope

b is the y-intercept

The formula for m is:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Given the two points:

[tex]\begin{gathered} (x_1,y_1)=(7,3.32) \\ (x_2,y_2)=(11,3.67) \end{gathered}[/tex]

Let's find m,

[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{3.67-3.32}{11-7} \\ m=\frac{0.35}{4} \\ m=0.0875 \end{gathered}[/tex]

We have

[tex]y=0.0875x+b[/tex]

Let's find b by using a point:

[tex]\begin{gathered} y=0.0875x+b \\ 3.32=0.0875(7)+b \\ 3.32=0.6125+b \\ b=2.7075 \end{gathered}[/tex]

Thus the linear model equation is:

[tex]P(t)=0.0875t+2.7075[/tex]

b)

Exponential function has a form:

[tex]y=ab^x[/tex]

Given the two points:

[tex]\begin{gathered} (x_1,y_1)=(7,3.32) \\ (x_2,y_2)=(11,3.67) \end{gathered}[/tex]

Let's substitute them into the equation:

[tex]\begin{gathered} y=ab^x \\ 3.32=ab^7 \\ and \\ 3.67=ab^{11} \end{gathered}[/tex]

Let's divide the 2nd equation by the firsst equation and solve for b:

[tex]\begin{gathered} \frac{3.67=ab^{11}}{3.32=ab^7} \\ 1.1054=\frac{b^{11}}{b^7} \\ 1.1054=b^{11-7} \\ b^4=1.1054 \\ b=\sqrt[4]{1.1054} \\ b=1.0254 \end{gathered}[/tex]

So the equation becomes:

[tex]y=a(1.0254)^x[/tex]

Let's use the first point and solve for a:

[tex]\begin{gathered} y=a(1.0254)^x \\ 3.32=a(1.0254)^7 \\ 3.32=a(1.1919) \\ a=\frac{3.32}{1.1919} \\ a=2.7855 \end{gathered}[/tex]

Thus the exponential model equation is:

[tex]P(t)=2.7855(1.0254)^t[/tex]