Respuesta :
Answer:[tex]h=45e^{\frac{t}{50} }[/tex] feet
Step-by-step explanation:
Let [tex]h[/tex] be the height of the tree at time [tex]t[/tex].
Given that the height of a tree increases by approximately [tex]2[/tex]% each year.
So,[tex]\frac{dh}{dt}=\frac{2}{100}h[/tex]
[tex]\frac{dh}{h}=\frac{2}{100}dt[/tex]
Given that when [tex]t=0[/tex],[tex]h=45 feet[/tex]
Integrating on both sides,
[tex]\int\limits^h_{45} {\frac{dh}{h} } =\int\limits^t_0 {\frac{2}{100} } \, dt[/tex]
[tex]ln(h)-ln(45)=\frac{2}{100}(t-0)[/tex]
[tex]ln(h)=\frac{2}{100}t+ln(45)[/tex]
applying anti-logarithm on both sides,
[tex]h=45\times e^{\frac{t}{50}}[/tex]
