# need to differentiate?

• Jun 27th 2006, 07:25 PM
freswood
need to differentiate?
I'm not really sure what to do here:

The area covered by a species of ground-cover plant has a rate of increase of ycm^2/week, given by:

y = -t^3 + 5x^2 + 6t

where t is the number of weeks after planting. According to the model, if the plant covered 100cm^2 when planted, what area will it cover after 4 weeks?

Thanks! :)
• Jun 27th 2006, 08:21 PM
Soroban
Hello, freswood!

Quote:

The area covered by a species of ground-cover plant has a rate of increase of y cm²/week,
given by: .$\displaystyle y \:= \:-t^3 + 5t^2 + 6t$ where $\displaystyle t$ is the number of weeks after planting.

According to the model, if the plant covered 100 cm² when planted.

What area will it cover after 4 weeks?

Since $\displaystyle y \:=\:-t^3 + 5t^2 + 6t$ is the rate of increase of ground cover,

the amount (area) of ground cover at time $\displaystyle t$ is given by:
. . $\displaystyle A(t) \;=\;\int(-t^3 + 5t^2 + 6t)\,dt \;= \;-\frac{1}{4}t^4 + \frac{5}{3}t^3 + 3t^2 + C$

We are told that, when $\displaystyle t = 0,\;A = 100.$

So we have: .$\displaystyle -\frac{1}{4}\cdot0^4 + \frac{5}{3}\cdot0^3 + 3\cdot0^2 + C \:= \:100\quad\Rightarrow\quad C = 100$

Hence, the area function is: .$\displaystyle A(t) \:= \:-\frac{1}{4}t^4 + \frac{5}{3}t^3 + 3t^2 + 100$

When $\displaystyle t = 4:\;\;A(4) \;=\;-\frac{1}{4}\cdot4^4 + \frac{5}{3}\cdot4^3 + 3\cdot4^2 + 100 \;= \;$ $\displaystyle \frac{572}{3}\;=\;190\frac{2}{3}$ cm².