Question

Suppose the height, in centimeters, of a growing plant t days after it breaks through the soil is given by $f\left(t\right)=0.36t^2$, as shown next at the left.

(a) What does the Mean Value Theorem from Section 3.1 say about the height of the plant during the first four days of its growth?

(b) What does the Mean Value Theorem for Integrals say about the height of the plant during the first four days of its growth?

Short answer

(a) The rate of growth of plant is 1.44 cm per second for some c in the first four seconds.

(b) The average growth is h(c) = 1.92 and c = 2.3094.

Step-by-step solution

Step 1. Given Information.

$f\left(t\right)=0.36t^2$

Step 2. part (a).

The mean value theorem says that there is some c in [0,4] for which

$$\begin{gathered} h\left(c\right)=\frac{h\left(4\right)-h\left(0\right)}{4-0}, \\ So, \\ h\left(c\right)=\frac{0.36{\left(4\right)}^2-0}{4-0}=\frac{0.36\left(4\right)\left(4\right)}{4}=0.36\left(4\right)=1.44. \\ Therefore,therateofgrowthofplantis1.44cmpersecondforsomec \\ inthefirstfourseconds. \end{gathered}$$

Step 3. part (b).

$$\begin{gathered} Averagegrowthis: \\ =\frac{1}{4-0}{\int }_0^{4}0.36t^{2}dt \\ =\frac{1}{4}{\left[\frac{0.36t^3}{3}\right]}_0^4 \\ =\frac{1}{4}\left[\frac{0.36{\left(4\right)}^3}{3}\right] \\ =1.92 \\ So,theaveragegrowthis1.92i.e.,h\left(c\right)=1.92. \\ Solvingforcwegetc=2.3094. \end{gathered}$$