Chapter 2.5, Problem 38E

(a)

To determine

The formula for the sum

Answer to Problem 38E

  $V\left(t\right)=\frac{t^2}{2}+2t+60$

Explanation of Solution

The above- ground volume (stern and leaves) of a plant represented by the following equation.

  $V_a\left(t\right)=3t+20+\frac{t^2}{2}\mathrm{.........................}\left(1\right)$

The below-ground volume (roots) of a plant represented by the following equation.

  $V_b\left(t\right)=-t+\mathrm{40.........................}\left(2\right)$

Here, the volumes are measured in ${\text{cm}}^{\text{3}}$ and t is measured in days.

Therefore, the total population is as follows:

  $\begin{array}{c}V\left(t\right)=V_a\left(t\right)+V_b\left(t\right)\\ =3t+20+\frac{t^2}{2}-t+40\\ =\frac{t^2}{2}+2t+\mathrm{60.........................}\left(3\right)\end{array}$

(b)

To determine

The derivative of each component and the units

Answer to Problem 38E

  $\begin{array}{l}{V}_a\text{'}\left(t\right)=3+t\\ V_b\text{'}\left(t\right)=-1\end{array}$

The units of rate of change are ${\text{cm}}^{\text{3}}$ of volume per day.

Explanation of Solution

The above- ground volume (stern and leaves) of a plant represented by the following equation.

  $V_a\left(t\right)=3t+20+\frac{t^2}{2}\mathrm{.........................}\left(1\right)$

The below-ground volume (roots) of a plant represented by the following equation.

  $V_b\left(t\right)=-t+\mathrm{40.........................}\left(2\right)$

The derivative of each type by differentiating equation (1) and equation (2) with respect to t,

  $\begin{array}{l}{V}_a\text{'}\left(t\right)=\frac{d}{dt}\left(3t+20+\frac{t^2}{2}\right)\\ V_a\text{'}\left(t\right)=\frac{d}{dt}\left(3t\right)+\frac{d}{dt}\left(20\right)+\frac{d}{dt}\left(\frac{t^2}{2}\right)\\ V_a\text{'}\left(t\right)=3+0+\frac{2t}{2}\\ V_a\text{'}\left(t\right)=3+t\\ \text{and}\\ V_b\text{'}\left(t\right)=\frac{d}{dt}\left(-t+40\right)\\ V_b\text{'}\left(t\right)=-\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(40\right)\\ V_b\text{'}\left(t\right)=-1\end{array}$

The units of rate of change are ${\text{cm}}^{\text{3}}$ of volume per day.

(c)

To determine

The derivative of the sum

Whether the sum rule worked or not

Answer to Problem 38E

  $V\text{'}\left(t\right)=V_a\text{'}\left(t\right)+V_b\text{'}\left(t\right)$

Explanation of Solution

Take the derivative of equation (3) with respect to t,

  $\begin{array}{l}V\text{'}\left(t\right)=\frac{d}{dt}\left(\frac{t^2}{2}+2t+60\right)\\ V\text{'}\left(t\right)=\frac{d}{dt}\left(\frac{t^2}{2}\right)+\frac{d}{dt}\left(2t\right)+\frac{d}{dt}\left(60\right)\\ V\text{'}\left(t\right)=\frac{2t}{2}+2\\ V\text{'}\left(t\right)=t+2\\ V\text{'}\left(t\right)=\left(t+3\right)+\left(-1\right)\\ V\text{'}\left(t\right)=V_a\text{'}\left(t\right)+V_b\text{'}\left(t\right)\end{array}$

(d)

To determine

The description of happenings in words

Explanation of Solution

From the rate of change of above-ground volume of a plant, it is observed that the volume is always increasing.

From the rate of change of below-ground volume of a plant, it is observed that the volume is always declining.

From the rate of change of total volume of a plant, it is observed that the volume is always increasing.

(e)

To determine

To sketch: A graph of each component and total as functions of time.

Explanation of Solution

The graph of each component along with the total volume is drawn as

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