Chapter 12.3, Problem 63E

(a)

To determine

To find: The rate of change of the calcium level with respect to time at $t=0$.

Answer to Problem 63E

The rate of change of the calcium level with respect to time at $t=0$ is $-0.5$.

Explanation of Solution

The amount of calcium remaining is $C\left(t\right)=\frac{1}{2}{\left(2t+1\right)}^{-\frac{1}{2}}$.

The rate of change of the calcium level with respect to time is given by,

$\begin{array}{c}{C}^{\prime }\left(t\right)=\frac{d}{dt}\left(\frac{1}{2}{\left(2t+1\right)}^{-\frac{1}{2}}\right)\\ =\frac{1}{2}\cdot \left(-\frac{1}{2}\right)\cdot {\left(2t+1\right)}^{-\frac{3}{2}}\cdot 2\\ =-\frac{1}{2}{\left(2t+1\right)}^{-\frac{3}{2}}\end{array}$

The rate of change of the calcium level at $t=0$ is,

$\begin{array}{c}{C}^{\prime }\left(0\right)=-\frac{1}{2}{\left(2\cdot 0+1\right)}^{-\frac{3}{2}}\\ =-\frac{1}{2}{\left(1\right)}^{-\frac{3}{2}}\\ =-\frac{1}{2}\\ =-0.5\end{array}$

Hence, rate of change of the calcium level at $t=0$ is $-0.5$.

(b)

To determine

To find: The rate of change of the calcium level with respect to time at $t=4$.

Answer to Problem 63E

The rate of change of the calcium level with respect to time at $t=4$ is $-0.02$.

Explanation of Solution

The amount of calcium remaining is $C\left(t\right)=\frac{1}{2}{\left(2t+1\right)}^{-\frac{1}{2}}$.

From part (a), the rate of change of the calcium level with respect to time is $C^{\prime }\left(t\right)=-\frac{1}{2}{\left(2t+1\right)}^{-\frac{3}{2}}$

The rate of change of the calcium level at $t=4$,

$\begin{array}{c}{C}^{\prime }\left(4\right)=-\frac{1}{2}{\left(2\cdot 4+1\right)}^{-\frac{3}{2}}\\ =-\frac{1}{2}{\left(9\right)}^{-\frac{3}{2}}\\ =-\frac{1}{54}\\ =-0.02\end{array}$

Hence, rate of change of the calcium level at $t=4$ is $-0.02$.

(c)

To determine

To find: The rate of change of the calcium level with respect to time at $t=7.5$.

Answer to Problem 63E

The rate of change of the calcium level with respect to time at $t=7.5$ is $-0.008$.

Explanation of Solution

The amount of calcium remaining is $C\left(t\right)=\frac{1}{2}{\left(2t+1\right)}^{-\frac{1}{2}}$.

From part (a), the rate of change of the calcium level with respect to time is $C^{\prime }\left(t\right)=-\frac{1}{2}{\left(2t+1\right)}^{-\frac{3}{2}}$

The rate of change of the calcium level at $t=7.5$,

$\begin{array}{c}{C}^{\prime }\left(7.5\right)=-\frac{1}{2}{\left(2\cdot 7.5+1\right)}^{-\frac{3}{2}}\\ =-\frac{1}{2}{\left(16\right)}^{-\frac{3}{2}}\\ =-\frac{1}{128}\\ =-0.008\end{array}$

Hence, rate of change of the calcium level at $t=7.5$ is $-0.008$.

(d)

To determine

Whether C is always increasing or decreasing; explain how to determine.

Answer to Problem 63E

The Calcium remaining in the blood stream is always decreasing.

Explanation of Solution

The amount of calcium remaining is $C\left(t\right)=\frac{1}{2}{\left(2t+1\right)}^{-\frac{1}{2}}$.

From part (a), the rate of change of the calcium level with respect to time is $C^{\prime }\left(t\right)=-\frac{1}{2}{\left(2t+1\right)}^{-\frac{3}{2}}$

From the parts (a), (b), and (c), observe that $C^{\prime }\left(t\right)<0\text{ for }t\ge 0$.

Thus, the amount of calcium in the blood will decrease with respect to time is decreasing gradually.