Chapter 2, Problem 28SP

a.

To determine

To graph:The secant line to the given curve from time $t=2$ min to $t=2.5$ min.

Explanation of Solution

Given:

The volume of a cell as a function of time is given by $V\left(t\right)=2-2t+t^2$ , $V\left(t\right)$ (in 1000 s of $\text{μ}{\text{m}}^{\text{3}}$ ) and t the time (in minutes).

Method used:

The graph is sketched using graphic calculator.

Graph:

The graph of function and the secant from $t=2$ min to $t=2.5$ min is shown in figure here.

  

Interpretation:

The curve represented by the function $V\left(t\right)=2-2t+t^2$ is parabola.

b.

To determine

To find:The equation of secant line at points $t=2$ min to $t=2.5$ min sketched in part (a).

Answer to Problem 28SP

The equation of secant joining the points $t=2$ and $t=2.5$ is $V(t)=2.5t-3$.

Explanation of Solution

Given:

The curve in part (a), and secant passes through points $t=2$ and $t=2.5$ .

Formula used:

The equation of a line to the curve $y=f\left(x\right)$ at point $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Calculation:

The given curve is $V\left(t\right)=2-2t+t^2$ . The points $t=2$ and $t=2.5$ correspond to $\left(2,V\left(2\right)\right)\equiv \left(2,2\right)$ , and $\left(2.5,V\left(2.5\right)\right)\equiv \left(2.5,3.25\right)$ .

Hence, the equation of secant joining the points $t=2$ and $t=2.5$ is $V\left(t\right)-2.0=\frac{3.25-2.0}{2.5-2.0}\left(t-2.0\right)$

  $\Rightarrow V(t)-2.0=\frac{1.25}{0.5}\left(t-2.0\right)$

  $\Rightarrow V(t)=2.5t-3$

Conclusion:

The equation of secant joining the points $t=2$ and $t=2.5$ is $V(t)=2.5t-3$.

c.

To determine

To find: The derivative of function $V(t)$ given in part (a) at point $t=2$ , expressed in differential and prime notations..

Answer to Problem 28SP

The derivatives of $V(t)$ is $V^{\prime }\left(t\right)=-2+2t$ and its value atpoint $t=2$ in differential and prime notation are $\frac{dV}{dt}(att=2)=2$ , and $V^{\prime }\left(t=2\right)=2$ respectively.

Explanation of Solution

Given:

The function $V\left(t\right)=2-2t+t^2$ .

Formula used:

The derivative of $y=x^n$ is $\frac{dy}{dx}=n{x}^{n-1}$ .

Calculation:

The given function is $V\left(t\right)=2-2t+t^2$ .

The derivative of $V\left(t\right)$ is

  $\frac{dV}{dt}=0-2\cdot t^{1-1}+2\cdot t^{2-1}=-2+2t$

  $\frac{dV}{dt}(att=2)=-2+2\cdot 2=2$  (Differential notation)

  $V^{\prime }\left(t=2\right)=-2+2\cdot 2=2$  (Prime notation)

Conclusion:

The derivatives of $V(t)$ is $V^{\prime }\left(t\right)=-2+2t$ and its value atpoint $t=2$ in differential and prime notation are $\frac{dV}{dt}(att=2)=2$ , and $V^{\prime }\left(t=2\right)=2$ respectively.

d.

To determine

To find: The cell volume $V(t)$ at $t=27$ min.

Answer to Problem 28SP

The volume of the cell at $t=27\min$ is measured $2621\text{μ}{\text{m}}^{\text{3}}$ .

Explanation of Solution

Given:

The expression for $V(t)$ in part (a).

Concept used:

Substitute $t=27$ in expression for $V(t)$.

Calculation:

The expression for $V(t)$ is $V\left(t\right)=2-2t+t^2$ .

Therefore, the volume of the cell at time $t=27\min =27\times 6\text{0}=1620$ seconds is

  $V\left(1620\right)=2-2\times 1620+{\left(1620\right)}^2=2621162\text{μ}{\text{m}}^{\text{3}}$ .

But volume is measured in 1000 s of $\text{μ}{\text{m}}^{\text{3}}$ .

Hence, the measured volume of the cell at $t=27\min$ is $V\left(t=27\right)=\frac{2621162}{1000}\text{μ}{\text{m}}^{\text{3}}=2621\text{μ}{\text{m}}^{\text{3}}$

Conclusion:

The volume of the cell at $t=27\min \to$ is measured $2621\text{μ}{\text{m}}^{\text{3}}$ .

e.

To determine

To graph: The derivative of function $V(t)$ in part (c).

Explanation of Solution

Given:

The derivative of $V(t)$ is $V^{\prime }\left(t\right)=-2+2t$ .

Concept used:

The graph is plotted using graphic calculator

Graph:

The graph of the derivative of function $V(t)$ as function of time is shown in figure here.

  

Interpretation:

The graph of the derivative of volume function $V^{\prime }\left(t\right)=-2+2t$ is positive $t>1$ that means after time $t>1$ the rate of increasing of volume increases.