Chapter 2.4, Problem 39E

a.

To determine

To graph: The secant line to the given function at base point $t=1$ with $△t=1$ .

Explanation of Solution

Given:

The given function is $y\left(t\right)=10t^{2.5}$ . Base point is $t=1$ and secant joins the base point $t=1$ and the point with $△t=1$ .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  

The given function is $y\left(t\right)=10t^{2.5}$ .

The coordinates of base point say, P are

  $\left(t,10t^{2.5}\right)\equiv \left(1,10\cdot {(1)}^{2.5}\right)=\left(1,10\right)$ .

The point at distance $△t=1$ from base point is

  $Q\left(t+△t,10{(t+△t)}^{2.5}\right)\equiv \left(1+1,10\cdot {(1+1)}^{2.5}\right)$

  $\Rightarrow Q=\left(2,10\cdot 2^{\frac{5}{2}}\right)=\left(2,40\sqrt{2}\right)$

The graph of the function and the secant is shown in Figure (a)

b.

To determine

To graph: The secant line to the given function at base point $t=1$ with $△t=-1$ .

Explanation of Solution

Given:

The given function is $y\left(t\right)=10t^{2.5}$ . Base point is $t=1$ and secant joins the base point $t=1$ and the point with $△t=-1$ .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  

The given function is $y\left(t\right)=10t^{2.5}$ .

The coordinates of base point say, P are $\left(t,10t^{2.5}\right)\equiv \left(1,10\cdot {(1)}^{2.5}\right)=\left(1,10\right)$ .

The point at distance $△t=1$ from base point is $Q\left(t+△t,10{(t+△t)}^{2.5}\right)$

  $\Rightarrow Q\equiv \left(1-1,10\cdot {(1-1)}^{2.5}\right)$

  $\Rightarrow Q\equiv \left(1-1,10\cdot {(1-1)}^{2.5}\right)=\left(0,0\right)$ , that is the origin.

The graph of the function and the secant is shown in Figure (b).

c.

To determine

To graph: The secant line to the given function at base point $t=1$ with $△t=0.1$ .

Explanation of Solution

Given:

The given function is $y\left(t\right)=10t^{2.5}$ . Base point is $t=1$ and secant joins the base point $t=1$ and the point with $△t=0.1$ .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  

The given function is $y\left(t\right)=10t^{2.5}$ .

The coordinates of base point say, P are $\left(t,10t^{2.5}\right)\equiv \left(1,10\cdot {(1)}^{2.5}\right)=\left(1,10\right)$ .

The point at distance $△t=1$ from base point is $Q\left(t+△t,10{(t+△t)}^{2.5}\right)$

  $\Rightarrow Q\equiv \left(1+0.1,10\cdot {(1+0.1)}^{2.5}\right)$

  $\Rightarrow Q=\left(1.1,10{(1.1)}^{5/2}\right)=\left(1.1,12.69\right)$

The graph of the function and the secant is shown in Figure (c). We see the secant line tends to tangent line at point P .

d.

To determine

To graph: The secant line to the given function at base point $t=1$ with $△t=0.1$ .

Explanation of Solution

Given:

The given function is $y\left(t\right)=10t^{2.5}$ . Base point is $t=1$ and secant joins the base point $t=1$ and the point with $△t=-0.1$ .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  

The given function is $y\left(t\right)=10t^{2.5}$ .

The coordinates of base point say, P are $\left(t,10t^{2.5}\right)\equiv \left(1,10\cdot {(1)}^{2.5}\right)=\left(1,10\right)$ .

The point at distance $△t=1$ from base point is $Q\left(t+△t,10{(t+△t)}^{2.5}\right)$

  $\Rightarrow Q\equiv \left(1-0.1,10\cdot {(1-0.1)}^{2.5}\right)$

  $\Rightarrow Q=\left(0.9,10{(0.9)}^{5/2}\right)=\left(0.9,7.68\right)$

The graph of the function and the secant is shown in Figure (d).

We see the secant line tends to tangent line at point P .

e.

To determine

To graph: The secant line to the given function at base point $t=1$ with $△t\to 0$ , let $△t=0.01$ .

Explanation of Solution

Given:

The given function is $y\left(t\right)=10t^{2.5}$ . Base point is $t=1$ and secant joins the base point $t=1$ and the point with assumed $△t=0.01$ .

Method used:

The function and the secant line is plotted using graphic calculator.

Calculations and Graph:

  

The given function is $y\left(t\right)=10t^{2.5}$ .

The coordinates of base point say, P are $\left(t,10t^{2.5}\right)\equiv \left(1,10\cdot {(1)}^{2.5}\right)=\left(1,10\right)$ .

The point at distance $△t=1$ from base point is $Q\left(t+△t,10{(t+△t)}^{2.5}\right)$

  $\Rightarrow Q\equiv \left(1+0.01,10\cdot {(1+0.01)}^{2.5}\right)$

  $\Rightarrow Q=\left(1.01,10{(1.01)}^{5/2}\right)=\left(1.01,10.25\right)$

The graph of the function and the secant is shown in Figure (e).

We observe that points $P\left(1,10\right)$ and $Q\left(1.01,10.25\right)$ almost coincide, therefore the secant line and tangent line at point $P\left(1,10\right)$ coincide. That we say, when $△t\to 0$ the secant line becomes a tangent line at base point. The tangent line at point $P\left(Q\to P\right)$ in this case looks right.

Here, the slope of the tangent by definition is $m=\tan \angle PLM=\frac{PM}{LM}=\frac{10}{.4}=25$ .

And from the formula as the tangent line is the limiting case of secant line $m=\frac{y_2-y_1}{x_2-x_1}=\frac{10.25-10}{1.01-1.00}=\frac{0.25}{0.01}=\frac{25}{1}=25$ .

Thus, the slope of the tangent line is 25.