Chapter D, Problem 11E

(a)

To determine

To calculate:

Radius produced by disk.

Answer to Problem 11E

Radius $17.84124cm$ produces such a disk.

Explanation of Solution

Given information:

Area $=1000c{m}^2$

Calculation:

Suppose that $\text{r}$ be radius of circular metal disk with area $A$ .

Then $A=\pi \text{r}$ .

Lets suppose area of disk is $A=1000c{m}^2$

Then, $\begin{array}{l}\pi r^2=1000\\ r^2=\frac{1000}{\pi }\\ r=\sqrt{\frac{1000}{\pi }}\\ =10\sqrt{\frac{10}{\pi }}\\ =17.84124\end{array}$

Radius $10\sqrt{\frac{10}{\pi }}cm\approx 17.84124cm$ produces such a disk.

(b).

To determine

To calculate:

tolerance that machinist will control the radius with to the ideal radius than part (a).

Answer to Problem 11E

Machinist must control the radius with tolerance $\delta =0.04494$ to the ideal radius.

Explanation of Solution

Given information:

Area $=1000c{m}^2$

Calculation:

Assume that machinist is allowed an error tolerance of $\pm 5c{m}^2$ . So need to find out the tolerance $\delta >0$ in radius such that,

  $\left|r-10\sqrt{\frac{10}{\pi }}\right|<\delta$ implies $\left|A-1000\right|<5$

We have,

  $\begin{array}{l}\left|A-1000\right|<5\\ -5<A-1000<5\\ 995<A<1005\end{array}$

Here drawn a horizontal lines $A=995,y=1005$ along with the curve $A=\pi r^2$ as given below:

From the graph it is observed that in case it is restrict the values of $r$ between $17.7963$ and $17.8857$ then graph of (1) lies between $A=995,y=1005$ .

The distance of exact radius $r_0=17.84124$ from $17.7963$ and $17.8857$ are

  $\begin{array}{l}17.84124-17.84124=0.04446\\ 17.84124-17.79630=0.04494\end{array}$

In case we select $\delta =0.04494$ then

  $\left|r-10\sqrt{\frac{10}{\pi }}\right|<\delta$ implies $\left|A-1000\right|<5$

Thus, machinist must control the radius with tolerance $\delta =0.04494$ to the ideal radius.

(c)

To determine

To calculate:

Value of $\epsilon ,\delta$ and $a,L$ .

Answer to Problem 11E

  $a$ is the ideal radius of disk with area $1000c{m}^2$ .The $L$ is the area $1000c{m}^2$ , $\epsilon =5$ and $\delta =0.0446$ .

Explanation of Solution

Given information:

Area $=1000c{m}^2$

Calculation:

In terms of $\epsilon ,\delta$ definition of limit $\underset{x\to a}{\lim }f\left(x\right)=L$

Here $\begin{array}{l}L=1000\\ x=r\end{array}$

  $\begin{array}{l}f\left(x\right)=\pi r^2\\ \delta =0.04446\end{array}$

The $x$ represents the radius of circular disk, the function $f\left(x\right)$ represents the area of circular disk, the $a$ represents the ideal radius of disk with area $1000c{m}^2$ .

The $L$ represents the area $1000c{m}^2$ .

The given value of $\epsilon =5$ and $\delta =0.0446$ .