Chapter 13.6, Problem 19PSC

(a)

To determine

To find: The distance that the marble travelled.

Answer to Problem 19PSC

The distance that the marble travelled is $\frac{16\pi }{3}$ .

Explanation of Solution

Given:

The given point at which the marble was placed is $\left(2,4\sqrt{3}\right)$ .

The equation for the graph is $x^2-12x+y^2=28$ .

Calculation:

Consider the simplified form for the given equation is,

  $\begin{array}{l}{x}^2-12x+y^2=28\\ {\left(x-6\right)}^2+y^2=64\end{array}$

As the hypotenuse is twice as long as one leg and the angle between them is $60°$ . Thus, the marble travelled $120°$ .

Thus, the radius is $r=8$ and the circumference is $C=16\pi$ .

Thus, the distance to which the marble travelled is,

  $16\pi \left(\frac{120°}{360°}\right)=\frac{16\pi }{3}$

(b)

To determine

To find: The distance that would have been saved if the marble had rolled in the straight line.

Answer to Problem 19PSC

The distance saved is $2.90$ and the distance travelled is $8\sqrt{3}$ .

Explanation of Solution

Calculation:

The path travelled is the straight line path from the point $\left(2,4\sqrt{3}\right)$ and $\left(14,0\right)$ . Then, the distance is calculated as,

  $\begin{array}{c}d=\sqrt{{\left(14-2\right)}^2+{\left(4\sqrt{3}\right)}^2}\\ =\sqrt{144+48}\\ =\sqrt{192}\\ =8\sqrt{3}\end{array}$

The distance saved is,

  $\frac{16\pi }{3}-8\sqrt{3}=2.90$