Chapter 6, Problem 15P

To determine

To Find: The way in which the $b$ is related to the value of $a$ .

Answer to Problem 15P

The relation of the video is $B=2^{4}a$ .

Explanation of Solution

Given:

The equation of the parabola is $y=x^2$ , when $x=0$ , $x=a$ and $x=b$ . Here, $0<a<b$ .

Calculation:

Consider the required diagram is shown in Figure 1

  

Figure 1

Consider the point $p$ in the first quadrant with the coordinates $\left(a,a^3\right)$ so the equation of the curve is,

  $y=x^3$

Then, the slope of the tangent is,

  $\frac{dy}{dx}=3x^2$

Then, the slope of the tangent that passes through the point P is,

  $3a^2$

Then, the equation of the tangent is,

  $\begin{array}{l}\left(y-a^3\right)=3a^2\left(x-a\right)\\ y=3a^{2}x-2a^3\end{array}$

Consider the other point of the intersections of the curve $y=x^3$ and the tangent line then,

  $\begin{array}{l}3a^{2}x-2a^3=x^3\\ 3a^{2}x-2a^3-x^3=0\end{array}$

Consider for $x=a$ then the equation is satisfied so $\left(x-a\right)$ then,

  $\begin{array}{l}\left(x-a\right)\left(x-a\right)\left(x+2a\right)=0\\ {\left(x-a\right)}^2\left(x+2a\right)=0\\ x=a,-2a\end{array}$

Thus, the other the points of the intersection is $\left(-2a,-8a^3\right)=Q$ then the equation of the tangent line is passing through the point Q is,

  $\begin{array}{l}\left(y+8a^3\right)=3{\left(-2a\right)}^2\left(x+2a\right)\\ y=12a^{2}x+16a^3\end{array}$

Then, by the symmetry this tangent line will intersect the curve again at,

  $\begin{array}{c}x=-2\left(-2a\right)\\ =4a\end{array}$

Consider the area of the curve $y=x^3$ is above the first tangent and the below the second tangent so the area is evaluated as,

  $\begin{array}{c}A={\displaystyle {\int }_{-2a}^a\left[x^3-\left(3a^{2}x-2a^3\right)\right]}dx\\ ={\left[\frac{x^4}{4}-\frac{3}{2}{a}^2{x}^2+2a^{3}x\right]}_{-2a}^a\\ =\left[\frac{3}{4}{a}^4-\left(-6a^4\right)\right]\\ =\frac{27}{4}{a}^4\end{array}$

The area of the B is,

  $\begin{array}{c}B={\displaystyle {\int }_{-2a}^{4a}\left[\left(12a^{2}x+16a^3\right)-x^3\right]}dx\\ ={\displaystyle {\int }_{-2a}^{4a}\left[12a^{2}x+16a^3-x^3\right]}dx\\ ={\left[6a^2{x}^2+16a^{3}x-\frac{1}{4}{x}^4\right]}_{-2a}^{4a}\\ =108a^4\end{array}$

Then,

  $\begin{array}{l}A=16\left(\frac{27}{4}\right)a^4\\ \left\{108=27\times 4\right\}\end{array}$

Then, the relation between B and A is,

  $\begin{array}{c}B=16A\\ =2^{4}a\end{array}$

Thus, the relation of the video is $B=2^{4}a$ .