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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

The figure shows a circle with radius $ 1$ touching the parabola

$\displaystyle P:\ y = x^2 $

at two points.

\includegraphics[width=.6\linewidth]{TdM_11_A1_bild_en}


Determine the center $ C=(0,c)$ of the circle, the point of tangency $ T=(t,t^2)$ as well as the areas of the two shaded segments $ A_1$ and $ A_2.$


Answer:

$ c = $  
$ t = $  
$ A_1 = $  
$ A_2 = $  

(The results should be correct to four decimal places.)

Solution:

The line $ l$ passing through the center $ C = (0,c)$ and the point of tangency $ T = (t,t^2)$ is perpendicular to the tangent line. Thus, the slopes satisfy

$\displaystyle -1 = s_l \cdot s_t = \dfrac{t^2 - c}{t} \cdot 2t . $

Solving for the coordinate of the center, we obtain

$\displaystyle c = t^2 + \dfrac{1}{2}. $

From

$\displaystyle 1 = \vert\overline{CT}\vert^2 = (0-t)^2 + {(c-t^2)}^2 = t^2 + \dfrac{1}{4} $

we get
$ t = \sqrt{3} / 2 \approx 0.8660$     and      $ c = 5/4 = 1.25.$

\includegraphics[width=.45\linewidth]{TdM_11_A1_Lsg_bild_en}

The triangle $ TC\tilde{T}$ has the height $ \vert\overline{SC}\vert=c-t^2=1/2$ and the length of the base equals $ 2t=\sqrt{3}.$ Therefore, its area is $ \sqrt{3}/4\,.$ It is an isosceles triangle with a base angle $ \alpha = \sphericalangle \, (C,T,S)$ measuring $ 30^\circ$ (since $ \sin\alpha = 1/2$). So, $ \sphericalangle \, (\tilde{T},C,T) = 120^\circ .$
Hence, the area of the sector of the circle is

$\displaystyle \dfrac{120}{360} \pi = \dfrac{\pi}{3}\,, $

and as a result the area of the segment of the circle $ A_3$ is

$\displaystyle A_3 = \dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}\,. $

The area $ A_1$ can be determined by subtracting the area of the circle's segment $ A_3$ from the area between the line $ y=t^2$ and the parabola $ y=x^2,$ $ -t \leq x \leq t\ .$

$ A_1$ = $ \displaystyle \int_{-t}^t t^2 -x^2\,dx - A_3 $
  = $ 2t^3 - \dfrac23 t^3 - \dfrac\pi3+\dfrac{\sqrt{3}}{4} $
  = $ \dfrac{4}{3} t^3 - \dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4}$
  = $ \dfrac{3}{4} \sqrt{3} - \dfrac{\pi}{3} \approx 0.2518 .$

Subtracting the area of the circle and $ A_1$ from the area bounded by the parabola and the line $ m: y = c + 1 = 9/4$, we obtain

$ 2A_2$ = $ \displaystyle 2 \int_{0}^{3/2} \dfrac{9}{4} - x^2dx - \pi - A_1$
  = $ 2 \left[\dfrac{3}{2} \cdot \dfrac{9}{4} - \dfrac{1}{3} \left(\dfrac{3}{2}\right)^3\right] - \pi - \left[\dfrac{3}{4} \sqrt{3} - \dfrac{\pi}{3}\right]$
  = $ \dfrac{9}{2} - \dfrac{2}{3}\pi - \dfrac{3}{4}\sqrt{3} $

$\displaystyle \Longrightarrow A_2 = \dfrac{9}{4} - \dfrac{1}{3}\pi - \dfrac{3}{8}\sqrt{3} \approx 0.5533 \ . $

Here, the upper limit of the integral was derived from the equation $ x^2 = \dfrac94 .$

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