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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

The shaded triangle is bounded by the $ y-$axis as well as the tangent and normal lines of the graph of the function $ y = x^4$ at the point $ P.$

\includegraphics[width=0.5\linewidth]{TdM_09_A1_bild1_en.eps}

Determine $ y_1$ and $ y_2$ in terms of the $ x-$coordinate of $ P.$ For which $ x_{\min} > 0$ is the area $ A$ of the triangle minimal, and what is the minimum value $ A_{\min}?$


Answer:

$ y_1 = $ $ x^4$    
$ y_2 = $ $ x^4$ + $ /x^2$    
$ x_{\min} = $    
$ A_{\min} = $    

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

Solution:

At the point $ P=(x,y)$ the slope of the tangent line is $ 4x^3$ and the slope of the normal line is $ - \dfrac{1}{4x^3}$.
Hence, we have

$\displaystyle y_1 = y - x (4x^3) = -3x^4 $

and

$\displaystyle y_2 = y + x \dfrac{1}{4x^3} = x^4 + \dfrac{1}{4x^2}\ . $

The area of the shaded triangle is

$\displaystyle A(x) = \dfrac{x}{2} (y_2 - y_1) = \dfrac{x}{2} \left( 4x^4 + \dfrac{1}{4x^2}\right) = 2x^5 + \dfrac{1}{8x}\ . $

From

$ 0 = A'(x) = 10 x^4 - \dfrac{1}{8x^2}$
we obtain
$ x_{\min} = \sqrt[6]{1/80} = \sqrt[6]{0.0125} = 0.4817. $

Since $ A(x) \rightarrow \infty $ as $ x \rightarrow 0$ or $ x \rightarrow \infty , $ we obtain the corresponding minimal area

$\displaystyle A_{\min} = \dfrac{1}{2x_{\min}} \left( 4 x^6_{\min} +\dfrac{1}{4} \right) = \dfrac{3}{20} \sqrt[6]{80} = 0.3114 \,. $

[problem of the week]