It's all Tangents

Calculus Level 4

(1): If A B \overleftrightarrow{AB} and A B \overleftrightarrow{A^{'}B^{'}} are tangent to the curves y = x 1 / 4 y = |x|^{1/4} and y = x 4 y = -x^4 at points A , B , A , B A,B,A^{'},B^{'} find the area of trapezoid A B B A AB^{'}BA^{'} above.

(2): If C D \overleftrightarrow{CD} and C D \overleftrightarrow{C^{'}D^{'}} are tangent to the curves y = x y = \sqrt{|x|} and y = x 2 y = -x^2 at points C , D , C , D C,D,C^{'},D^{'} find the area of the trapezoid D C C D DC^{'}CD^{'} above.

Express the answer as the difference of the area of the trapezoids A A B B A A D C C D A_{AB^{'}BA^{'}} - A_{DC^{'}CD^{'}} to seven decimal places.

Extra: Find the perpendicular distance between the parallel lines above, say the perpendicular distance between A B \overleftrightarrow{AB} and C D \overleftrightarrow{CD} .


The answer is 0.0575785.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rocco Dalto
Apr 27, 2018

For (1):

Using the symmetry about the y y axis let f ( x ) = x 4 { f(x) = -x^4 } and g ( x ) = x 1 / 4 { g(x) = x^{1/4} \implies }

d d x ( f ( x ) ) x = a = 4 a 3 \dfrac{d}{dx}(f(x))|_{x = a} = -4a^3 and d d x ( g ( x ) ) x = b = 1 4 b 3 4 \dfrac{d}{dx}(g(x))|_{x = b} = \dfrac{1}{4b^{\frac{3}{4}}} \implies 4 a 3 = 1 4 b 3 4 a = 1 1 6 1 3 b 1 4 -4a^3 = \dfrac{1}{4b^{\frac{3}{4}}} \implies a = \dfrac{-1}{16^{\frac{1}{3}}b^\frac{1}{4}}

Using A : ( a , a 4 ) = ( 1 1 6 1 3 b 1 4 , 1 1 6 4 3 b ) A: (a,-a^4) = (\dfrac{-1}{16^{\frac{1}{3}}b^\frac{1}{4}},\dfrac{-1}{16^{\frac{4}{3}}b}) and B : ( b , b 1 4 ) B: (b,b^{\frac{1}{4}}) \implies

The slope m = 1 6 4 3 b 5 4 + 1 ( 16 b 3 4 ) ( 1 6 1 3 b 5 4 + 1 ) = 1 4 b 3 4 m = \dfrac{16^{\frac{4}{3}} b^{\frac{5}{4}} + 1}{(16b^{\frac{3}{4}})(16^{\frac{1}{3}} b^{\frac{5}{4}} + 1)} = \dfrac{1}{4b^{\frac{3}{4}}} \implies 4 8 3 b 5 4 + 1 = 4 5 3 b 5 4 + 4 4^{\frac{8}{3}}b^{\frac{5}{4}} + 1 = 4^{\frac{5}{3}}b^{\frac{5}{4}} + 4 \implies 4 5 3 b 5 4 = 1 b = 1 4 4 3 a = 1 4 1 3 4^{\frac{5}{3}}b^{\frac{5}{4}} = 1 \implies b = \dfrac{1}{4^{\frac{4}{3}}} \implies a = \dfrac{-1}{4^{\frac{1}{3}}}

A : ( 1 4 1 3 , 1 4 4 3 ) \implies A: (\dfrac{-1}{4^{\frac{1}{3}}}, \dfrac{-1}{4^{\frac{4}{3}}}) and B : ( 1 4 4 3 , 1 4 1 3 ) B: (\dfrac{1}{4^{\frac{4}{3}}},\dfrac{1}{4^{\frac{1}{3}}})

Using the symmetry about the y y axis A : ( 1 4 1 3 , 1 4 4 3 ) \implies A^{'}: (\dfrac{1}{4^{\frac{1}{3}}}, \dfrac{-1}{4^{\frac{4}{3}}}) and B : ( 1 4 4 3 , 1 4 1 3 ) B^{'}: (\dfrac{-1}{4^{\frac{4}{3}}},\dfrac{1}{4^{\frac{1}{3}}}) \implies

B B = 2 4 4 3 , A A = 2 4 1 3 BB^{'} = \dfrac{2}{4^{\frac{4}{3}}}, AA^{'} = \dfrac{2}{4^{\frac{1}{3}}} and using point P : ( 1 4 4 3 , 1 4 4 3 ) P: (\dfrac{-1}{4^{\frac{4}{3}}},\dfrac{-1}{4^{\frac{4}{3}}}) the height h h of the given trapezoid is h = B P = 5 4 4 3 A A B B A = ( 5 4 4 3 ) 2 h = B^{'}P = \dfrac{5}{4^{\frac{4}{3}}} \implies A_{AB^{'}BA^{'}} = \boxed{(\dfrac{5}{4^{\frac{4}{3}}})^2}

In addition:

Using points A A and B m A B = 1 y = x + 3 4 4 3 B \implies m_{AB} = 1 \implies y = x + \dfrac{3}{4^{\frac{4}{3}}}

Using points A A^{'} and B m A B = 1 y = x + 3 4 4 3 B^{'} \implies m_{A^{'}B^{'}} = -1 \implies y = -x + \dfrac{3}{4^{\frac{4}{3}}}

For (2):

Again using the symmetry about the y y axis let f ( x ) = x 2 { f(x) = -x^2 } and g ( x ) = x { g(x) = \sqrt{x} \implies }

d d x ( f ( x ) ) x = a = 2 a \dfrac{d}{dx}(f(x))|_{x = a} = -2a and d d x ( g ( x ) ) x = b = 1 2 b \dfrac{d}{dx}(g(x))|_{x = b} = \dfrac{1}{2 \sqrt{b}} \implies

2 a = 1 2 b a = 1 4 b -2a = \dfrac{1}{2 \sqrt{b}} \implies a = -\dfrac{1}{4 \sqrt{b}}

D : ( a , a 2 ) = ( 1 4 b , 1 16 b ) D: (a,-a^2) = (-\dfrac{1}{4 \sqrt{b}} , -\dfrac{1}{16 b}) and C : ( b , b ) C: (b, \sqrt{b} ) \implies

slope m = 1 2 b = 1 4 b ( 16 b 3 2 + 1 4 b 3 2 + 1 ) m = \dfrac{1}{2 \sqrt{b}} = \dfrac{1}{4 \sqrt{b}} (\dfrac{16 b^\frac{3}{2} + 1}{4 b^\frac{3}{2} + 1}) \implies

16 b 3 2 + 1 = 8 b 3 2 + 2 8 b 3 2 = 1 b = 1 4 16 b^\frac{3}{2} + 1 = 8 b^\frac{3}{2} + 2 \implies 8 b^\frac{3}{2} = 1 \implies b = \dfrac{1}{4}

\implies slope m = 1 m = 1 and a = 1 2 a = -\dfrac{1}{2}

Using D : ( 1 2 , 1 4 ) y = x + 1 4 D: (-\dfrac{1}{2},-\dfrac{1}{4}) \implies y = x + \dfrac{1}{4}

y = x + 1 4 y = x + \dfrac{1}{4} is tangent to g ( x ) g(x) at C : ( 1 4 , 1 2 ) C: (\dfrac{1}{4} , \dfrac{1}{2}) and f ( x ) f(x) at D : ( 1 2 , 1 4 ) D: (-\dfrac{1}{2},-\dfrac{1}{4})

Using the symmetry about the y y axis D : ( 1 2 , 1 4 ) D : ( 1 2 , 1 4 ) D:(\dfrac{-1}{2},\dfrac{-1}{4}) \rightarrow D^{'}:(\dfrac{1}{2},\dfrac{-1}{4}) and C : ( 1 4 , 1 2 ) C : ( 1 4 , 1 2 ) C:(\dfrac{1}{4},\dfrac{1}{2}) \rightarrow C^{'} :(\dfrac{-1}{4},\dfrac{1}{2}) .

Using points D : ( 1 2 , 1 4 ) D^{'} :(\dfrac{1}{2},\dfrac{-1}{4}) and C : ( 1 4 , 1 2 ) y = 1 4 x C^{'} :(\dfrac{-1}{4},\dfrac{1}{2}) \implies y = \dfrac{1}{4} - x and D D = 1 , C C = 1 2 DD^{'} = 1, CC^{'} = \dfrac{1}{2} and the height of the trapezoid C Q = 3 4 C^{'}Q = \dfrac{3}{4} \implies the area of the trapezoid A D C C D = 1 2 ( 3 4 ) ( 3 2 ) = 9 16 = ( 3 4 ) 2 A_{DC^{'}CD^{'}} = \dfrac{1}{2}(\dfrac{3}{4})(\dfrac{3}{2}) = \dfrac{9}{16} = \boxed{(\dfrac{3}{4})^2}

A A B B A A D C C D = ( 5 4 4 3 ) 2 ( 3 4 ) 2 0.0575785 \implies A_{AB^{'}BA^{'}} - A_{DC^{'}CD^{'}} = (\dfrac{5}{4^{\frac{4}{3}}})^2 - (\dfrac{3}{4})^2 \approx \boxed{0.0575785}

Extra:

To find the perpendicular distance between A B \overleftrightarrow{AB} and A B \overleftrightarrow{AB} .

For A B \overleftrightarrow{AB} , y = x + 3 4 4 3 y = x + \dfrac{3}{4^{\frac{4}{3}}}

For C D \overleftrightarrow{CD} , y = x + 1 4 y = x + \dfrac{1}{4}

Let D F \overleftrightarrow{DF} be the line perpendicular to y = x + 1 4 y = x + \dfrac{1}{4} and y = x + 3 4 4 3 y = x + \dfrac{3}{4^{\frac{4}{3}}}

D F \overleftrightarrow{DF} pass thru the point ( 1 2 , 1 4 ) (\dfrac{-1}{2},\dfrac{-1}{4}) with slope m = 1 m_{\perp} = -1 \implies

y = x 3 4 y = -x - \dfrac{3}{4} represents line D F \overleftrightarrow{DF}

Find the point of intersection for the lines:

y = x + 3 4 4 3 y = x + \dfrac{3}{4^{\frac{4}{3}}}

y = x 3 4 y = -x - \dfrac{3}{4}

x = 3 8 ( 1 + 1 4 1 3 ) \implies x = \dfrac{-3}{8}(1 + \dfrac{1}{4^{\frac{1}{3}}}) and y = 3 8 ( 1 1 4 1 3 ) y = \dfrac{-3}{8}(1 - \dfrac{1}{4^{\frac{1}{3}}})

\implies the distance D F = 1 8 4 4 3 + 9 4 2 3 24 2 4 1 3 . 157310 DF = \dfrac{1}{8}\sqrt{\dfrac{4^{\frac{4}{3}} + 9 * 4^{\frac{2}{3}} - 24}{2 * 4^{\frac{1}{3}}}} \approx \boxed{.157310} after simplifying.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...