A calculus problem by Rocco Dalto

Calculus Level pending

Let a , {\bf a, } b , {\bf b, } and c {\bf c } be integers.

If the equation of the tangent line to the curves f ( x ) = x 2 {\bf f(x) = -x^2 } and g ( x ) = x {\bf g(x) = \sqrt{x} } can be expressed as a x + b y + c = 0 {\bf ax + by + c = 0 } , where g c f ( a , b , c ) = 1 {\bf gcf(|a|,|b|,|c|) = 1 }

Find: a + b + c . {\bf |a| + |b| + |c| .}


The answer is 9.

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1 solution

Rocco Dalto
Oct 7, 2016

f ( x ) = x 2 {\bf f(x) = -x^2 } and g ( x ) = x {\bf g(x) = \sqrt{x} \implies }

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

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

A : ( a , a 2 ) = ( 1 4 b , 1 16 b ) {\bf A: (a,-a^2) = (-\frac{1}{4 \sqrt{b}} , -\frac{1}{16 b}) } B : ( b , b ) {\bf B: (b, \sqrt{b} ) \implies }

slope m = 1 2 b = 1 4 b ( 16 b 3 2 + 1 4 b 3 2 + 1 ) {\bf m = \frac{1}{2 \sqrt{b}} = \frac{1}{4 \sqrt{b}} (\frac{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 {\bf 16 b^\frac{3}{2} + 1 = 8 b^\frac{3}{2} + 2 \implies 8 b^\frac{3}{2} = 1 \implies b = \frac{1}{4} }

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

Using A : ( 1 2 , 1 4 ) y + 1 4 = x + 1 2 {\bf A: (-\frac{1}{2},-\frac{1}{4}) \implies y + \frac{1}{4} = x + \frac{1}{2} \implies }

4 x 4 y + 1 = 0 {\bf 4x - 4y + 1 = 0 }

a = 4 , b = 4 , c = 1 {\bf \therefore |a| = 4, |b| = 4, |c| = 1 } and g c f ( a , b , c ) = 1 {\bf gcf(|a|,|b|,|c|) = 1 \implies }

a + b + c = 9. {\bf |a| + |b| + |c| = 9. }

4 x 4 y + 1 = 0 {\bf 4x - 4y + 1 = 0 } is tangent to g ( x ) {\bf g(x)} at B : ( 1 4 , 1 2 ) {\bf B: (\frac{1}{4} , \frac{1}{2}) } and f ( x ) {\bf f(x) } at A : ( 1 2 , 1 4 ) {\bf A: (-\frac{1}{2},-\frac{1}{4}) }

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