Tangential Curve

Calculus Level 5

Let t ∈ [ 0 , 1 ] t \in [0, 1] be a parameter. Let ℓ t \ell_t be the line passing through the points ( t , 0 ) (t,0) and ( 0 , 1 − t ) (0,1-t) .

For each value of a a we can define the point P t P_t where ℓ t \ell_t intersects the neighboring lines ℓ t + d t \ell_{t+dt} .

The collection of points P t P_t traces out a curve, as suggested in the drawing above. This curve satisfies the equation x n + y n = 1 x^n + y^n = 1 . Determine the value of n n .


The answer is 0.5.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

Arjen Vreugdenhil
Feb 26, 2016

The equation for ℓ t \ell_t is ℓ t : ( 1 − t ) x + t y = t ( 1 − t ) . \ell_t:\ (1-t)x + ty = t(1-t).

Point P t = ( x , y ) P_t = (x, y) is defined by the fact that this equation remains true under small changes of t t . Thus we take the derivative of the equation above, relative to t t : − x + y = 1 − 2 t . -x + y = 1-2t. Substituting this in the original equation for ℓ t \ell_t gives x + t ( 1 − 2 t ) = t ( 1 − t ) ∴ x = t 2 . x + t(1-2t) = t(1-t)\ \ \therefore\ \ x = t^2. It follows that y = 1 − 2 t + x = t 2 − 2 t + 1 = ( 1 − t ) 2 . y = 1 - 2t+x = t^2 - 2t + 1 = (1-t)^2.

Clearly, x + y = t + ( 1 − t ) = 1 \sqrt x + \sqrt y = t + (1-t) = 1 ; this means that n = 1 2 n = \boxed{\tfrac12} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Great problem! I was struggling with this problem a few months ago. Thanks for the proof.

A Former Brilliant Member - 5 years, 3 months ago

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