The Vector Jogger

Calculus Level 2

A jogger is jogging around a circular park of radius 1 kilometer at a speed of 6.28 kilometers per hour. The jogger starts at some point A A , and jogs around the park to a point B B , diametrically opposite to A A . During his jog, how many minutes after starting from A A , does the ratio between the jogger's distance and the jogger's displacement reach a maximum?

Details:

Take π ≈ 3.14 \pi\approx 3.14

The ratio to be maximized is D i s t a n c e D i s p l a c e m e n t \frac{Distance}{Displacement}


The answer is 30.

Nanayaranaraknas Vahdam by Nanayaranaraknas Vahdam , 22, India

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2 solutions

Let t be the time in hours. Distance moved is proportional to t. Displacement = sin ⁡ 2 t + ( cos ⁡ t − 1 ) 2 =\sqrt {\sin^2 t + (\cos t -1)^2} . The zero of d d x \dfrac{d}{dx} displacement/distance which corresponds to a minimum and has t in [ 0 , 1 2 ) [0, \frac{1}{2}) is at t = 1 2 t=\frac{1}{2}

1 Helpful 1 Interesting 0 Brilliant 1 Confused

This is a very different and interesting approach! It puts my method to shame.

Nanayaranaraknas Vahdam - 7 years ago
Sagar Pradhan
Jun 6, 2014

Ratio is r(radius)*angle divided by radius r Giving the ratio to be max. At the maximum angular position. Therfore angle =180 degrees, I.e, time taken for half the circle.

1 Helpful 0 Interesting 0 Brilliant 1 Confused

Could you specify how the distance is r ∗ θ r*\theta , if θ \theta is the angle?

Nanayaranaraknas Vahdam - 7 years ago

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