That way and this

Calculus Level 5

A particle starts at the origin of 3 − D 3-D space and moves x x units in the direction of x x -axis, y y units in the direction of y y -axis and z z units in the direction of z z -axis. Where x , y , z x,y,z are non-negative real numbers, x + y + z = 10 x+y+z=10 . The value of x x is chosen randomly from [ 0 , 10 ] [0,10] , but the value of y y is chosen randomly from [ 0 , 10 − x ] [0,10-x] . Find the expected value of the square of the final displacement of the particle from the origin. If the answer is of the form a b \frac{a}{b} , where a , b a,b are co prime natural numbers, input a + b a+b .

Inspiration


The answer is 509.

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Raghav Vaidyanathan by Raghav Vaidyanathan , 23, USA

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1 solution

First we choose x x from [ 0 , 10 ] [0,10] . The probability of doing that is d x 10 \large \frac{dx}{10}

Then we choose y from [ 0 , 10 − x ] [0,10-x] . the probability of doing that is d y 10 − x \large \frac{dy}{10-x}

Now z z has to satisfy x + y + z = 10 \large x+y+z =10 so z = 10 − x − y z=10-x-y .

our problem reduces to finding the double integral.

1 10 ∫ 0 10 ∫ 0 10 − x x 2 + y 2 + ( 10 − x − y ) 2 10 − x d y d x \large \frac{1}{10}\int_{0}^{10}\int_{0}^{10-x}\frac{x^{2}+y^{2}+(10-x-y)^{2}}{10-x} dydx

The integral is not hard to evaluate but rather requires patience. At the end hard work pays off and we arrive at the answer 500 9 \large \frac{500}{9} .

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