Strolling Along the River

Calculus Level 3

Janice is taking a walk along a river that runs horizontally from west to east. There is a lamp post 50 meters north of the river that remains there for the entirety of Janice's walk.

Janice starts 50 meters west of the lamp post. Then, she walks in the eastern direction in such a way that her distances from the river and the lamp post are equal at every moment.

At the moment when Janice is closest to the river, the total distance she had walked is equal to a ( b + ln ⁡ ( c + d ) ) a (\sqrt{b} + \ln (c + \sqrt{d})) meters, where a , b , c , d a, b, c, d are positive integers such that b b and d d are not divisible by the square of any prime number.

Find a + b + c + d . a + b + c + d.


The answer is 30.

Steven Yuan by Steven Yuan , 20, USA

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1 solution

Steven Yuan
Jul 8, 2017

The path that Janice takes is actually a parabola. To see this, we will put the problem on the coordinate plane.

Let the river be represented by the equation y = 0 y = 0 and the lamp post be located at ( 0 , 50 ) . (0, 50). Janice's starting point is ( − 50 , 50 ) . (-50, 50). For any point ( x , y ) (x, y) on Janice's path, the distance from her to the river is y y meters, while the distance from her to the lamp post is x 2 + ( y − 50 ) 2 . \sqrt{x^2 + (y - 50)^2}. Setting these two equal to each other, we have

y = x 2 + ( y − 50 ) 2 y 2 = x 2 + ( y − 50 ) 2 0 = x 2 − 100 y + 2500 y = x 2 100 + 25. \begin{aligned} y &= \sqrt{x^2 + (y - 50)^2} \\ y^2 &= x^2 + (y - 50)^2 \\ 0 &= x^2 - 100y + 2500 \\ y &= \dfrac{x^2}{100}+ 25. \end{aligned}

Thus, Janice's path is the parabola represented by the equation y = x 2 100 + 25. y = \dfrac{x^2}{100} + 25.

Since Janice is moving in the eastern direction, she will be closest to the river when her x-position is 0, when she is at ( 0 , 25 ) . (0, 25). Thus, we must find the arc length of the parabola representing Janice's path from x = − 50 x = -50 to x = 0. x = 0.

Let L L represent the length of Janice's path. Recall that the arc length of the graph of y = f ( x ) y = f(x) from x = a x = a to x = b x = b is ∫ a b 1 + ( d y d x ) 2 d x . \displaystyle \int_a^b \sqrt{1 + \left (\dfrac{dy}{dx} \right)^2} \, dx. Since d y d x = x 50 \dfrac{dy}{dx} = \dfrac{x}{50} in this case, we have

L = ∫ − 50 0 1 + ( x 50 ) 2 d x = 1 50 ∫ − 50 0 2500 + x 2 d x = 1 50 [ 1 2 ( x 2500 + x 2 + 2500 ln ⁡ ( x + 2500 + x 2 ) ) ] − 50 0 = 1 100 [ 2500 ln ⁡ 50 − ( − 2500 2 + 2500 ln ⁡ ( 50 2 − 50 ) ) ] = 25 [ ln ⁡ 50 − ( ln ⁡ 50 + ln ⁡ ( 2 − 1 ) − 2 ) ] = 25 ( 2 − ln ⁡ ( 2 − 1 ) ) = 25 ( 2 + ln ⁡ ( 1 + 2 ) ) . \begin{aligned} L &= \int_{-50}^0 \sqrt{1 + \left (\dfrac{x}{50} \right)^2} \, dx \\ &= \dfrac{1}{50} \int_{-50}^0 \sqrt{2500 + x^2} \, dx \\ &= \dfrac{1}{50} \left [ \dfrac{1}{2} \left ( x \sqrt{2500 + x^2} + 2500 \ln \left ( x + \sqrt{2500 + x^2} \right ) \right) \right ]_{-50}^0 \\ &= \dfrac{1}{100} \left [ 2500 \ln 50 - \left ( -2500 \sqrt{2} + 2500 \ln \left ( 50 \sqrt{2} - 50 \right ) \right )\right ] \\ &= 25 \left [ \ln 50 - \left (\ln 50 + \ln \left (\sqrt{2} - 1 \right) - \sqrt{2} \right ) \right ] \\ &= 25 \left ( \sqrt{2} - \ln \left ( \sqrt{2} - 1 \right ) \right ) \\ &= 25 \left ( \sqrt{2} + \ln \left ( 1 + \sqrt{2} \right ) \right ). \end{aligned}

Thus, a + b + c + d = 25 + 2 + 1 + 2 = 30 . a + b + c + d = 25 + 2 + 1 + 2 = \boxed{30}.

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