I wish the car stayed still: part II

Calculus Level pending

A ladybug is climbing on a Volkswagen Beetle (a type of car). In its starting position, the surface of the car is represented by the unit semicircle x 2 + y 2 = 1 , y ⩾ 0 x^{2} + y^{2} = 1, y \geqslant 0 in the x y xy -plane. The x x -axis is the road. At time t = 0 t = 0 the ladybug starts at the front bumper ( 1 , 0 ) (1, 0) , and crawls counterclockwise on the surface of the car at a unit speed relative to the car. At the same time, the car moves to the right at speed 10.

What is the total distance traveled by the bug on the interval of time 0 ⩽ t ⩽ π 4 0 \leqslant t \leqslant \dfrac{\pi}{4} ? If total distance is E \mathfrak{E} , find ⌊ 1000 E ⌋ \left \lfloor 1000 \mathfrak{E} \right \rfloor .

Credit : Problem information is adapted from MIT OCW.


The answer is 7594.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Hobart Pao
May 15, 2016

You can represent the ladybug's position as a vector. You can decompose that vector into two vectors: one vector that connects the origin and the midpoint of diameter on the x x -axis on the Volkswagen Beetle, and another that connects the aforementioned midpoint and the ladybug's position.

The vector connecting the origin and aforementioned midpoint is ⟨ 10 t , 0 ⟩ \left \langle 10t, 0 \right \rangle because it is mentioned that its speed is 10. The vector connecting aforementioned midpoint and the ladybug's position is ⟨ cos ⁡ t , sin ⁡ t ⟩ \left \langle \cos t, \sin t \right \rangle because it's a unit semicircle. Adding the vectors together to get the vector giving the ladybug's position gives ⟨ 10 t + cos ⁡ t , sin ⁡ t ⟩ \left \langle 10t + \cos t, \sin t \right \rangle .

Let's call the vector that gives the ladybug's position p ⃗ \vec{p} . Then , d p ⃗ d t = v ( t ) = ⟨ 10 − sin ⁡ t , cos ⁡ t ⟩ \dfrac{d\vec{p}}{dt} = v(t) = \left \langle 10 - \sin t, \cos t \right \rangle .
We wanted to know the total distance on the interval of time 0 ⩽ t ⩽ π 4 0 \leqslant t \leqslant \dfrac{\pi}{4} . We know that ∫ α β ∣ ∣ v ( t ) ∣ ∣ d t \displaystyle \int_{\alpha}^{\beta} \left| \left| v(t) \right| \right| \, dt gives the distance. In this case, this is ∫ 0 π / 4 ( 10 − sin ⁡ t ) 2 + cos ⁡ 2 t d t \displaystyle \int_{0}^{\pi/4} \sqrt{(10 - \sin t)^2 + \cos^2 t} \, dt . To three decimal places, that is 7.594 7.594 .

Finally, the problem statement asks for the floor of 1000 times the total distance E \mathfrak{E} , which turns out to be 7594 \boxed{7594} .

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What please is a "unit speed relative to the car"???

Andreas Wendler - 5 years, 1 month ago

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The bug is covering the car surface distance at a rate of 1 unit per unit time, not including any distance gained from the car's movement. You know this because on the car, the ladybug's position vector is ⟨ cos ⁡ t , sin ⁡ t ⟩ \left \langle \cos t, \sin t \right \rangle and the velocity vector (on the car, not counting the car's movement) is ⟨ − sin ⁡ t , cos ⁡ t ⟩ \left \langle -\sin t, \cos t \right \rangle . To get the speed of the ladybug on the car, you take the norm of the velocity vector which happens to be sin ⁡ 2 t + cos ⁡ 2 t \sqrt{\sin^{2} t + \cos^{2} t} , which is 1, meaning that with respect to the car and not counting the car's movement, the ladybug is always moving at a speed of 1 on the car's surface.

By the way, "unit speed" was the exact wording given by the MIT exam including this topic, so it should be clear. I hope this helps.

Hobart Pao - 5 years, 1 month ago

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