Jake rides Trevor Piggyback IV

Calculus Level 4

It's a beautiful Friday morning. The wind is gently rolling over the hills and the view from the course does not disappoint.

Starting at ( 0 , 0 ) (0,0) , Jake tells Trevor to move in a spiral with the nth move being of length 1 n 2 \dfrac{1}{n^{2}} starting at step n = 1 n = 1 and continuing to step n n \rightarrow \infty and turning 9 0 90^{\circ} clockwise after every move.

What is the straight line distance (magnitude) of the final displacement in meters Trevor and Jake travelled from the origin? Answer to 3 decimal places.


The answer is 0.9386.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Without loss of generality we can assume that Trevor first heads east, (i.e., in the positive x x -direction), then north, (i.e., in the positive y y -direction), then west, south, east, north, and so on ad infinitum.

The net distance traveled in the positive x x -direction is then

X = k = 0 ( 1 ) k 1 ( 2 k + 1 ) 2 = K , X = \displaystyle\sum_{k=0}^{\infty} (-1)^{k} \dfrac{1}{(2k + 1)^{2}} = K, Catalan's constant , which has a value of

0.915965594177.... 0.915965594177.... (Interestingly, it is not known if this number is irrational.)

The net distance traveled in the positive y y -direction is

Y = k = 1 ( 1 ) k + 1 1 ( 2 k ) 2 = 1 4 k = 1 ( 1 ) k + 1 1 k 2 = A 4 Y = \displaystyle\sum_{k=1}^{\infty} (-1)^{k+1} \dfrac{1}{(2k)^{2}} = \dfrac{1}{4} \sum_{k=1}^{\infty} (-1)^{k+1} \dfrac{1}{k^{2}} = \dfrac{A}{4}

where A = k = 1 1 k 2 2 k = 1 1 ( 2 k ) 2 = B B 2 = B 2 , A = \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^{2}} - 2*\sum_{k=1}^{\infty} \dfrac{1}{(2k)^{2}} = B - \dfrac{B}{2} = \dfrac{B}{2},

where B = k = 1 1 k 2 = π 2 6 , B = \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^{2}} = \dfrac{\pi^{2}}{6}, known as the solution to Basel's problem .

Thus Y = B 8 = π 2 48 , Y = \dfrac{B}{8} = \dfrac{\pi^{2}}{48}, and so the magnitude of Trevor's displacement is

X 2 + Y 2 = K 2 + ( π 2 48 ) 2 = 0.939 \sqrt{X^{2} + Y^{2}} = \sqrt{K^{2} + \left(\dfrac{\pi^{2}}{48}\right)^{2}} = \boxed{0.939} to 3 decimal places.

@Trevor Arashiro You two must be getting tired with all this piggybacking. :P Just thought I should mention that there is some un-Latexed text in the question. You may also want to mention that the n n th move has the given length.

Brian Charlesworth - 6 years, 1 month ago

Log in to reply

Thanks for catching that. :)

Btw, we switched positions. In his problems, he was piggybacking me. Now, I'm piggybacking him since he ran out of energy. Haha.

I'll post my solution later as our solutions differ by a relatively noticeable margin. I didn't know about Catalan's constant so I couldn't do it by your methods.

Trevor Arashiro - 6 years, 1 month ago

Log in to reply

@Trevor Arashiro Just a quick note about your new "mini-problem". I got an answer, (which I shall not mention), which was considered incorrect. I then checked WA, which confirmed my answer. Would you mind having a second look at your posted answer? Thanks. :)

Brian Charlesworth - 6 years, 1 month ago

Log in to reply

@Brian Charlesworth Once again, I dont know what I was doing. Lol, I've been posting some erroneous problems recently due to how busy Ive been with Brilliant and school and golf.

Trevor Arashiro - 6 years, 1 month ago

Log in to reply

@Trevor Arashiro No worries! I'm amazed that you have any time at all for Brilliant given everything else that's going on for you right now. :)

Brian Charlesworth - 6 years, 1 month ago
Brock Brown
May 13, 2015

Python 2.7:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
# JAKE AND TREVOR'S PIGGYBACK SIMULATION v1.0
from math import sqrt
infinity = 100000
class Piggyback():
    def __init__(self):
        self.x = 0
        self.y = 0
        self.direction = "up"
        self.step = 1
        self.turn = {\
            "right":"down",
            "down":"left",
            "left":"up",
            "up":"right"}
    def dist_from_origin(self):
        return sqrt(self.x**2 + self.y**2)
    def walk(self):
        if self.direction == "up":
            self.y += 1.0/self.step**2
        elif self.direction == "right":
            self.x += 1.0/self.step**2
        elif self.direction == "down":
            self.y -= 1.0/self.step**2
        elif self.direction == "left":
            self.x -= 1.0/self.step**2
    def go(self):
        while self.step <= infinity:
            self.walk()
            self.step += 1
            self.direction = self.turn[self.direction]
jake_and_trevor = Piggyback()
jake_and_trevor.go()
print "Answer:", round(jake_and_trevor.dist_from_origin(),3)

Eyyy, glad to see you back Brock. Haven't seen you on for a few weeks or so. I see you still haven't changed a bit, solving everything through CS :P

Trevor Arashiro - 6 years, 1 month ago

Log in to reply

I didn't know you played Hat Fortress 2.

My Steam name is Jeezy Creezy Lemon Squeezy if you want to add me. I must warn you however I have vast hat wearing capabilities.

Brock Brown - 6 years, 1 month ago

Log in to reply

Hahah, well in that case I guess you're a better player than I if you have more hats. I haven't played in a while because Ive been so busy lately.

Of course, I'll add you when I get a chance.

Trevor Arashiro - 6 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...