Month: July 2019

Baffling Bungee

This week’s science challenge requires the principals of physics and chemistry to solve.

Diagram of Baffling Bungee Jump.

Bob the bungee jumper goes bungee jumping over water on a cord made out of C5H8. The thickness of the rope is 3 centimetres. It’s natural length, the length it will have if not acted on by a force, is 1 metre. Bob starts the jump with a velocity vector with a magnitude of 1 metre per second and an angle of 45°. The lowest point of the jump is 1 metre above the water. Afterwards, the cord pulls Bob up to a stable position higher above the water. The challenge is to calculate the elasticity of the rope to figure out exactly how high.

Solution

Solution to Recursive Renewals

We want to find the optimal strategy for renewing a book at the Leibniz Library. First, let’s look at the maximum number of days you can keep a book before renewing it. 8 at first, then 7, 7, 6, and then 4. And that’s it, you cannot renew the book again. The fine equation will start above 0 when you renew the book a fifth time. But the hold equation will max out at 100% after only 10 days. The best is clearly 16 days. But it’s very likely it will be put on hold and you’ll be stuck with 14, or worse, 8. After 2 days, the hold equation ≈22%, and the loss would be 1 day if you don’t know. But if you don’t renew, you will probably gain a day. So don’t renew then. But on day 3, you can lose 2 days, and the hold equation now ≈24%. It’s less than 33%, so the day gained would still be worth it. On day 4, the hold equation is now at ≈29%, more than 25%, so you renew. We start again on day 5, with ≈40%, it’s less than 50%, so we don’t renew. But on day 6, we have ≈51%, far more than 33%. Although we lose the possibility of an extra day, it’s quite unlikely, and not nearly enough to make it worth not renewing. So we ought to renew. Not until day 8 can we gain another day. By then the hold equation is already ≈79%. It doesn’t matter if it has been renewed yet, if it hasn’t yet we do. By the time we can gain another day, the hold equation will have reached 100%. So our answer is to renew on days 4, 6, and 8.

f(x)=d^2+5d-7cos(r)+r^3-100\\[10pt] h(x)=\frac{t^2+3sin(t)+15}{100}

Solution to Spinning Stick

Diagram of the Spinning Stick.

We wanted to find the velocity of the ball when it rolls off of the spinning stick. The velocity of a point stick will be 2π metres per minute times it’s distance from the origin, in whatever direction it’s traveling in. This is 90° clockwise of the direction of the stick. This vector makes a circle with radius 2π metres, and it completes a rotation at the same speed. Knowing this, we find the centripetal force the stick experiences to be 4π2 metres per minute squared multiplied by the distance, another 90° clockwise of the stick. This makes centripetal force towards the origin, as you probably already know. If to spin in a circle the ball must accelerate, then it wont. The stick does not hold it in, and so it will quickly fly away. Since acceleration is proportional to distance, velocity increases acceleration. If you try solving Captivating Circles 5, you will see that such relationships relate to e. The rotation doesn’t interfere at all, though you’d think it would. When you have they are the same size and the proportions of change equal, their sum increases exponentially. Every unit of time when one would double the other if the other wasn’t returning the favour, they both multiply by e. They multiply at the same rate even if they aren’t the same size. The difference will decrease by a factor of e in the same duration of time. If one increases the other faster by a factor of n, the other will approach being larger by a factor of √n. The distance from the correct ratio will also decrease by a factor of e in the same duration of time as before. Using this knowledge, we can calculate the velocity, using the duration of time shown before as the unit for time, and take it’s derivative to get distance. Solve this equation to equal 2, and we find the time when the ball leaves the stick. Put time back into the equation for velocity, and we have our answer. 12.3134 metres per minute.

\frac d{dx}\frac{e^t-e^{-t}}{2\sqrt{2\pi}}=\frac{e^t+e^{-t}}{2\sqrt{2\pi}}\\[16pt] t=log(2\sqrt{2\pi}+\sqrt{8π-1})\approx2.295\\[16pt] \overrightarrow v\approx\frac{e^{2.295}-e^{-2.295}}{2\sqrt{2\pi}}\approx12.313

Recursive Renewals

You will need to do algebra, combinatorics, calculus, and trigonometry, and also use multiple matrices, to solve today’s challenge.

When you take a book out at the Leibniz Library, you must return it before the fine equation, f(x), rises above zero. If not, your fine is determined by the fine equation. The variable r represents the number of times you have renewed the book, beggining at zero. The variable d counts days and increases by 1 every day, beggining at zero. The variable t counts days and increases by 1 each day, beggining at 0. Each time you renew the book, the variable r in the fine equation, which begins at zero, increases by 1, and the variable t is reset to 0. You may not renew the book if someone else has it on hold or the fine equation is above 0. The probability that someone puts your book on hold on any given day is determined by the hold equation, h(x). How can you maximize the number of days you have with your book without paying anything?

f(x)=d^2+5d-7cos(r)+r^3-100\\[10pt] h(x)=\frac{t^2+3sin(t)+15}{100}
Solution

Spinning Stick

You will need an understanding of physics and calculus to solve this science challenge.

Diagram of the Spinning Stick.

A uniform stick with length 2 metres rotates at a speed of 1 rotation per minute on a flat plane. It comes into contact with a stationary ball. The stick is kept spinning about it’s origin at the exact same speed by a motor. The ball is only affected by the stick’s rotation, and does not experience friction or roll over the stick. The ball will eventually escape the stick from centrifugal force. The challenge is to find the velocity of the ball when it is flung off of the stick.

Solution

Ridiculous Radical

This challenge covers the topics of calculus and algebra.

In calculus we often find the derivatives of things. The derivitave of 1, π, of any other constant c, is 0. The derivative of x is 1, and that of any constant multiplied by x is that constant. And then the derivative of x2 is 2. All common knowledge. But what about the derivative of √x? That would be √x / 2x, We multiply by the exponent and then decrease it by 1. This does work with x2. But what about the derivative of √x+1? Try to factor the 1 out of the radical. That is today’s challenge. Good luck!

\frac d {dx} \sqrt {x+1}
Solution

Solution to Machiavellian Matrix

We wanted to solve the Machiavellian Matrix. We can began by dividing the top row by xyz to simplify it to ( 1 1 1 1 ). Then we subtract row 1 from row 3 to fin that x = -1/3. This makes x2 = 1/9. Then we can get a system of equations for y and z, that y + z = 4/3, and y2 + z2 = 8/9. If y + z = 4/3 then z = 4/3 – y, and we can substitute to get the quadratic equation. Solve to get (y – 2/3) * (y – 2/3) = 0, so y = 2/3. Our answer is thus that x = -1/3, y = 2/3, and z = 2/3.

\left(\begin{matrix} 1 & 1 & 1 & 1 \\ x & y & z & 1 \\ 4 & 1 & 1 & 0 \end{matrix}\right) \overrightarrow{R_3-R_1→R_1} \left(\begin{matrix} 3 & 0 & 0 & -1 \\ x & y & z & 1 \\ 4 & 1 & 1 & 0 \end{matrix}\right) \\[10pt] y^2+\left(\frac43-y\right)^2-\frac89=0

Machiavellian Matrix

You will need serious matrix solving skills to solve today’s challenge.

Todays challenge is strait forward: solve the Machiavellian Matrix. Machiavellian means crafty and duplicitous, like a trickster. Sometimes this means breaking the rules or finding loopholes. With matrices, machiavellian often means horrifying. Variables are x, y, and z, sum on the right.

\left(\begin{matrix} xyz & xyz & xyz & xyz \\ x & y & z & 1 \\ 4 & 1 & 1 & 0 \end{matrix}\right)
Solution