Calculus – Page 2 – Zachary's Challenges

Gangbuster Geometry

Geometry, trigonometry, and calculus are required for this challenge.

Diagram of pond after being hit by rock. waves are just for illustration purposes and are not part of the problem.

We left an unsolved puzzle at the end of Fascinating Frequencies 3. We want to find the time it takes for a stone to stop after hitting water. The stone has an initial velocity of 1 m/s, a mass of 1 kilogram, and radius 10cm. The force acting on it from gravity is 1kg*g down. The buoyant force acting on it is the weight of the water displaced by the stone. The water has density 0.9982 kg/m³. The friction from the water in any direction acts perpendicular to the surface, inwards, with a strength equal to the sine of the angle multiplied by the density of the water.

\vec F_G=1kg g\ down\\[8pt] \vec F_B=\frac{998.2kg gV_w}{m^3}\ up\\[8pt] \vec F_d=\frac{469.2kgA_f\vec V^2}{m^3}\ up

solution

Sophisticated Shapes

In this challenge, you will need to use geometry, calculus, algebra, and angle theorems.

A roulette is a shape created by rolling things. If you roll circles, you have a trochoid. If you draw the path of a point on the circumference, you have a cycloid. And if you roll that circle inside another circle, you have a hypocycloid. Simple enough. Say you have a circle with radius 25 centimetres and a circle with radius 1 metre. When you rotate the smaller circle inside the bigger one, the path traveled by a point on the circumference is a hypocycloid. It will touch the bigger circle 4 times every rotation. At these points, the instantaneous velocity of the point is zero. The challenge is to find the instantaneous acceleration if the small circle completes 1 revolution inside the bigger one each second. This is different from rolling completely over the inside, which it will do every 4.

solution

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.

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

Captivating Circles 6

This is the final Captivating Circles challenge, this challenge requires the concepts of geometry and trigonometry.

We begin this challenge with 2 concentric circles. The smaller circle divides the larger 1 into 2 equal pieces. We then draw 2 parallel chords in the large circle, which divide it evenly into three. They do not divide the small circle into three equal pieces. The challenge is then to find the area of the biggest piece in the smallest circle if the large circle has radius 1.

Solution