Rectangles, Squares, Areas, and Calculus

Rectangles, squares, areas, and calculus are closely related, but do you know how much? Well, see this question. It's very common with numbers, I'm just bringing the general version.

A rectangle with sides of lengths x and y has perimeter P. Prove that the largest area of this rectangle is given when the rectangle is a square.

Well, the question already gives us the variables. All we need is to use them, combine all four terms in the question (rectangles, squares, areas, and calculus) and prove what we need to prove.

So, the perimeter is P. This means that 2(x + y) = P. To get an equation with one variable, let's find y in terms of the other numbers:
x + y = P/2
y = P/2 - x

Now we know two sides of the rectangle: x and P/2 - x. Let's multiply them to find the area function:
A = x(P/2 - x)
A = xP/2 - x²

To find the maximum value for A, let's differentiate:
A' = P/2 - 2x

Equate A' to zero:
0 = P/2 - 2x
2x = P/2
x = P/4

To make sure it's a maximum, let's find the second derivative of A and see if it's negative:
A'' = -2

That's definitely negative, so x = P/4 is the point where the area is maximal. As you can see, that's a quarter of the perimeter. When one side of a rectangle is a quarter of the perimeter, it's a square. To prove that, plug P/4 for x in the equation for y:
y = P/2 - P/4
y = P/4

It's now proven. The biggest area of a rectangle with a constant perimeter is a square.
Nadav

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Proof of Trigonometric Inequalities

After yesterday's inequality induction, it's time for some trigonometric induction. The proof is a little hard to understand at first, but it's definitely worth seeing.

Prove:
xsin(1/x) < 1 when x > 0

This question can definitely fit in the short question-long answer template. However, it doesn't necessarily fit short question-hard answer.

First, we know that x is positive, so we can divide by it without changing the inequality sign:
sin(1/x) < 1/x

Now the argument of the sine function and the right side of the inequality are the same, so let's call them y. All we have to prove now is that sin y < y for every positive y.

First, when y is greater than 1, this is obvious. A sine can never be greater than 1. When y is 1, there is no problem either, because sin 1 is about 0.84 (the argument is in radians, don't forget). The problem begins when 0 < y < 1.

To solve that, let's derivate the inequality we want to prove. We get cos y < 1. As you can see, the rate of growth of sin y is cos y, and the rate of growth of y is linear. When y goes up by 0.5, sin y goes by less than that. This means that sin y is always less than y when 0 < y < 1.

Since y = 1/x, we can conclude that:
sin(1/x) < 1/x (when 1/x > 0, which means x > 0)
xsin(1/x) < 1

Problem solved.
Nadav

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Parabolic Minimums

Today's question deals with parabolas, lines, and some calculus. It's not very hard, but, as always, it's very interesting.

Two parabolas, y = x²/x + 7 and y = -x²/4 + 3x, are drawn on the same coordinate plane. A vertical line is going through the parabolas. P and Q are the points of intersections with the first and second parabolas, respectively. Find the vertical line and the shortest distance between P and Q.

The first thing to do when you face such a question is to find an equation that represents the quantity you want, this time the distance PQ. Since the two points are on the same vertical line, they both have the same x-coordinate. Let's call it x. That makes the points:
P(x, x²/2 + 7)
Q(x, -x²/4 + 3x)

The distance between them is just the difference in the y-coordinates (because they lie on a vertical line). The distance is:
d = x² + 7 - (-x²/4 + 3x) = x²/2 + 7 + x²/4 - 3x = 3x²/4 - 3x + 7

We want the shortest d possible, and that's achieved when the derivative of d is zero:
d' = 3x/2 - 3
0 = 3x/2 - 3
3 = 3x/2
6 = 3x
x = 2

Now we know the vertical line, x = 2. Plug this value in d and find the shortest distance between P and Q:
d = 3 * 2²/4 - 3 * 2 + 7 = 3 - 6 + 7 = 4

The shortest distance between P and Q is 4.

Have a great week,
Nadav

nadavs

Sinusoid Temperature

Temperatures are very hard to predict. Even weathermen get them wrong most of the times. However, there are some paradises which have a constant temperature range, such as Brazzaville (which is in Congo). Here is the question with the amazing details:

The temperature in Brazzaville is given by the following function:
C = 17 - 6cos(πt/12 + 5)
0 <= t < 24

1. What are the highest and lowest temperatures on Brazzaville?

2. What is the biggest rate of change (biggest drop/rise) in temperature?

To show how people complicate questions, I'm going to answer it using calculus, and then using a much simpler method.

The highest and lowest temperatures are achieved when the derivative of the function is 0, so let's do it:
C' = 6π/12 * sin(πt/12 + 5)
0 = πsin(πt/12 + 5) / 2

Divide by π / 2:
0 = sin(πt/12 + 5)
πt/12 + 5 = π*k (where k is an integer)
πt + 60 = 12πk
πt = 12πk - 60
t = 12k - 60/π

Now all we need to do is find the k values which give t's that match our requirement. Since 60/π is about 19, we need k's that are bigger than 1, so:
t₁ = 24 - 60/π = 4.901
t₂ = 36 - 60/π = 16.901

To verify which number is a maximum and which is a minimum, we can use the second derivative:
C'' = 6π²/144 * cos(πt/12 + 5)

Plug the t's:
t₁ makes it positive, so it's a minimum.
t₂ makes it negative, so it's a maximum.

Plug the two t's in the original equation:

C = 17 - 6cos(πt₁/12 + 5) = 17 - 6 = 11°C
C = 17 - 6cos(πt₂/12 + 5) = 17 - 6(-1) = 23°C

The temperature in Brazzaville ranges between 11°C - 23°C for the entire year. We can only be jealous of them...

To get the biggest rates of change, we need to find the maximum of the derivative, so we'll set the second derivative to zero and find the two points. One will give out a negative change and the other one will give a positive change. They are opposites of each other, since the cosine function is symmetrical on a vertical line going through a maximum or minimum.
C'' = 6π²/144 * cos(πt/12 + 5)
0 = 6π²/144 * cos(πt/12 + 5)
cos(πt/12 + 5) = 0
πt/12 + 5 = ±π/2 + 2πk (again, k is an integer)
πt + 60 = ±6π + 24πk
πt = ±6π + 24πk - 60
t = ±6 + 24k - 60/π

The t values in bounds are:
t₃ = 10.901
t₄ = 22.901

Plug them in C':
C' = π/2 * sin(πt₃/12 + 5) = π/2 = 1.57 °C/hour
C' = π/2 * sin(πt₄/12 + 5) = -π/2 = -1.57 °C/hour

So the sharpest temperature changes are 1.57 °C per hour.

This question can be solved using a much easier way: common sense.

The t inside the function is multiplied by π and divided by 12, so plugging in t values from 0 to 24 will give values from 0 to 2π inside the cosine function. This means it goes through an entire cycle from t = 0 to t = 24. With that, we can figure out that the maximum value for the cosine function is 1 and the minimum value is -1. Plug these two numbers and you get:
C = 17 - 1*6 = 11°C
C = 17 - (-1)*6 = 23°C


Also, the cosine function is the "steepest" when it goes through the x-axis, or more accurately, when the argument makes the cosine zero. All we need to do is find when the argument is zero and plug it in the derivative:
πt/12 + 5 = ±π/2 + 2πk

I already solved that, so I'm not going to do it again.

As you can see, thinking outside the box can really help when it comes to such long and tiring functions.

Yours,
Nadav

nadavs

Boating Calculus

Today's question is dedicated to all boating lovers and calculus fans. It is not very hard, it has no use in real life, and yet it is a very beautiful question (like the ones on super math tips).

A boat leaves the dock at 12:00pm and heads west at 40km/h. Another boat travels north at 20km/h and enters the dock at 1:00pm. When is the distance between the boats the shortest and what is that distance?

Nice, isn't it? We have an optimization question here, so we need to create a function to find a variable to optimize. We're dealing here with given speeds, variable times and varying distances, so let's call the time variable t (in hours, to match the speeds) and the distance variable d (in km).

Now, the distance from the first boat to the dock is 40t (it starts at the dock). The distance from the second boat to the dock is 20 when it starts, and 20 - 20t total (since it travels towards the dock).

The boats are heading in perpendicular directions, so they form a right triangle. Using the Pythagorean theorem, we can find the distance between the boats, d:
d² = (40t)² + (20 - 20t)²
d² = 1600t² + 400 - 800t + 400t²
d² = 2000t² - 800t + 400

Since we want to find the minimum value for d and d must be positive, we can also find the minimum value of d² (also must be positive). Let y = d²:
y = 2000t² - 800t + 400

Now differentiate this function:
y' = 4000t - 800

To find an extremety, we need to set y' to zero:
0 = 4000t - 800
4000t = 800
t = 1/5

To see that it is indeed a minimum, let's find the second derivative of y:
y'' = 4000
Since it is positive, y is minimal when t = 1/5

Now calculate y when t = 1/5:
y = 2000 * (1/5)² - 800 * (1/5) + 400
y = 2000 * (1/25) - 160 + 400
y = 80 + 240 = 320

However, y is the distance squared, so let's take the square root of this number:
d = sqrt(y) = sqrt(320) = 8sqrt(5) = 17.888... km.

This means that after 1/5 of an hour, which is 60 / 5 = 12 minutes, the boats will be the closest to each other at 8sqrt(5) (or 17.888) km apart.

Hope you enjoyed.
Nadav

nadavs

Trigonometric Derivation

Today I have a question about differentiation with the definition of differentiation, which means using the lim_{h -> 0} (f(x + h) - f(x)) / h. However, today it is going to be about trigonometric functions (it's always about trigonometry, isn't it?).

Differentiate cos(x) using the definition of differentiation.

First, the most important limit to remember is this: lim_{x -> 0} sin(x) / x = 1. When you have this in mind, you can solve anything (well, many things).

Let's write what we need to find:
lim_{h -> 0} (cos(x + h) - cos(x)) / h

Using the identity of the cosine of sums of angles, we can get:
lim_{h -> 0} (cos(x)cos(h) - sin(x)sin(h) - cos(x)) / h

Now follow closely:
lim_{h -> 0} ((cos(x)cos(h) - cos(x)) / h - sin(x)sin(h) / h)
lim_{h -> 0} (cos(x)cos(h) - cos(x)) / h - lim_{h -> 0} sin(x)sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1) / h - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1)(cos(h) + 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos²(h) - 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(-sin²(h))/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h

Now watch the beauty of mathematics reveal itself:
L'Hopitals rule says that to find lim_{x -> a} f(x)/g(x), you can also find lim_{x -> a} f'(x)/g'(x). Let's do that on the first limit:
lim_{h -> 0} -sin²(h) / h
= lim_{h -> 0} (-sin²(h))' / h'
= lim_{h -> 0} -2sin(h)cos(h) / 1
As h approaches 0, sin(h) also approaches zero, so the entire limit is zero, and thus the whole first part of the first big limit.

This leaves us with:
- sin(x) * lim_{h -> 0} sin(h) / h
As you can see, we have lim_{h -> 0} sin(h) / h = 1
That means:
-sin(x) * lim_{h -> 0} sin(h) / h = -sin(x) * 1 = -sin(x)

So the entire derivative equals -sin(x).

That's why super math tips exists. Or you can simply use Wikipedia's list of trigonometric derivatives (but where is the fun in that?).

Nadav

nadavs

Buggy Calculus

Today I have a question of a higher level from the previous two, but it's much easier from the previous ones for people who know calculus (seriously, much easier).

So, here is the question:

A bug is walking on a wooden log. Its distance from the right edge of the log after t minutes is given by the function x(t) = t³ - 9t² + 500 inches.

A. What is the bug's velocity after 4 minutes?
B. When does the bug change direction?
C. When does the bug's acceleration equal zero?

If you don't know calculus you'd probably say "There's no way to know that", but there is a way, and it's simple.

The velocity function of the bug is the derivative of its distance function. The acceleration of the bug is the derivative of its velocity function. When we get those two functions, the question is very easy.

A. Let's find v(t), the bug's velocity function:
x'(t) = v(t) = 3t² - 18t .
To find the velocity after four minutes, we just plug in 4 instead of t:
4² - 18 * 4 = 16 - 72 = -56 inches/minute. A negative velocity means that the bug is going "backwards", meaning towards the right edge (because we defined a positive distance as going away from the right edge, meaning left).

B. To find when the bug changes direction, we need to find out when its velocity is zero. That means the velocity is changing signs, and by changing signs it means a change of direction. We already have the function, so let's make it equal zero:
3t2 - 18t = 0
t2 - 6t = 0
t(t - 6) = 0
t = 0, 6

The bug cannot change direction at minute zero, since it did not have a direction before. That means the bug changes direction on the 6th minute.

C. To find when the acceleration is zero, we first have to find the acceleration function. We do that by derivating the velocity function:
v'(t) = a(t) = 6t - 18.
Now let's make it equal to zero:
6t - 18 = 0
6t = 18
t = 3

That means the bug's acceleration is zero after three minutes.

Don't understand a word of this post? Don't be shy to ask anything on this blog's comments or the super math tips.

Hope you learned something.
Yours,
Nadav

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