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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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

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Towns With Angles

Today's question also involves the law of cosines, but with a different purpose. It is a very neat question from a test which is considered "hard". Try it before looking at the answer.

A biker travels from town A to town B at 10 km/h. Another biker is traveling from town B to town C at 12 km/h. The distance from town A to town B is d. Angle ∠ABC = 120°. The bikers are closest after 2.5 hours of riding. Find d.

First, try to draw the situation. Draw the three towns, draw d, draw the angle, and place two points on segment AB and segment BC, which are the bikers.

Let t be the time the bikers are traveling. This means the first biker's distance from town A is 10t and the second biker's distance from town B is 12t. Since the first biker's distance from town A is 10t, his distance from town B is d - 10t.

We need a variable that represents the distance between the bikers. Let's call it x.

As you can see now, we have a triangle with three sides (x, 12t, d - 10t) and an angle of 120° which is opposite to the side with measure x. This situation calls for the law of cosines to find the relation of all variables:
x² = (12t)² + (d - 10t)² - 2(12t)(d - 10t)cos 120°

cos 120° = -0.5, so the relation becomes (after expanding):
x² = 144t² + d² - 20dt + 100t² + 12dt - 120t²

Add like terms and take the square root:
x = √(124t² + d² - 8dt)

We know that x has a minimum when t = 2.5, so if we derive x and plug 2.5 for t, we can find d.

As you should know, when you derive √f(x), you get f'(x)/2√f(x). Since we equate the derivative to zero (at a maximum or a minimum, the derivative is zero), we can now ignore the denominator (since it's always positive and we can multiply by it). Don't ignore it on a test, I'm doing it to save time.

Since we derive by t (dx/dt), the derivative of d² is 0 (d is a constant, not a variable).
0 = 248t - 8d

We know that t = 2.5, so:
0 = 620 - 8d
8d = 620
d = 77.5 km

Towns A and B are 77.5 km apart.

Have a great weekend,
Nadav

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Differential Equations

Today's question deals with differential equations. It's not very difficult, but it's very nice. Also, it provides a reminder for those who forgot how to solve those equations.

Solve: dy/dx = (2tan y) / x

In order to solve that and get a function, we first need to get each variable to where it belongs, and the appropriate "d notation" on its side. To do that, divide by tan y (to move it to the dy side) and multiply by dx (to move it to the x side):

cot y dy = 2/x dx

Now all the variables are sorted out, and we can easily integrate each side to get a function:
ln sin(y) = 2ln x + c (c is the arbitrary constant of integration)

This also equals:
ln sin(y) = ln x² + c

Since we want a function in the form of y = something x, let's make everything an exponent of e (to eliminate the ln's):
sin y = x² * e^c

Since c is an arbitrary constant, we can call it k and make it simpler:
sin y = kx²

Now, take the arcsin of both sides to eliminate the sine:
y = arcsin(kx²)

Have a great weekend,
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

Two Way Limits

It does not happen often, but when it does, it's simply beautiful: a question with two completely different yet right answers. I saw such question on Yahoo Answers today. Although I didn't answer it, the two different answers were so interesting and different in their approach, I had to bring it here.

The question is quite simple for people who know limits, but the two different answers are just amazing.

Find the limit:
lim_{x -> 4} (3 - (5 + x)^1/2) / (1 - (5 - x)^1/2)

As you can clearly see, plugging x = 4 gives a zero in the denominator, which is why the question asks for a limit. You can plug in numbers and get a rough estimation, but we want a definite answer.

The first method is using L'Hopital's rule. This rule says that in order to find lim_{x -> a} f(x)/g(x), you can also find lim_{x -> a} f'(x)/g'(x). All we need now is to differentiate the numerator and denominator and see what we get:
lim_{x -> 4} (-1/2 * (5 + x)^-1/2) / (1/2 * (5 - x)^-1/2)

The halves cancel, and by using the law that says aⁿ / bⁿ = (a/b)ⁿ, we can show that:
lim_{x -> 4} -((5 + x) / (5 - x))^-1/2

Now plug in 4 and you will get -((5 + 4)/(5-4))^-1/2
Which is -9^-1/2 = -1/3

This is one very good solution, and most people who know calculus would choose that. However, there is another solution to this problem. To do that, multiply the numerator and denominator by the conjugate of the denominator:
lim_{x -> 4} (3 - (5 + x)^1/2) / (1 - (5 - x)^1/2) * (1 + (5 - x)^1/2) / (1 + (5 - x)^1/2)

Using the law that says (a + b)(a - b) = a² - b² and the distributive property, we can conclude that:
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (1 - (5 - x))
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (x - 4)

Now multiply both parts of the fraction by the conjugate of the original numerator:
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (x - 4) * (3 + (5 + x)^1/2) / (3 + (5 + x)^1/2)
lim_{x -> 4} (9 - (5 + x)) * (1 + (5 - x)^1/2) / ((x - 4) * (3 + (5 + x)^1/2))
lim_{x -> 4} (4 - x) * (1 + (5 - x)^1/2) / ((x - 4) * (3 + (5 + x)^1/2) )

That is really nice. (4 - x) / (x - 4) = -1, so we can cancel out two terms and turn them into a nice little minus sign:
lim_{x -> 4} -(1 + (5 - x)^1/2) / (3 + (5 + x)^1/2)

Now we can safely plug 4 for x and we get:
-(1 + (5 - 4)^1/2) / (3 + (5 + 4)^1/2)
-(1 + 1^1/2) / (3 + 9^1/2)
-(1 + 1) / (3 + 3)
-1/3

Again, we get -1/3 as an answer.

Math can be very easy if you think outside the box. Try doing it as often as you can.
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

Integration and Substitution

Today I have a simple question of integration by substitution. Although it's simple, it is a question many people find difficult.

Integrate (5x + 10)/(3x² + 12x - 7) by substitution

Unlike most short questions, this one also has a short answer. First, we need to define a variable to be one part of the function. This part, when differentiated, must be divisible by another part, or we'll be left with two variables, which is not fun.

As you can see, when you differentiate the denominator, you get 6x + 12, which is similar to the numerator, 5x + 10 (take 6 and 5 as a common factor, respectively). Now all we need is to define a variable and we're set to go:
z = 3x² + 12x - 7
z' = 6x + 12

This can also be written as:
dz/dx = 6x + 12

Make dx the subject:
dx = dz/(6x + 12)

We want to find:
integral((5x + 10)/(3x² + 12x - 7) dx)

Substitute z for the denominator and also replace dx with what we found:
integral((5x + 10)/z dz/(6x + 12))

Take out common factors:
integral(5(x + 2)/z * dz/6(x + 2))

Cancel (x + 2):
integral(5/6z dz)
5/6 * ln(z) + c

Now substitute back the value of z:
5/6 * ln(3x² + 12x - 7) + c

And that's the integral.

Simple, yet somewhat long, isn't it?
Yours,
Nadav

nadavs

More Trigonometric Identities Revealed

For today there is another question from Yahoo Answers, also about trigonometric identities. This time it's not about simple ones like before, but a more complex one, like the ones I sometimes answer on super math tips. Here it is:

Prove:
(cos 4x + cos 2x) / (sin 4x + sin 2x) = cot 3x

At first glance this seems irritating and impossible, but with some creative thinking it's very possible.

Now, what do 2, 3, and 4 have in common? Of course, 3 is the average, and we're going to use that property heavily.

Since we want to have cot 3x at the end, let's change all sines and cosines to 3x + x or 3x - x and hope they cancel:
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x.

Well, they don't cancel so fast, so we'll have to expand all expressions (remember what's sin (A + B)?) and then hope that we can cancel out some terms. Let's start (I put the arguments in parentheses so you can see them better):
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x
Let's start from the numerator:
cos(3x)cos(x) - sin(3x)sin(x) + cos(3x)cos(x) + sin(3x)sin(x)
2cos(3x)cos(x)

Now, the denominator:
sin(3x)cos(x) + cos(3x)sin(x) + sin(3x)cos(x) - cos(3x)sin(x)
2sin(3x)cos(x)

Now divide those two terms and see what happens:
2cos(3x)cos(x) / 2sin(3x)cos(x)
The 2 and cos(x) cancel:
cos(3x) / sin(3x) = 1/(sin(3x) / cos(3x)) = 1/tan(3x) = cot (3x)

Job complete.

Besides solving these questions, here is a challenge for you: try to solve the previous question (with the towns) using calculus. I haven't solved it myself yet (except for the geometry part, which is easy).

The first one to solve (on comments or email) will be written in the hall of fame on the blog's list (which I'll create) with a link of his choice and the right to post here any question.

Go on to solving!
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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