Radioactive Decay Problem

Radium decomposes at a rate proportional to the amount at any instant. In 100 years, 100 mg of radium decomposes to 96 mg. How many mg will be left after 100 years?

Since it is a Radioactive decay problem we will use the formula

dQ⁄dt = kQ

where:

           dQ⁄dt = rate of change of the substance

Q = amount of the substance present at any time t

k = constant of proportionality

The equation is separable so we need to separate the Differential Equation first.

dQ⁄Q = kdt

Now, we will be going to integrate both sides of the equation.

∫dQ⁄Q = ∫kdt

∫1⁄Q dQ = k ∫dt

After integrating, we’ve got

ln Q =kt + C

This will be our general equation.

In our given, when our t = 0 the amount of substance present is 100 mg. So here we will use the given to find for the value of our constant. By substituting the given to the equation we’ve got

ln Q = kt + C

ln 100 = k(0) + C

ln 100 = 0k + C

C = ln 100 or C = 4.60517

Now that we’ve got the value of C, we next need to find the value of k. In order to see that, we let out time equal to 100 years and the amount of substance present is 96 mg. substituting from our derived equation, we’ve got

ln Q = kt + C

ln 96 = k(100) + ln 100

100k = ln 100 – ln 96

k = (ln 100 – ln 96) ⁄ 100

k = (ln 100⁄96)⁄100

k = -0.0004082

Now that we’ve found the value of our C and k, we will be able to find the amount of mg after another 100 years. So here is our t=200. By substituting the given value to the equation we have

ln Q = kt + C

ln Q = (-0.0004082)(200) + (4.60517)

ln Q = (-0.08164) + (4.60517)

ln Q = 4.52353

e^ln Q = e^4.52353

Q = 92.16

Therefore, the amount of mg left after 200 years is 92.16 mg.

EXACT DIFFERENTIAL EQUATION

 

Newton’s Law of Cooling

Newton’s Law of Cooling states that the rate of change in temperature should be proportional to the difference between the temperature of the object and its surrounding temperature.

Let’s write the statement in mathematical terms.

If the temperature of the object and the surrounding temperature are pretty close then the rate of change is not so big. If the temperature of the object is bigger than the surrounding temperature, then it should be cooling. The temperature should be decreasing and a decreasing temperature would imply a negative rate of change. So, how could be the rate of change have a negative value if the temperature of the object is hotter than the surrounding temperature? To have a negative rate of change when the temperature of the object is bigger than the surrounding temperature is that the k should be negative. So we write it as:

So this is a Separable Differential Equation. Now we are going to find the solution of the Separable Differential Equation.

Substitute the value of C in the given equation.

Where:

Ts = surrounding temperature

T_o = initial temperature or the temperature of the object

t = time

This will give us the formula for Newton’s Law of Cooling.

Let’s have this example.

The rate at which the substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is 30°k and the substance cools from 370°k to 340°k in 15 minutes, when will be the temperature be 310°k?

Let’s first identify the given.

30°k = temperature of the air

340°k = temperature of the substance after 15 minutes

370°k = original temperature of the substance

310°k = temperature of the substance at t

After we identify the given, we need first to know the value of k. So substitute the given to the derived formula.

 

What we are finding here is the value of k so we are going to get its natural logarithm.

Or

Therefore the value of k = 0.00616.

So since we are looking for the time of the substance to have a temperature of 310°k, then using the formula again ad together with the value of k to determine the time at which the temperature of the substance is at 310°k.

So we started at a temperature of 370°k and the temperature decreased to 340°k in 15 minutes and then the temperature decreased to 310°k.

Remember that our surrounding temperature is 30°k. Now, to go from 340°k to 310°k, it took 16.52 minutes. Now, when will the temperature be in 310°k, or how long will it go from 370°k to 310°k? What do you think? We know that it would be the sum of 15 minutes and 16.52 minutes which is basically 31.52 minutes. But we will use the equation to use that answer. By the way, whenever the temperature gets closer to the surrounding temperature, the rate of temperature decreases. The temperature decreases more slowly. In the first 15 minutes, the temperature went down by 30°k, and in the next 18.06 minutes, the temperature also went down 30°k. As the difference between the object and the surrounding temperature decreases, the rate at which the temperature changes decreases as well. So the temperature changes much faster when the difference is larger. But when the difference between the object and the surrounding is small, the temperature doesn’t change that quickly anymore.

Now let’s prove that from 370°k to 310°k is all about 33.04 minutes.

So when we start from 370°k as our initial temperature of the substance at 30°k air temperature, it takes 31.52 minutes for the temperature of the substance to be at 310°k. And when we started at 340°k, it takes 16.52 minutes for the temperature of the substance to be at 310°k.

Find the Differential Equation of the Family of Parabolas with vertex and focus on the x-axis

Given the equation of a parabola

  

We let (h,0)  be the location of the vertex, and  (h+p,0)be the location of the focus where h will be any number along the x-axis and p is the distance of the focus added to the vertex’s distance h. This is to represent the distance traveled by the focus from the point of origin.

The curves are then sideways parabolas with equation

Now we need to find a differential equation that does not depend on arbitrary constants  h and p. Since there are two arbitrary constants to deal with, we need to find its second differentiation. In other words, the highest order derivative that should appear in the differential equation is y”, i.e. the second derivative of y with respect to x.

Now let us differentiate the equation using implicit differentiation.

That gives

which still depends on the constant p.

Since there is still an arbitrary constant which is p , we need to get its second derivative using implicit differentiation for the second time

This last equation does not depend on the constants h  and p.
The desired differential equation is

 

The Importance of Differential Equations

We live in the age of astonishing advancement. Engineers can create robots, physicist can describe the motion of waves, pendulums or chaotic systems, while we communicate wirelessly in a vast world wide network. But underline this modern wonders, are deep and mysteriously powerful, called Differential Equations. But what are differential equations? Where does differential equations come from? And why does they work so well in a vast discipline such as biology, physics, chemistry, economics, engineering, etc… Why are they not generally observed in our daily lives?

Differential equations are equations that contains one or more terms involving derivatives of one variable (dependent variable) with respect to another variable (independent variable) or we can say that these are equations involving derivatives of a function or functions. They have a remarkable ability to predict the world around us. They are used to describe exponential growth and decay, population growth of species or the change in investment return over time, bank interest, even in solving radioactive decay problems, continuous compound interest problems, flow problems, cooling and heating problems, orthogonal trajectories, and also in investigating problems involving fluid mechanics, circuit design, heat transfer, population or conservation biology, seismic waves. They are used in specific field such as, in the field of medicine, where differential equations are used for modelling cancer growth or the spread of disease. In chemistry, they are used for modelling chemical reactions and to computer radioactive half-life. In economics, they are used to find optimum investments strategies. In physics, they are used to describe the motion of waves, pendulums or chaotic systems. They are also used in physics with Newton’s Second Law of Motion and the Law of Cooling that pertains to the temperature of objects and its surroundings. In engineering, they are used for describing the movements of electricity. Differential equations are also used in creating software to understand computer hardware belongs to applied physics or electrical engineering. They are also used in game features to model velocity of a character in games. They are essential tools for describing the nature of the physical universe and naturally also an essential part of models for computer graphics and vision. Differential equations are also used as aspect of algorithm on machine learning which includes computer vision. Also involves solving for optimal certain conditions or iterating towards a solution with techniques like gradient descent or expectation maximization. In Mother Nature, differential equations are essential tool for describing the nature of the physical universe. Even in networking, they are used to understand an outcome of an edge creation model like preferential attachment which says the nodes with probability proportional to their existing degrees. And also used in theories and explanations like using a determinants to estimate the area of the Bermuda triangle. Even in Bots (short for robots), partial and ordinary differential equations helps to provide shapes and interior and exterior designs of machine.

With their broad and advance uses, no wonder why some people found it difficult. However, with their remarkable ability, people spend their time and effort analyzing and describing their behaviors. Some people spend their youths exploring and mastering its concepts and digging dipper on it, exploring its mysteries. It is indeed challenging and worthwhile knowing its characteristics and behaviors, playing a major role in our environment, that’s why everyone of us should appreciate them.

The Journey Begins

Thanks for joining me!

Good company in a journey makes the way seem shorter. — Izaak Walton