What we usually think when we hear the word supernova is a massive explosion of a star that can be seen from a large distance because of its huge scale. This is actually a type II supernova, where a massive star, more than 1.4 solar masses (the Chandrasekhar limit), violently explodes from the red supergiant state to become either a neutron star or a black hole.

However there is a different type of supernovae, named type Ia, which is a very different phenomenon. It occurs when a white dwarf accumulates mass from a nearby star (usually in a binary system) and when that mass exceeds the Chandrasekhar limit, a supernova explosion is observed.

Science

What is the difference between a type I and type II supernovas?

The life of a star depends heavily on the size (the mass) of the star. All stars are initially formed by clouds of dust, that come together by forces of gravity to form a protostar. Once temperature is high enough, the star starts to fuse hydrogen into helium in what is called the “main-sequence” phase of the star’s life.

However, after all hydrogen is fused into helium the path of stars’ lives diverges in two. Smaller stars, that have a mass up to approximately 1.4 times the mass of our sun will become red giants and start fusing higher elements until they obtain a carbon core, where fusion will stop and the star will expel its outer layers of hydrogen and helium in an explosion called “planetary nebula”. What remains from this is a white dwarf that still emits light until its temperature drops and it becomes a black dwarf. 

However, if a star is heavier than the 1.4 solar masses limit (known as the Chandrasekhar limit) the path differs significantly. The star becomes a red supergiant, where temperatures are significantly increased which allows the fusion of higher mass elements, until iron. The star will then explode in a massive explosion called a supernova. What remains after that also depends on the mass of the star. If its core is less than about 2-3 solar masses (known as the Oppenheimer-Volkoff limit) the star will become a neutron star, but if it is heavier, the star will become a black hole.

What is the life cycle of a star?

This is a functions transformation question. What you need to know to solve this is how the addition or subtraction of a number, either within or outside of the function parentheses, i.e. f(x+...) or f(x)+..., affects the function. 

When you’re simply adding numbers to the function, you’re just causing a shift of the whole function in one direction. When you add or subtract a number within the parentheses, you cause a horizontal shift in the function, where adding a number means a shift to the left and adding a negative number means a shift to the right.

When adding or subtracting a number to the whole function, i.e. outside the parentheses, you cause a shift in the vertical direction, and when you add a number you shift upwards and when you subtract a number you shift downwards.

The answers to the questions above are therefore:
a) Max at (4,3); this is simply the original function, no working is needed.
b) Max at (4,-1); here you subtract 4 from the function so you shift the whole function 4 units downwards so the y coordinate changes from 3 to -1 and the x coordinate remains unaffected.
c) Max at (6,4); here you subtract 2 inside the parentheses which results to a rightward shift of 2 units changing the x coordinate from 4 to 6 and then you add 1 to the whole function resulting in an upward shift of 1 unit changing the y coordinate from 3 to 4.

Maths

Can someone please explain to me how to do these questions ? The coordinates of the maximum point of the graph y = f(x) are (4,3). Give the coordinates of the maximum point of the graph with equation: a) y = f(x), b) y = f(x) - 4, c) y = f(x-2) + 1

The short answer is 2e^2.

However to reach this solution you need to work a bit. This is a second order ordinary differential equation, which means that the solution will be a function that comprises of two parts, the particular integral and the complementary function. 

The complementary function is the easiest to find. It is the solution of the differential equation when the left hand side is 0, i.e. y”-2y’+y=0. This you will find by guessing the solution y=e^αx, where α is the constant to be found. By successive differentiation you find that y’=αe^αx and y”=α^2 e^αx. Substitute this in the equation above and you will find: (α^2 -2α +1)e^αx =0. The exponential cannot be 0 therefore the polynomial in the parentheses must be 0, where when you solve that equation you find the double solution α=1. When you have a double solution, your complementary function will be of the form Ae^αx +Βxe^αx. Therefore in this case the complementary function is Ae^x +Bxe^x.

Now, the particular integral will be any function that satisfies the original differential equation. Here you basically have to guess which function may be suitable, but it is most likely going to be one that contains an exponential. In fact it is bx^2 e^x, where b is a constant to be found. You find the constant b using the same method as before, that is successive differentiation to find y’ and y”, which are 2xbe^x +bx^2 e^x and 2be^x +4bxe^x +bx^2 e^x respectively. Plug these answers into the original differential equation and you should get 2be^x=e^x which gives b=1/2. Therefore the particular integral is (x^2 e^x)/2. 

The full solution is the sum of the complementary function and the particular integral so:
y(x)=Ae^x +Bxe^x +(x^2 e^x)/2. Taking the boundary condition y=0 gives A=0 and then by differentiating y(x) once and applying the condition y’=0 gives B=0. The solution is therefore simply y(x)=(x^2 e^x)/2 which means that y(2)= 2e^2.

Y''-2y'+y=e^x，y(0)=y'(0)=0. Solve y(2)

The process of balancing chemical equations is done so that the law of conservation of mass is obeyed. What you need to do when balancing equations is to count how much of each element you have on the left side and how much you have on the right side and add a number (coefficient) so that the numbers on both sides are the same.

For example in the simple equation for the formation of water you know that the process taking place involves a chemical reaction between hydrogen and oxygen. The first step is to write an equation with chemical elements i.e:
H2 +O2 -> H2O

You see then that on the left hand side you have 2 atoms of oxygen and on the right hand side you only have 1 atom. To balance this you add a coefficient on the right hand side like this:
H2 +O2 -> 2(H2O)   *the parentheses are not needed when you write this in proper chemical notation

But then you realise that you have 4 atoms of hydrogen on the right hand side and 2 on the left so to fix this you add a coefficient of two on the left hand side in hydrogen:

2(H2) +O2 -> 2(H2O)

And this is the final balanced equation.


Chemistry

How do I balance equations?

Integration, or anti-differentiation, is one of the two major operations in calculus, the opposite of differentiation. An integral is the mathematical expression where integration is performed and a function is derived, where the original function is the derivative of the integrated function. 

What are integrals?

Express 24 as a product of prime factors

The formula for the area is l•w so your data gives you the exponential equation: 125^(2n) • 25^(n-1) = 625. 

Notice now that every term in the equation has a base that is a power of 5: 5^2= 25, 5^3=125 and 5^4=625.

Using power rules, you can write your equation as: 5^(3•2n) • 5^(2•(n-1))=5^4 or more simply: 5^(6n+2n-2)=5^4

In the equation, since the bases of both sides are the same, it means that the exponents should be equal. Therefore you get: 8n-2=4, or 8n=6 so n=6/8 or 3/4.


How do I find n?                                 There is a rectangle and l=125^2n and w=25^n-1 and there area in 625cm^2. I know n=3/4 and I know you have to multiply l and w. What steps are required after that?

You find the area as usual, by using one of the formulas you already know. The only trick here is to use the lengths in square root format and not as decimal approximations until the end of calculations, as this will give you the most accurate result. 

Sometimes you may also notice that the surds may simplify to an exact number. For example if your base is sqrt(2) units and your height is sqrt(8) units, you’ll find the area by base times height divided by 2, and you’ll get sqrt(16)/2 or 4/2 or simply 2 square units. Notice therefore that leaving the square root intact and not using an approximated decimal gives you a very accurate and exact answer. 

This rule applies in general in mathematics: decimals are usually not useful within calculations so prefer exact numbers, square roots or fractions, and approximate to decimal form after calculations are done.

How do you find the area of a triangle with surds as lengths?

Hello Ambar,

Algebra, and maths in general, can be quite challenging but there are tricks to get through the difficulties. 

What works in every case is practice, because the more problems you solve, the better you’ll understand the underlying theory and you’ll be able to solve many more problems.

However, algebra is a huge area of mathematics, so if you need help in any topic in particular, I’d be happy to help :)

Fotios Tsitsos

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