Surds: Rationalizing the Denominator

In this video, I demonstrate how to rationalize the denominator of a fractional surd with a rule / technique learned in basic algebra called "the difference of 2 squares" - which is the result of multiplication by the conjugate.

We start with the fraction:

\frac{1}{2\sqrt5-\sqrt3}

Now, the difference of 2 squares a² - b² can be expressed as (a-b)(a+b). So if you apply the same principle here and multiply the denominator by 2√5 + √3, we can turn it into an integer.

However, since we multiply the denominator by 2√5 + √3, we also need to multiply the numerator by this as well, so that we are effectively multiplying the entire expression by 1.

Hence:

\frac{1}{2\sqrt5-\sqrt3}=\frac{1}{2\sqrt5-\sqrt3}\times \frac{2\sqrt5+\sqrt3}{2\sqrt5+\sqrt3}

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How to Integrate ∫sin^3(x)cos^3(x)dx - odd powers

Integrals of the product of the powers of sine and cosine come in 4 permutations:

1. The powers m and n are both even
2. The powers m and n are even and odd respectively
3. The powers m and n are odd and even respectively
4. The powers m and n are both odd

In this video, we explore case 4 where both powers are odd. The example we use to demonstrate the methodology in this case is

\int \sin^3x\cos^3x\mathrm{d}x

Remember that cos(x) is the first derivative of sin(x), so we reserve one and write the integrand as...

\sin^3x\cos^3x=\sin^3x\cos^2x\cos x

Then if we write the cos²(x) term as...

\cos^2x=1-\sin^2x

we have...

\sin^3x\left [ 1-\sin^2x \right ]\cos x=\left [ \sin^3x-\sin^5x \right ]\cos x

Then using the substitution u = sin(x), we can easily evaluate the integral.

Please ask me a maths question by commenting below and I will try to help you in future videos.

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Sin, Cos, Tan and the Unit Circle

With sin30˚ = 1/2, cos45˚ = √2/3, and tan60˚ = √3, how do these results relate to the unit circle and how can we use these results to find the sine, cosine and tangent of other angles?

Firstly, what is a unit circle? A unit circle is simply a circle, centred about the origin (0,0), with a radius of 1 unit.

Sine, cosine and tangent, as well as being called trigonometric functions, are also called circular functions. This is because every point on a unit circle can be described in terms of these functions.

In this video, I show you how the x and y-coordinates of a point on the unit circle cosine and sine of the angle of the point with respect to the horizontal.

Using this principle, we then find the sine and cosine of angles that are larger than 90˚ (i.e. obtuse and reflex angles)

Application of Trigonometry and Pythagoras' Theorem

In this video, I demonstrate how to use trigonometry and Pythagoras' Theorem for find the long side of a triangle, with the 2 shorter sides of 2.0m and 3.0m and the angle between them of 135˚ given.

The triangle in this problem is a scalene triangle (3 different angles, and 3 unequal angles). Our task is to find the length of the longest side.

Without knowing, the law of cosines, we need to convert this triangle into a right-angle triangle . We can do that by augmenting it with a 45˚-45˚-90˚ triangle.

Why sin(30˚) equals 1/2

Using some simple geometry, Pythagoras' Theorem and laws of trigonometry, we can easily derive the exact values for sine, cosine and tangent of the angles of 30˚, 45˚ and 60˚.

You may have been expected to remember values such as sin(30˚) = 1/2, sin(45˚) = √2/2, sin(60˚) = √3/2. Unfortunately, this is pretty boring stuff to try and remember.

What's far more interesting however, is where these values come from. In deriving the results, you will also gain a better understanding of how trigonometry works.

For the angle of 45˚, we use a 45˚-45˚-90˚ triangle to derive the results.

For the angles of 30˚ and 60˚, we use a 30˚-60˚-90˚ triangle to derive the results.

Converting Complex Numbers from Cartesian to Polar Form

Complex numbers can be expressed either in Cartesian form

z = x + iy

or in polar form

z = r\textup{cis}\theta

wherever the situation is convenient.

In this video, I demonstrate how to convert a complex number from its Cartesian form to its polar form.

This is done simply by plotting the complex number as a point on the Argand plane, and then drawing a vector from the origin to that point. The angle that the vector makes with the horizontal real axis is the angle θ.

The magnitude of the vector |z| determines r.

The Perpendicular Distance from a Point to a Line

In this video, I demonstrate how to calculate the perpendicular distance from a point (2, -1) to the line described by the equation y = 3x + 1.

In order to calculate this distance, we use the "Perpendicular Distance Formula", given by:

d=\frac{\left | Am+Bn+C \right |}{\sqrt {A^2+B^2}}

where:

• A, B and C are the constants of the standard form linear equation: Ax + By + C = 0
• m is the x-coordinate of the point of interest
• n is the y-coordinate of the point of interest.

Thus, rearranging the equation of our line from y = 3x + 1 to 3x - y + 1 = 0 gives the values (A, B, C) = (3, -1, 1) and our point of interest (m, n) = (2, -1). We can substitute these values into the perpendicular distance formula to calculate the perpendicular distance.

Evaluating a Definite Integral - Example 1/(9+x^2)

In this video we evaluate the definite integral of 1/(9+x²) from x=0 to x=3. An function must be continuous over the domain of integration for a definite integral to work. Definite integrals always result in a number or a value, rather than an anti-derivative function (as in the case of indefinite integrals).

Firstly, we can rewrite the integrand as:

\frac{1}{3^2+x^2}

The anti-derivative of this integrand is:

\frac{1}{3}\arctan \left ( \frac{x}{3} \right )

For definite integrals, we don't need to include the integration constant "+C", because it will cancel itself out when we evaluate the numerical values.

We still need to apply the bounds of the upper limit of x=3 to the lower limit of x=0.

Suggested video:

https://youtu.be/Xi3GdKPK63I

Integral of ∫sec^2(x)dx

The integral of sec²(x) should be a very straight forward operation, because early in our calculus studies, we would have learned that the derivative of tan(x) is equal to sec²(x). That is...

\frac{\mathrm{d} }{\mathrm{d} x}\tan x=\sec^2x

And since integration is the reverse of differentiation, it follows then that...

\int \sec^2x=\tan x+C

However, what if you didn't know this result in the first place?

Well, ironically, you would need to know quite a higher level of trigonometric identities and calculus to complete this integral.

In this video, we explore 2 ways of performing this integral:

1. Integration by parts
2. Using the t-substitution and partial fraction decomposition.

How to Sketch a Parabola - Example 3 (y = x^2 + 6x + 10)

Graphing the function

y = x² + 6x + 10

we go through the normal process of finding:

• its concavity (whether it smiles or frowns)
• y-intercept
• turning point / vertex
• any x-intercepts

But this particular example does not have any x-intercepts because as a positive parabola, the y-coordinates for the vertex and intercepts are positive, and therefore, the parabola will not cross the x-axis.

And also, by the quadratic formula, which is used to solve for the x-intercepts:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

The term b² - 4ac evaluates to a negative number, which cannot be calculated in terms of real numbers.

Conic Sections: The Hyperbola - Part 2

With the standard form of the hyperbola being

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

and given the eccentricity, e > 1, we can define and determine all of the features of the hyperbola.

For instance, the...

• vertices are located at coordinates (a,0) and (-a,0)
• conjugate axis is formed by the coordinates (0,b) and (0,-b)
• foci are located at (ea,0) and (-ea,0)
• directrices have equations: x = a/e and x = -a/e
• asymptotes have the equations: y = (b/a)x and y = -(b/a)x

Derivative of tan(x)

In this video, we find the derivative of tan(x) by considering it as the quotient of sin(x)/cos(x). To find this derivative, we can use a method called the Quotient Rule, which is given by:

y' = \frac{vu'-uv'}{v^2}

So if we let:

u = sin(x)
v = cos(x)

Then...

u' = du/dx = cos(x)
v' = dv/dx = -sin(x)

And substituting these derivatives into the Quotient Rule, we get...

y' = \frac{\cos x\cdot \cos x-\sin x\cdot -\sin x}{\cos^2x}

Conic Sections: The Hyperbola - Part 1

A hyperbola is a type of conic section formed by the locus of a moving point P, such that the ratio (eccentricity, e) of its distance to a fixed focal point F, to a fixed imaginary line called the directrix D is a positive constant that is greater than 1.

Like the ellipse, the hyperbola is bifocal and therefore also has 2 directrices.

In this video, we formulate the standard equation of the hyperbola with the help of 2 geometric properties:

Simplifying Powers / Indices / Exponents - Example 2

In this video, I show you how to turn a long, ugly series of multiplications into a simple elegant expression using powers and index laws.

We start with the expression:

64\times \sqrt[3]{125}\times \frac{1}{16}\times 5^{-2}

We will use the following index laws:

\begin{matrix} 1. &a^{-n} &= &\frac{1}{a^{n}} \\ 2. &a^{\frac{n}{m}} &= &\sqrt[m]{a^{n}} \\ 3. &a^{n}\times a^{m} &= & a^{n+m} \end{matrix}

Our goal is to try and turn this into an expression of 2ⁿ x 5^m (a power of 2 multiplied by a power of 5)

Derivative of (x - 1)ln(x)? - Applying the Product Rule

In this video, I demonstrate how to find the derivative of f(x) = (x - 1)ln(x) by noting that the function is a product of 2 smaller functions - this means the product rule can be applied to find the answer.

The product rule is defined as:

\frac{\mathrm{d} }{\mathrm{d} x}\left ( u\cdot v \right )=v\frac{\mathrm{d} u}{\mathrm{d} x}+u\frac{\mathrm{d} v}{\mathrm{d} x}

Or in shorthand form:

f' = \left (u\cdot v \right )'=vu'+uv'

So if we let...

u = x - 1
v =ln(x)

The derivatives of u and v are...

du/dx = u' = 1
dv/dx = v' = 1/x

We can then substitute these terms according into the product rule to find the derivative of (x-1)ln(x)

The Product Rule of Differentiation - Example x^2 tan(x)

In this video, we find the differential of the function x²tan(x) by considering it as a product of 2 functions, x² and tanx. This allows us to apply a method called the Product Rule, which helps us to find the derivative with relative ease.

The product rule is defined as:

\frac{\mathrm{d} }{\mathrm{d} x}\left ( u\cdot v \right )=v\frac{\mathrm{d} u}{\mathrm{d} x}+u\frac{\mathrm{d} v}{\mathrm{d} x}

Or in shorthand form:

\left (u\cdot v \right )'=vu'+uv'

So if we let...

u = x²
v = tan(x)

The derivatives of u and v are...

du/dx = u' = 2x
dv/dx = v' = sec²(x)

We can then substitute these terms accordingly into the product rule to find the derivative of x²tan(x)

How to differentiate cos(x)/x^2? - Applying the Quotient Rule

Suppose we have the function

f(x) = \frac {\cos x}{x^2}

We note that f is composed of the division of 2 different functions of x, and as such, we have to use the quotient rule in order to find the derivative.

The Quotient Rule is given by:

f' = \frac {vu' - uv'}{v^2}

So if we let...

u = cos(x)
v = x²

then...

u' = -sin(x)
v' = 2x

And applying the Quotient Rule, we get...

f' = \frac {x^2\cdot -\sin x - \cos x\cdot 2x}{(x^2)^2}

The integral of (4x+3)/(x^2+1) - Version 1

In this video, I demonstrate how to integrate (or find the antiderivative of) the expression...

\int \frac{4x+3}{x^2+1} \mathrm{d}x

by separating the numerator into 2 components. This leads performing two separate, but common integrals of:

\int \frac{4x}{x^2+1} \mathrm{d}x + \int \frac {3}{x^2+1} \mathrm{d}x

The first integral can be performed in the same fashion as shown in the video tutorial on how to integrate 4x/(x^²+6), by using a u-substitution. Watch the video here: https://www.youtube.com/watch?v=c_Wgh6F5d5Y

And the second integral can be performed according to the video tutorial on how to integrate 1/(x²+a²), by using a trigonometric substitution. Watch the video here:  https://www.youtube.com/watch?v=Xi3GdKPK63I 

Equation for ellipse with vertical Major Axis

The equation of an ellipse that has its major axis oriented vertically differs slightly from the standard form in that the terms a and b, which represent the lengths of the semi-major and semi-minor axes respectively, is swapped around.

It is given by...

\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1

Notice that the semi-major axis a, now sits under y, the vertical coordinate.

Similarly, the coordinates and relationships for the features of the ellipse are also switched around:

The x-intercepts are given by: (±b, 0)

The y-intercepts are given by: (0, ±a)

The foci are given by: (0, ±ae)

The directrices are given by: (0, ±a/e)

In this video, we go through the following examples:

1. Sketching the ellipse x²/49 + y²/64 = 1 and defining all features
2. Sketching the ellipse (x-1)²/4 + (y+2)²/16 = 1 and defining all features