Calculus (Part-2)-geometric explanation of differentiation

Almost all online maths tutors like the following  concept very much

Geometric explanation of differentiation-  If we find the derivative of function f(x) at x = x0 then it is equal to the slope of the tangent  to the graph of given function f(x) at the given point [(x0, f(x0))].

But what is a tangent line?

It is not merely a simple line that joins the graph of given function at one point.

It is actually the limit of the secant lines joining points P = [(x0, f(x0)] and Q on the graph of f(x) as Q moves very much close to P.

The tangent line contacts the graph of given function at given point [(x0, f(x0)] the slope of the tangent line matches the direction of the graph at that point. The tangent line is the straight line that best approximates the graph at that point.

<img src="derivative and slope.png" alt="derivative and slope">

 

As we are given the graph of given function, we can draw the tangent to this graph easily. Still, we’ll like to make calculations involving the tangent line and so will require a calculative method to explore the tangent line.

 

We can easily calculate the equation of the tangent line by using the slope-point form of the line. We slope of a line is m and its passing through a point (x0,y0) then its equation will be

                                                              y − y0 = m(x − x0)

 

So now we have the formula for the equation of the tangent line. It’s clear that to get an actual equation for the tangent line, we should know the exact coordinates of point P. If we have the value of x0 with us we calculate y0 as

  y = f(x0)

The second thing we must know is the slope of line

  m = f’(x0)

Which we call the derivative of given function f(x).

 

Definition:

The derivative f’(x0) of given function f at x0 is equal to the slope of the tangent line to y = f(x) at the point P = (x0, f(x0).

All online maths tutors suggest solving fair amount of questions based on this concept

ExampleFind the tangent line to the following function at z=3 for the given function

Solution

We can find the derivative of the given function using basic differentiation as discussed in the previous post            

We are already given that z=3 so

Equation of tangent line is

                                                    y − y0 = m(x − x0)      here y0=R(3)=√7

Putting these values we get equation of tangent line

 

In my online maths tutors series, I will continue discussing calculus and its concepts

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IB Mathematics Tutors-Part 1 (Calculus)

In my IB Mathematics Tutors series, I will explain different topics taught at HL and SL levels of IB Mathematics. Calculus is the first one.

<img src="ib mathematics tutors.jpg" alt="ib mathematics tutors">

If we want to understand the importance of Calculus in IB Diploma Programme, We should have a look at the number of teaching hours recommended for it. It’s 40 hours in SL(out of 150 total hours) and 48 (out of 240 total hours) hours in HL. This makes calculus the most important topic for IB Mathematics Tutors as well as for IB students.

What is Calculus- Calculus is an ancient Latin word. It means ‘small stones’ used for counting. In every branch of Mathematics, we study of something specific, like in Geometry, we study about shapes, in Algebra, we study about arithmetic operations, in coordinate, we study about locating a point. In calculus, we do the mathematical study of continuous change. It mainly has two branches- Read more

How to find sum and product of zeros of equations

In my previous post on IB Mathematics, I have discussed how to solve a quadratic polynomial using Quadratic formula. Here I will tell you about different relationships based on sum and product of quadratic polynomial, cubic polynomial and bi-quadratic polynomials.

<img src="ib mathematics.png" alt="ib mathematics">

Quadratic:

ax2 + bx + c = 0

                                         Sum of the roots = −b/a

                                          Product of the roots = c/a

If we know the sum and product of the roots/zeros of a quadratic polynomial, then we can find that polynomial using this formula

x2 − (sum of the roots)x + (product of the roots) = 0

Cubic:

Now let us look at a Cubic (one degree higher than Quadratic):

ax3 + bx2 + cx + d=0

 if α, β and γ are the zeros of this cubic polynomial then

                                                       α+β+γ=-b/a
             
                                                   αβ+βγ+γα=c/a
           
                                                           αβγ=-d/a
If we know these relationships of polynomials then we cal also calculate the polynomial using this formula:
                         
                                       x³-(α+β+γ)x²+(αβ+βγ+γα)x-(α+β+γ)=0

Bi-Quadratic:

If we are given a bi-quadratic polynomial with degree 4 like:

                                             ax 4+bx³+cx²+dx+e=0
 
                                and its roots/zeros are α, β, γ, and δ then
 
                                           α+β+γ+δ=-b/a
 
                                          αβ+αγ+αδ+βγ+βδ+γδ=c/a
 
                                          αβγ+βγδ+γδα+αβδ=-d/a
 
                                           αβγδ=e/a
using these formulas of sum and product of zeros of polynomials, we can find a lot of relationships in zeros of polynomials. Usually, we are asked to find these types of relationships in zeros.
Question: If α and β are the zeros of polynomial x²-px+q=0 then find the following relationships.
i) 1/α+1/β   ii) α²β+αβ²   iii) α²+β²    iv) α/β+β/α   v) α³+β³
Ans: To find these relationships we convert every value either in sum (α+β)  or in the product (αβ) of zeros. For this conversion, we use following Mathematical Tricks
1)Try to take common
2) Try to take L.C.M
3) Try to make a Perfect Square
4) Use algebraic identities wherever required
If we use above steps properly, in most of the cases we are able to convert everything either in sum or in product of zeros
If we compare the given equation with std. form ax²+bx+c=0 then
                                                                       a=1, b=-p and c=q
                             sum of zeros α+β=-b/a=-(-p/a)=p

                             product of zeros    αβ=c/a=q/1=q                                                     (i) 1/α+1/β=β+α/αβ              [By L.C.M]

               
                 =p/q
(ii) α²β+αβ²  = αβ(α+β)               [By common]
                      =q.p
 (iii) α²+β² = (α²+β²+2αβ)-2αβ          [add and subtract 2αβ, make it a perfect square ]
          
          =(α+β)²-2αβ
 
          =p²-2q
(iv) α/β+β/α = β²+α²/βα          [By taking L.C.M]
 
 
we have already found β²+α² that is p²-2q so
 α/β+β/α=(p²-2q)/q

(v) (α+β)³=(α+β)³-3αβ(α+β)                [Direct algebraic identity]

                =p³-3qp
You can further read about quadratics in PDF(quadratics ) given here. There are a lot of practise questions given in this PDF
<img src="demo.png" alt="demo">