The Distance & Section Formulae

These two formulae form the basis of coordinate geometry. The distance formula enables us to find the distance between two points a & b, and the section forula helps us to find the coordinated of a point P which divides a line segment AB by some ratio.

The Section Formula

Here, we have a point P which lies on the line joining A & B. It also divides the line segment AB in some ratio.
Lets assume that the ratio here is m:n. The section formula gives us the scope of finding the coordinates of this point P using the coordinates of A & B. We need the ratio m:n and the coordinates of the points A & B to find the coordinates of P.

We observe in the above figure that AHP & PCB are similar triangles,
Hence, we have AH/PJ=PH/BJ=AP/AB

The Distance Formula

The distance formula gives us the distance between two points in space. We establish the distance formula using the cartesian system here. It is a direct result of the Pythagorean theorem. It is the most important formula and forms the basis of coordinate geometry.

The Co-ordinate Systems

The two major co-ordinate systems are The Cartesian & The Polar co-ordinate systems.

When using the cartesian system, we need to specify the abscissa(x) & the ordinate(y) to denote a particular point while in case of the polar system, we need to specify the distance of a point from the origin & the slope of the line joining the origin to that point(the position vector of that point).

So, we have two axes(OX & OY) in case of the cartesian co-ordinates & we have a pole(O) & an initial line(OX) in case of the polar co-ordinates.

The Most Beautiful Questions from Integration

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Differentiation Rules

If two vector functions are in the sum or difference, the sum property holds true, same as what we have in single variable scalar functions.

(u + v)' = u' + v'

To find the derivative of the product of a scalar & a vector function.

[Here, f is a scalar function while u is a vector function]

(ƒu)' = ƒu' + u dƒ/dt

If we are given two vector functions, & we need to find the derivative of their dot product.

u(t)=(5t2)i + (2t)j + k

Examples of Implicit Differentiation

1. sin(x+y)=x

cos(x+y)(1+y')=1

[Applying the chain rule, since, y=ƒ(x)

y'{cos(x+y)}=1-cos(x+y)

y'=[1-cos(x+y)]cos(x+y)-1

2. exy=(x-y)

exy[x(dy/dx)+y] = 1-(dy/dx)

xexy(dy/dx) + yexy = 1-(dy/dx)

(dy/dx)[1+xexy]=1-yexy

dy/dx=[1-yexy]{1+xexy}-1

3. x2+y3=siny

2x + 3y2(dy/dx) = cosy(dy/dx)

2x = (dy/dx){cosy-3y2)

dy/dx = 2x{cosy-3y2)-1

The Arc Length

We define the arc length for vector functions to be the total length of the arc from the corner point a to b ie the interval for which the curve is defined. It is calculated for both the 2d & 3d vector functions in the same manner.

The Product & Quotient Rules

If we have two or more functions in the product & we need to find the derivative, we use the product rule.
The rule says:

(uv)' = uv' + vu'

& If we have two functions in the form (u/v), we need the quotient rule or we can simply use the product rule replacing (u/v) by u.(1/v)which is now a procut of two functions u & 1/v.

Both the rules have been explained with examples in the videos...

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