Volume of sphere

Introduction to Sphere : A tennis ball and a fully blown football  are some familiar objects which bring to our mind the concept of a sphere .A sphere is a three dimensional geometrical object  which can be defined as follow The set of all points in space which are equidistant from a fixed point , is called a sphere.

The fixed point is called the center of the sphere and the constant distance is called its radius. A line segment through the center of a sphere, and with the end points on the sphere . All diameters of a sphere  are of constant length , being equal to twice the radius of the sphere .Thus , if d is the length of a diameter of a sphere   of radius r  then d= 2r . The length of diameter is also called the diameter of sphere .The solid sphere is the region in sphere, bounded by sphere .

Also every point whose distance from the center is less than or equal to the radius is a point of the solid sphere .A sphere can also be considered as a solid obtained on rotating a circle about its diameter.

Volume of a sphere: The volume of a sphere (v) of radius r is given by v = 4/3 ∏r3 cubic units .Let us take an example how to volume of sphere.

Find the sphere volume  of radius 7 cm We know that formula for volume of a sphere of radius r is given by v = 4/3 ∏r3  cubic units here r = 7 cm  therefore v = 4/3 x 22/7 x 7 x 7 x 7 cm3 another example of volume of a sphere.

Calculate volume of a sphere whose surface area is 154 square cm. Let the radius of the sphere be r cm .then ,
Surface area = 154 cm2 => 4∏r2  = 154 => 4 x 22/7 x r2 = 154 => r2 = 154 x 7 / 4 x 22 = 49/4 => r = 7/2 cm , so let v be the volume of sphere .
We will use volume of a sphere equation

That is , v= 4/3∏r3   = 4/3 x 22/7 x 7/2 x 7/2 x 7/2 cm3 = 179.66cm3.

Let us take more example volume of a sphere. A sphere of diameter 6 cm is draped in a right circular cylinder vessel partly filled with water .The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?

We have radius of sphere = 3cm, volume of a sphere formula = 4/3 ∏ r3  cm3 = 4/3 ∏ (3)³ cm³ = 36cm³, radius of cylindrical vessel = 6cm.
Suppose of water level rises by h cm and radius 6 cm = (∏x 62 x h) cm 3 = 36∏h cm3 , clearly volume of the water displaced by the sphere is equal to the volume of  the  sphere =36∏h = > h=1 cm , hence water level rises by 1 cm

Difference (Newton) Quotient Made Simple

To set up a difference quotients for a given function requires an understanding of a function notation. Given the function f(x)=4x^2-3x-7. This notation is read as “f of x equals..” This implies that the value of the function, that is the y-value depends upon the replacement for x. We get the numerical value for the function by substituting a number for ‘x’. If a non-numerical quantity is substituted for ‘x’, we get an expression rather than a numerical value. One important point to be remembered is careful use of parenthesis which is essential. For instance, f(x)=4x^2-3x-7; f(3)=4(3)^2 – 3(3)-7=36-9-7=20

Derivative Quotients  at x for a function f is given by, [f(x+h)-f(x)]/h. Sometimes it is written using delta(x) for the change in x and delta(y) for change in y; delta(x)=h and delta(y)=f(x+delta(x))- f(x). The Difference Quotient is so called as it involves the operations subtraction and division. The common forms of Difference Quotient are as follows:
1. [f(x+h) – f(x)]/h
2. [f(a+h) – f(a)]/h
3. [f(x+delta(x)) – f(x)]/delta(x)

Simplifying Difference Quotient
The difference quotient is simplified to get a h or delta(x) in the denominator which can be canceled to get the final value. Let us consider a difference quotient example to understand the step involved in simplifying difference quotient ; f(x) = 4x^2-3x-7. First we need to find the function f(x+h),which we can get by substituting (x+h) in all x in the given function.  f(x+h)= 4(x+h)^2 – 3(x+h) – 7 = 4(x^2+2xh+h^2) – 3x – 3h – 7= 4x^2+8xh + 4h^2-3x – 3h – 7. Subsituting f(x+h) in the difference quotient, we get,
Difference Quotient = [f(x+h) – f(x)]/h
       = {[4x^2+8xh+4h^2-3x-3h-7] – [4x^2-3x-7]}/h
        = 4x^2+8xh+4h^2-3x-3h-7 -4x^2+3x+7]/h  (opening the parenthesis)
On simplification, we get
        = [8xh+4h^2-3h]/h  
        = h[8x +4h-3]/h            (taking h common)
        = (8x +4h-3)     (canceling h)

Difference Quotient Example
Given function, f(x) = 2x^2-1
First we need to calculate f(x+h) which is got by substituting (x+h) in all x of the function
f(x+h) = 2(x+h)^2 -1 = 2(x^2+2xh+h^2) -1 = 2x^2+ 4xh+2h^2 -1
Next we substitute f(x+h) and f(x) in difference quotient
 [f(x+h) – f(x)]/h ={[2x^2 +4xh +2h^2-1]- [2x^2 -1]}/h
= [2x^2+4xh +2h^2-2x^2+1]/h
= [4xh+2h^2]/h  (combining like terms)
= h[4x +2h]/h      (taking h common)
= [4x +2h] (canceling h)

Newton Quotient
The difference quotient is attributed to Sir Isaac Newton and hence given the name Newton Quotient. The slope of a line through the points [(x+h),f(x+h)] and [x, f(x)] is given by [f(x+h) – f(x)]/h. This expression is the Newton Quotient or Newton’s difference quotient.
Newton Quotient  =[f(x+h) – f(x)]/h

Anti Derivatives and their Rules

What are Anti derivatives?
Anti derivative is nothing but indefinite integral or primitive integral in calculus. If there is a function h, then the anti-derivative of this function will be a differential function, say H. The derivative of H will be equal to h.
H’ = h

The anti derivates are solved by a process called indefinite integration or anti differentiation, which is the opposite process of differentiation that finds the derivative.

Rules of Anti derivatives
The rules of anti derivatives are generally the reverse of the rules of the derivatives. We can say that the anti derivative of a function is equal to the sum of the derivative of the function and a constant, so it is just one step and simple.

Constant Rule
Consider a constant “a”, which has to be multiplied with the function g(y). The value obtained by multiplying the constant with the anti derivative of the function g(y) for all values of y will be equal to the anti derivative of the function g(y), which is calculated after multiplying the constant with the function g(y) for all values of y.

Sum Rule
If there are two functions f(y) and g(y), the anti derivative of the sum of the two functions will be equal to the sum of the anti derivative of the function f(y) and the anti derivative of the function g(y).

Difference Rule
If there are two functions f(y) and g(y), the anti derivative of the difference of the two functions will be equal to the difference of the anti derivative of the function g(y) from the anti derivative of the function f(y).

Reverse Rules
Listed below are the anti derivative rules that form the reverse of derivative rules:
• The anti derivative of cos y is given by sin y + a, where “a” is a constant. Thus we can say, the anti derivative of the derivative of y will result in y + a.
• The anti derivative of sin y is the negative of the sum of cos y and a.
• The anti derivative of the square of sec y is the sum of tan y and a.
• The anti derivative of the square of cosec y is the negative of the sum of cot y and a.
• The anti derivative of the product of sec y and tan y is given by the sum of sec y and a.
• The anti derivative of the product of cosec y and cot y is given by the negative of the sum of cosec y and a.

Power Rule
The power rule of the anti derivative is the reverse of the power rule of the derivative. The power rule of the derivative usually comprises of two steps as the power is brought in the front to be multiplied with the derivative and then the power is reduced by 1 and then it is simplified.

In case of power rule for anti derivative, the power rule comprises of the following two steps.
Step 1: The value of the power in the function is increased by 1. Say for example, g(y) = 6y^2. Then by increasing the power value by 1, the function will become g(y) = 6y^3.

Step 2: Divide the function g(y) obtained from step 1 by the new power value. In this example, it will be: (6 divided by 3) (y^3), which will result in g(y) = 2y^3.

Thus anti derivative of this example is given by 2y^3 + a, where “a” is a constant, as any anti derivative is the sum of the derivative and a constant “a”.

Calculus: Rules of Integration

Calculus Integration Rules
Following are the Rules of Integration where a, b, c, n are some constants and u=f(x);v=g(x) and w=h(x)
General Integration Rules
1.Integral [a dx]= ax +c
2. Integral [a f(x) dx]= a Integral [f(x) dx]
3.Integral [x^n dx]= x^(n+1)/(n+1) +c
4.Integral [f(x)+g(x)+h(x)]dx = Integral[f(x)dx]+ Integral[g(x)dx]+ Integral[h(x)dx]
5. Integral[f(x)-g(x)-h(x)]dx = Integral[f(x)dx] – Integral[g(x)dx] – Integral[h(x)dx]
6. Integration by parts: Integral [u dv] = uv – Integral [v du]
7.Integral[F(u)dx] = Integral [F(u)/u’] du
8. Integral[1/x dx] =ln|x| +c
9. Integral [1/(x^2+a^2)]dx =1/a tan^-1[x/a] +c
10.Integral [1/(x^2-a^2)]dx = [1/2a ]ln |x-a/x+a| +c

Integral Rules of Exponential Functions
1.Integral [e^x dx]= e^x +c
2. Integral [a^x dx]= a^x/ln a +c
3.Integral [ln x dx]= x(ln x -1) +c
4.Integral[log base a of x]dx= (x/ln a)(ln x -1) +c
5. Integral[x e^(ax)]dx= [e^(ax)/a^2](ax-1) +c
6. Integral[e^(ax)/x] dx = ln|x| + summation(i=1 to infinity) [(ax)^i/i.i!] +c
7.Integral[x^2 e^(ax)]dx= e^(ax)[(x^2/a – 2x/a^2 + 2/a^3)] +c
8. Integral[x^n e^(ax)]dx = (1/a)x^n e^(ax) – (n/a)Integral [x^(n-1) e^(ax)]dx
9.Integral[e^(ax)/x^n] dx = [1/(n-1)][- e^(ax)/x^(n-1) + a Integral e^(ax)/x^(n-1)]dx
10.Integral[x^n ln x] dx= [x^(n+1)]/(n+1)^2 [(n+1)lnx – 1] + c

Integration Rules of Trigonometric Functions
1.Integral [sin x dx]=  - cos x +c
2. Integral [cos x dx]= sin x +c
3. Integral [tan x dx]= ln |sec x|+ c
4. Integral [cot x dx]= ln|sin x| + c
5. Integral [sec^2(x) dx]= tan x +c
6. Integral [csc^2(x) dx]= - cot x +c
7. Integral [tan^2(x) dx]= tan x – x +c
8. Integral [cot^2(x) dx]= cot x – x +c
9.Integral [sec x tan x dx]= sec x +c
10. Integral [csc x cot x dx] = - csc x +c
11. Integral [sec x dx]= ln |sec x +tan x| +c
12.Integral [cos^2(x) dx] = x/2 + ¼(sin 2x) +c
13. Integral[sin ^n(x) dx]= (-1/n)sin^(n-1) x cos x + (n-1)/n .Integral [sin^(n-2) x dx]
14. Integral [cos^n(x)dx]= (1/n)cos^(n-1) x sin x + (n-1)/n. Integral[cos^(n-2) dx]

Integration Rules of Hyperbolic Functions
1.Integral[sinh x dx] = cosh x +c
2.Integral[cosh x dx]= sinh x +c
3. Integral[tanh x dx]= ln cosh x +c
4. Integral[coth x dx]= ln |sinh x| + c
5. Integral[sech x dx] = sin^-1[tanh x] +c
6. Integral[csch x dx] = ln tanh (x/2) +c
7. Integral[sinh^2(x)dx] = (sinh 2x)/4 – (x/2) + c
8. Integral[cosh^2(x)dx]= (sinh 2x)/4 + (x/2) +c
9. Integral[sech^2 (x) dx]= tanh x +c
10. Integral[csh^2(x)dx]= -coth x +c
11. Integral[tanh^2(x) dx]= x – tanh x +c
12. Integral[coth^2(x)dx]= x – coth x +c
13.Integral[sechx tanh x] dx =  - sech x +c
14. Integral[csch x coth x]dx = -csch x +c

Derivatives of Exponential functions of e

Derivative is the rate of change at a point which gives the slope of the curve at that point. When the given equation is y=f(x), the derivative is written as dy/dx or d[f(x)]/dx.  To find the derivatives of exponential functions, let us take a quick look at them. Exponential functions are the functions written in the form y = b^x, where b is a positive number that does not equal 1 and x is any real number.  They have a constant base and the exponent is a variable. The most important exponential function is e as the base, which is an irrational number. The function is written as, e(x) and is called the natural exponential function.  Now that we learnt about the natural exponential function e^x, let us learn more about the Derivatives of E.

The natural exponential function is remarkable and so are its derivatives. Let us first find the derivative of E, where E is f(x)=e^x :
As per the definition of derivatives, we get,
d[f(x)]/dx = lim(delta(x)?0) e^[x+delta(x)-e^x]/delta(x)
=lim(delta(x)?0) [e^xe^delta(x)- e^x]/delta(x)
=lim(delta(x)?0) e^x[e^delta(x)-1]/delta(x)
=lim(delta(x)?0) e^x[1+delta(x)-1]/delta(x)
= lim(delta(x)?0)e^xdelta(x)/delta(x)
=e^x
If f(x) = e^x then f’(x) = e^x. This means that slope is the same as the given function value or value of y for all the points on the graph. The other Derivatives of E or derivatives of e^x are as given below:
If u is a function of x, the derivative of an expression in the form e^u can be obtained and is given by d(e^u)/dx = e^u. du/dx

If an exponential function with base b is given, then the derivative of that expression is given by
d(b^u)/dx = b^u.ln b.du/dx

Let us take an example, derivative of E 2 which is derivative of e^x where x=2.  At this point x=2, the value of y=e^x  is approximately 7.39. We know that the derivative of e^x is e^x. So, the slope of the tangent, that is the derivative of e^2 at x=2 is also 7.39 approximately.

Derivative of E 2x will be the derivative of e^2x. To find the derivative of this exponential function, let us take y= E 2 or y= e^2x where u=2x. Using the chain rule, we get  dy/du = de^u/du , where du/dx equals 2.  So, d/dx of [e^2x] is (e^u). du/dx = 2. e^u , substituting u=2x, the derivative of e^2x is 2e^2x

Derivative of E 3x will be the derivative of e^3x. Using d(e^u)/dx = e^u. du/dx where u=3x, we get, e^3x. du/dx which will be 3e^3x as du/dx = 3

Law of Cosines Explained

Trigonometry Law of cosines:
Trigonometry is a field of study relating the angles and sides of a triangle. However, the fundamental ratios are derived easily from a right-angled triangle and are as identified Pythagorean ratios, yet the definite correspondence between the sides and angles can be established using the law of cosines. It is the relation between the sides and cosine of angle.


Convention:
All the angles are depicts upper case letters and sides are depicted by lower case letters. The side opposite to a vertex is represented by the corresponding lower case letter. In a triangle ABC, the side AB= c, BC = a, and AC = b.


Trigonometry law of cosines:
a^2= b^2+c^2 - 2bc cos (A)
b^2 = a^2 +c^2 – 2ac cos (B)
c^2 = a^2 + b^2 – 2ab cos (C)


Law of cosines Example Problems: 
For example, let us consider a triangle ABC, in which a= 3, b= 4 and c= 5.
To evaluate the angle C,
Substitute the values of a, b and c in
c^2 = a^2 + b^2 – 2ab cos (C)
5^2 = 3^2 + 4^2 – 2(3) (4) cos (C)
25= 9+16 – 24 cos(C)
25 = 25 – 24 cos(C)
Solving for cos (C), we get cos(C) =0. Hence, C = 90?.


Prove Law of Cosines:
We always rely on the principles of geometrical principles to prove the laws in trigonometry. To prove the law of cosines we use Pythagorean principle.

 In the above figure, CP is perpendicular to AC extended to P. Hence, BP = a sin(C) and CP = a cos(C).
Applying Pythagoras theorem in the right-angled triangle APB we get,
(AP)^2 + (BP)^2 = (AB)^2
(b – a sin(c))^2 + (a sin(C))^2 = c^2
Expanding        b^2 -2ab cos(C) + b^2 cos2(C) + b^2sin2(C) = c^2
b^2 +a^2 – 2ab cos(C) = c^2


Law of Cosines Problems
Law of cosines, in Physical sciences and technology, has very wide applications. The law of cosine gives us the magnitude of the difference of two vectors acting at an angle. For example, to evaluate the magnitude of the difference of two vectors of magnitude 100 units each acting at an angle of 120?, we get
c^2= 1002+1002 – 2(100) (100) cos (120)
c^2= 1002+1002 -1002
Solving for c, we get c= 100 units.

Derive Law of Cosines
Though there are so many methods by which one can prove the law of cosines, we stick to the use of Pythagorean principles.

In the above figure we have, when CP is perpendicular to AB,
c= a cos (B) + b cos (A)
c^2= ac cos (B) +bc cos (A)
Similarly
b^2=bc cos (A) + ba cos(C)
a^2= ab cos(C) + ac cos (B)
Adding the above two equations we have
b^2 + a^2 = ac cos (B) + bc cos (A) + 2 ab cos(C)
Comparing the above equation with (1), we get
b^2 + a^2 = c^2 + 2ab cos(C)
b^2 + a^2 -  2ab cos(C) = c^2

Derivatives and Graphs of Exponential and Logarithmic Functions

Exponential and Logarithmic Functions
The Logarithmic function with base b is a function, y = logb x. Here b is greater than zero and the function x is defined for all x greater than zero. An Exponential function with base b is a function, y=bx, defined for every real number.
Inverse Function: To find an inverse function (f-1), we need to interchange x and y and then solve for y. Example: f -1( x)  of 2x +1 will be, y=2x+1 (interchange x and y and solve for y)
x =2y+1
y = (x-1)/2 = f-1(x)
The Exponential functions and Logarithmic functions are inverse functions, that is, for any base b, the functions f(x) = logb x, g(x) = bx  are inverses.
Example: let f(x) =ln x and g(x) = ex then f and g satisfy the inverse functions. f(g(x) = ln ex=  x and g(f(x) = eln x= x, f(g(x) = g(f(x) and hence the functions f(x) and g(x) are inverses

Derivatives of Logarithmic and Exponential Functions
The most common exponential and logarithmic functions are natural exponent function ex, and the natural logarithm function, ln(x). The derivatives of Exponential and Logarithmic Functions are:
Exponential Functions derivative: d/dx (ex) = ex   d/dx(ax) =ax ln a
Logarithmic Functions derivative: d/dx(ln x) =1/x    d/dx(loga x)= 1/x ln a
Example: Derivative of f(x) = e3x+2  is given by d/dx (e3x+2) = e3x+2. 3 = 3e3x+2
   Derivative of f(x) = ln (3x+2) is given by d/dx[ln (3x+2)] = [1/(3x+2)]. 2 = 2/(3x+2)

Graphing Exponential and Logarithmic Functions
Exponential functions play a large role in real life. From science to money, graphing these exponential functions provide a visual representations to many applications in real life. Graphs of Exponential and Logarithmic functions using examples are as follows,
Let us now graph an exponential function, f(x) = 2x. First we evaluate f(x) using the integers -3, -2, -1, 0, 1,2,3 and tabulate the values.
x   -3          -2        -1        0 1 2        3
f(x)    1/8 1/4 1/2 1 2 4 8


(x,y)   (-3,0.125)      (-2, 0.25)   (-1,0.5)   (0,1)       (1,2)         (2,4)       (3,8)
Once we get the ordered pairs (x,y) plot the points which gives us the graph of the exponential function f(x)=2x.

Graphing Logarithmic Functions: There are several ways to graph logarithmic functions. The easiest way to graph them is to re-write them in exponential form.
Example:  Graph the logarithmic function, f(x) = log5 x. Re-writing f(x) = y =log5 x in exponential form we get x = 5y, choose values for y and then compute corresponding values for x.  Tabulating the values,
y   -1         0 1        2
x = 5^y 1/5         1             5       25
(x,y) (0.2,-1)     (1,0)           (5, 1)      (25, 2)
Plot the (x,y) values. The graph we get is the graph of the logarithmic function, f(x)= log5 x.