The differentiation is the subfield of Calculus and there are various application of differentiation in real world. The differentiation is very important part of Math as it is used in many scientific fields. Differentiation can be defined as the process of finding the Derivatives of the Functions. Differentiation can be used as a tool to calculate or study the rate of change of a quantity with respect to change in some other quantity. The most common example is calculation of velocity and acceleration. Velocity is given by v = dx / dt, where ' x ' is the distance covered by a moving body in time ‘t’.
Similarly acceleration can be given by can be given by a = dv / dt as acceleration is rate of change of velocity with respect to time. Here ' a ' is the acceleration ' v ' is the velocity and ' t ' is time. Now we will see some other applications of differentiation- 1 ) Normal’s and tangents- Differentiation can be used to find the tangents and normal’s of curve we are studying the different forces acting on a body. Tangent- Tangent can be defined as a straight line that touches the curve at a Point and the Slope of curve and line is same. Normal- The perpendicular line to the Tangent of a curve is known as normal.
Slope can be calculated by using dy / dx or Slope= dy / dx. 2 ) Curvilinear motion- As we can calculate the velocity and acceleration of a moving body we can also use differentiation in curvilinear in which object moves along a curved path. Here we express x and y as function of time and it is known as parametric form. Here horizontal component of velocity is given by v_{x} = dx / dt, vertical component of velocity is given by v_{y} = dy / dt. Magnitude is calculated by v = √ ( v^{2}_{x} + v^{2}_{y} ).
Direction ⊖ of an object can be calculated by tan ⊖_{v }= v_{y} / v_{x}
3 ) Related rates- When two are varying with respect to time and if they are related, then they can be expressed in terms of each other. We will have to differentiate both sides with respect to time d / dt. 4 ) Drawing a curve- We can sketch a curve using differentiation, we can find the Maxima and Minima using given data by finding the first derivative that is dy / dx or y ' and putting it equals to 0 that is y ' = 0 if value of ' x ' is positive then function has local minima and otherwise function has a local maxima. Then we calculate second derivative that is d^{2 }( y ) / dx that is y' ' and if value of y' ' is greater than 0 or y' ' > 0 then curve has minimum type shape otherwise curve has a maximum type shape. |
Slope in Derivatives is a simple and very useful concept in Calculus. We will learn here that how to find the Slope in the different type of Differential Equations by the help of derivative. We will go through the several ways for finding the Slope of a derivative and also solves some of the problems related to the evaluation of the slopes in the derivatives.
When we talk about the topic, first we have to understand the derivative of any expression. Derivative of a function generally shows the change in the function as the input of the function changes. So we can say that a derivative of a function is a quantity that shows that how much a quantity changes when any change occurs in the response of change in some other related quantity. For example the change in the Position of a particle occurs according to the change in the velocity of that object.
The derivative of a function for a chosen value gives the best linear approximation in the function nearby of that input changing value. Now just talking about any of the real value function of a single variable, derivative at any of the Point of the function is the slope of the Tangent line to that graph at that point. In some of the higher dimensions, the derivative of a function at any particular point is the linear transformation of that function at that point. This linear transformation is also called as linearization of that function for that point. The concept of the slope in derivatives is totally same as the concept of the derivative in the differential equations. We simply calculate the differentiation of any of the function to get the derivative of that function. So, slope in derivatives is the process of finding differentiation of the function. In the process of differentiation, we simply finds the rate of change in the value of 'y' in the comparison of the change in the value of the other independent input variable 'x'. This rate of change in the function is called as the derivative of 'y' with respect to the other independent variable 'x'. In some simple words we can say that “y dependence on x” also means that the “y is a function of x” and this functional relationship is given by “ y = f (x)”. Here f is the function of x in the terms of y and this y is dependent to the value of x. If both of the variables are the real Numbers then we can plot a graph for this function and this graph will have a number of tangents for different points on the graph, for a particular point we can find a tangent for that graph and this tangent will show the slope of the function f(x) on that point. We can find the slope of the function at any of the point on the graph. The slope of any curve at any point on that curve is the slope of the tangent line of the curve on that point. Here are a number of types of the function in the calculus mathematics where we calculate the slope of the function for any point of that function. The simplest case is the linear types of Functions where we have to find the slope. For general Linear Function say y = f(x) = mx + c, is a function of line, where m is the slope of linear function and c is the intercept of the line made by line on the co-ordinates. Now here the slope 'm' of the line can be given as the:
m = (change in y) / (change in x),
m = Δy/Δx.
The sign of the delta 'Δ' (delta) shows the changes occurred in the quantities.
We can also get this formula by some of the calculation on the function. As Δ is the change in the quantities then we can write function as:
y = f(x + Δx) = mx + c,
y + Δ y = m (x + Δx) + c,
y + Δy = mx + mΔx + c,
As y = mx + c then it can be written as:
Δy = mΔx,
So, m = Δy/Δx.
In the case if function is not linear, means it is not a Straight Line and the value of m changes accordingly. Differentiation is a perfect method to find the rate of change in any of the quantity.
Here are several Notations for the slope in derivatives. In the Leibniz's notation, s we gives the slope in derivatives as dy/dx and this derivative is read as “derivative of y with respect to the x” or “dy by dx”. To understand the concept of the slope in derivative, if we zoom the graph at any of the point such that the graph looks like a straight line on that point then the derivative at that is the slope of the line. Now talking about a real life example for change in the derivatives, say a velocity of a bike changes as the driver changes the speed and change in the distance occurs when change in the velocity occurs. The acceleration is the example of double derivative.
Now just talking about the some of the examples,
1. Derivative of the expression: f(x) = 2x2 – 5x + 3.
Now, the dy/dx = differentiation of (2x2 – 5x + 3),
dy/dx = 2*2x – 5 = 4x – 5.
2. Calculating the slope of function y = x3, for the points x = 1 to x = 3. So, we have to find the derivative of the function at that point.
dy/dx = 3x2,
dy/dx at point x = 1 is 3 and dy/dx at x = 3 is 27.
3. y = 5x2 + 3x so tangent can be given as dy/dx = 10x + 3.
Similarly, we can calculate the slope in derivatives all type of equations.
When we make an angle with the positive direction of x axis in anticlockwise sense is called as a tangent of line or Slope or gradient of line. So, tangent is a trigonometric angle, which is called as a Slope or gradient of the line. The slope of line is generally denoted by m, where
m = tan t, here t is the angle which makes with the positive direction of x axis in anticlockwise sense and derivative of tangent is sec^{2} x. Now we discuss how we calculate tangent angle-
Parallel to x axis: when a line parallel to x axis makes an angle of 0 degree with x axis. Therefore its slope or tangent is tan 0 = 0.
Perpendicular to x axis: perpendicular to x axis or parallel to y axis is same situation. So, when we calculate an angle from x axis, it produces 90 degree. Therefore its slope or tangent is tan 90 = ∞.
Inclined with axis: The Slope of a line, equally inclined with axis, is 1 or -1 as it makes 45 degree or 135 degree with x axis and we know that tan 45 = 1 and tan 135 = -1.
Now we discuss tangent in terms of coordinates of two points-
Let two points A(x_{1},y_{1}) and B(x_{2},y_{2}) are represents a line, then tangent or slope of line is-
m = (y_{2} – y_{1})/(x_{2} – x_{1}) = difference of ordinates/difference of Abscissa.
For example a line represents two points (1,2) and (3,4), then tangent or slope of line is-
m = (y_{2} – y_{1})/(x_{2} – x_{1}) = (4 – 2)/(3 – 1) = 2/2 = 1
So, tangent of line is m = 1.
Now we discuss how we calculate a tangent from a given equation:
Let a line whose equation is ax + by + c = 0, then tangent or slope of that line is -
m = -a/b = -coefficient of x/coefficient of y
For understand this property, we take an example -
Example: Find the tangent of line, whose equation is 3x – 2y + 5 = 0.
Solution: given line equation is 3x – 2y + 5 = 0, then tangent or slope of line is-
m = -a/b = -(3/-2) = 3/2,
So, tangent of line is 3/2.
Now we discuss how we calculate tangent between two lines:
If t is a Angle between two lines whose slope m_{1} and m_{2}, then-
tan t = (m_{1} – m_{2})/(1 + m_{1}m_{2}),
If both lines are parallel, then angle between both line is 0 degree.
Then tan 0 = 0 => m_{1} – m_{2} /(1 + m_{1}m_{2}) = 0 => m_{1} = m_{2}.
Now we can say that if slope of two lines are equal than both lines are parallel.
Similarly there is a condition which shows perpendicularity:
When two lines are perpendicular, then angle between both lines are 90 degree. So,
tan 90 = ∞ = (m_{1} – m_{2})/(1 + m_{1}m_{2}) => m_{1}m_{2 }= -1,
Thus if two lines are perpendicular, then product of their slope is -1 and if m is the slope of a line, then slope of a line perpendicular to it -1/m.
In differential Point of view, a tangent is equal to dy/dx, where y is a function
So, m = tan t = dy/dx,
And differentiation of tangent is sec^{2} x.
We take some example which shows how we calculate tangent of a given curve by using differentiation -
Example 1: Find the tangent of curve x^{2} + 3y + y^{2} = 5 at (1,1).
Solution: The equation of curve is x^{2} + 3y + y^{2} = 5.
Now we differentiate the given curve with respect to x -
2x + 3*(dy/dx) + 2y*(dy/dx) = 0,
=> dy/dx = -2x/(2y+3),
=> (dy/dx)_{(1,1)} = -2/(2+3) = -2/5,
So, slope or tangent of curve at (1,1) is (dy/dx)_{(1,1) }= -2/5.
Example 2: Prove that tangent to the curve y = 2x^{3} – 3 at the points x=2 and x=-2 are parallel.
Solution: the equation of given curve is y = 2x^{3} – 3 …........equation(1)
Now we differentiate this curve with respect to x -
dy/dx = 6x^{2}
at x = 2, dy/dx = 6*(2)^{2 }= 6*4 = 24,
at x = -2, dy/dx = 6*(-2)^{2 }= 6*4 = 24,
So, at both point x=2 and x = -2, tangent are same. Therefore m_{1} = m_{2,}
Thus we can say that tangent to the curve y = 2x^{3} – 3 at the points x=2 and x=-2 are parallel.
Example 3: Prove that the tangents to the curve y = x^{2} – 5x + 6 at the points (2,0) and (3,0) are perpendicular to each other.
Solution: the equation of the curve is y = x^{2} – 5x + 6 …..........equation(1)
Now we differentiate this curve with respect to x-
dy/dx = 2x – 5,
Now, we calculate the slope-
Slope at (2,0) = m_{1} = dy/dx = 2*2 – 5 = 4 – 5 = -1,
Slope at (3,0) = m_{2} = dy/dx = 2*3 – 5 = 6 – 5 = 1,
Here m_{1}*m_{2} = -1,
So, we can say that the tangents to the curve y = x^{2} – 5x + 6 at the points (2,0) and (3,0) are perpendicular to each other.
These are example which shows us procedure to calculate tangents of curve and properties of tangents.
Now we discuss how we derive equation of a line from a tangent: If a Straight Line is passing through a point P(x_{1},y_{1}) and m is slope of line, then equation of line is -
y – y_{1 }= m(x – x_{1}),
Now we discuss how we calculate equation of a given tangent:
We use following steps to calculate equation-
Step 1 : First of all we assume a given curve as a variable like y = x^{2}.
Step 2 : After assuming the variable, we calculate dy/dx.
Step 3 : Find the value of dy/dx at the given point P(x_{1},y_{1}).
Step 4 : if dy/dx at that point is a non-zero finite number, then the equations of tangent is -
y – y_{1 }= (dy/dx) (x – x_{1}).
Step 5: if dy/dx = 0, then the equation of tangent y – y_{1} = 0,
and if dy/dx = ∞, then the equation of tangent x – x_{1} = 0,
We take some examples to understand the procedure to the equation of tangent-
Example 1: Find the equation of the tangent to the curve y = -5x^{2} + 6x + 7 at point (½,35/4).
Solution:
Step 1: First of all, the equation of given curve y = -5x^{2} + 6x + 7.
Step 2: now we calculate dy/dx of given curve with respect to x-
dy/dx = -10x + 6
Step 3: the value of dy/dx at (½,35/4) is-
(dy/dx)_{(1/2,35/4)} = -10(1/2) + 6 = -5 + 6 = 1,
Step 4: the required equation of the tangent at (½, 35/4) is-
y – 35/4 = (dy/dx) (x - ½),
=> y – 35/4 = (1) (x - ½),
=> y = x + 33/4.
So, equation of tangent is y = x + 33/4.
Example 2: Find the equation of the tangent to the Parabola y^{2} = 4ax at point (at^{2},2at).
Solution:
Step 1: First of all, the equation of given curve y^{2} = 4ax.
Step 2: now we calculate dy/dx of given curve with respect to x-
2ydy/dx = 4a => dy/dx = 2a/y,
Step 3: the value of dy/dx at (at^{2},2at ) is-
(dy/dx) = 2a/2at = 1/t,
Step 4: the required equation of the tangent at (at^{2},2at) is-
y – 2at = (dy/dx) (x - at^{2}),
=> y – 2at = (1/t) (x – at^{2}),
=> y + tx = 2at + at^{3}.
So, equation of tangent is y + tx = 2at + at^{3}.
Example 3: Find the equation of the tangent to the curve y = 2x^{2} + 3sin x at x=0.
Solution:
Step 1: First of all, the equation of given curve y = 2x^{2} + 3sin x,
Step 2 : now we calculate dy/dx of given curve with respect to x-
dy/dx = 4x + 3 cos x,
Step 3: the value of dy/dx at x = 0 is-
(dy/dx) = 4(0) + 3 cos(0) = 0 + 3 = 3,
Step 4: the required equation of the tangent at x = 0 is-
y – 0 = (dy/dx) (x - 0),
=> y – 0 = (3) (x - 0),
=> y = 3x.
So, equation of tangent is y = 3x.
These are example which shows procedure to calculate equation of tangent and shows how to calculate differentiation tangents.
Normal Differentiation is a method of obtaining the rate at which a dependent variable or a dependent output say ‘y’ changes with respect to the change in a independent variable or input. This rate of change is called the derivative of y with respect to x. However physical meaning of normal differentiation says that if a graph is plotted between a dependent variable y and a independent variable x, then Slope of the graph gives the derivative of variable 'y' with respect to independent variable 'x'. This Ratio of change or derivative is represented by Δy/Δx. Where symbol ‘Δ’ represents the change in variable whereas, a great mathematician Leibniz represented this ratio of change as dy/dx which also called the derivative of y with respect to x. Now, let’s derive the formula for normal differentiation of a function.
Let f(x) is a function of x then, change in this function can be represented by f(x+Δx) and the derivative of this function is given by the formula:
Δf(x)/ Δx = [f(x+Δx) – f(x)]/ Δx = df(x)/dx,
This expression is called Newton’s difference quotient. Friends, limit of a function play an important role while calculating the Derivatives of different Functions. For example, the derivative of a function f(x) at some Point a can be computed by the relation.
F'(a) = lim_{h→0} [f(a+h)-f(a)]/[(a+h)-a], further we can say that if h approaches to zero and if limit exists then, we can say f(x) is a differentiable function at a point of its interval ‘a’. Friends, concept of differentiation is widely applicable in many calculations in Calculus differentiation and is mostly used to find the equations of Tangent normal and slopes of different curves. Study of differentiation leads to the study of derivability, continuity and differentiability of a function.
In Geometry, a technique that defines basic concept of shape in a plane is called as curve sketching. So, curve sketching Calculus is basically used for solving a mathematical problem about shapes in geometry and for solving the typical mathematical problems like area, maximum, minimum value of certain equation or curve. For sketching a curve, we use following steps -Step 1 : Firstly we Find out the ‘x’ and ‘y’ intercepts of the curve, for finding the ‘x’ intercept, we put ‘y’ equals to 0 and for finding the ‘y’ intercept, we put ‘x’ equals to 0 means
If y = 0, then result = x
If x = 0, then result = y
Step 2: After intercepts, we find out symmetry of curve by putting ‘x’ as a (-x), in given equation where ‘x’ is an assumed variable -
If in the given equation of the curve, power of the ‘x’ is even, then in this situation y-axis is an Axis of Symmetry in curve and if in the given equation of the curve, power of the ‘y’ is even, then in this situation x-axis is an axis of symmetry in curve and if the sum of the degree of ‘x’ and ‘y’ is even or odd, then in this situation, the curve is symmetric about the origin, where origin is called center of the curve.
Step 3: After completion of above two steps, we calculate the limits of ‘x’ and ‘y’.
Step 4: After above three steps, we check that curve is passing through origin or not and if the curve passes through the origin, then we calculate Tangent lines. For calculation of tangent lines, we solve lowest order term from the equation and remove all other terms of equation in algebraic curves.
Step 5: Next we solve highest order term of equation and calculate the points where the curve meets the line at infinity in algebraic curve.
Step 6: After all these steps, we calculate asymptotes of curve and also calculate where the asymptotes of the curve intersect the curve, means we calculate from which side the curve approaches to the asymptotes.
In above 6 steps, we discussed all the processes, but we have not discussed, what are asymptotes of curve and how we sketch asymptotes of a curve. Knowledge of asymptotes is important for sketching a curve, so we will discuss asymptotes of a curve:
An Asymptote of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity. The word asymptote is derived from the Greek word asymptoto which has the meaning “not falling together” means any line which doesn’t intersects the curve is called as an asymptote of curve. Apollonian of Persia introduced this term.
Three kinds of asymptotes are used in mathematics named as a horizontal asymptotes, vertical asymptotes and oblique asymptotes. Horizontal asymptotes of a function are the horizontal lines such that the graph of the function approaches to +∞ or - ∞. Vertical asymptotes are vertical lines according to which the function grows without any bound.
The asymptotes are generally encountered in study of calculus and for the curve y = f(x).
Horizontal asymptotes are defined as,
Lim _{x→∞} f( x ) = r or,
Lim _{x→-∞ }f( x ) = p,
y = p is called the Horizontal Asymptote of graph of function ‘f’.
Vertical asymptote of a function is defined as,
Lim _{x →a+} f ( x ) = ± ∞ or,
Lim _{x →a-} f ( x ) = ± ∞,
Where ‘a’ is a real number and the vertical line ‘x = a’ is called the Vertical Asymptote.
Other asymptotes can be defined as
Lim _{x→∞} [f ( x )] – ( mx + b )] = 0 or,
Lim _{x→-∞} [ f (x )] – ( mx + b )] = 0,
The line y = mx + b is an asymptote of the graph. Here if m ≠ 0; y = mx + b is called an oblique asymptote.
If f( -x ) = f( x) then function ‘f’ is said to be symmetric about the y axis,
If f( - x) = - f ( x ) then the function ‘f’ is symmetric about the origin.
This is all about asymptotes of a curve.
Now we take an example which describes how we use above 6 steps -
Example 1: By using above six steps sketch the curve of following function:
f(x) = x^{3}/(9 - x^{2})?
Solution: We use above six steps to sketch curve of function f(x) = x^{3}/(9 – x^{2}) :
Step 1: First of all we evaluate ‘x’ and ‘y’ intercepts -
For ‘x’ intercept, we put ‘y’ equals to 0
Means y = x^{3}/(9 – x^{2}) = 0 = > x^{3}/(9 – x^{2}) = 0 = > x = 0,
For ‘y’ intercept, we put ‘x’ equals to 0
Means y = x^{3}/(9 – x^{2}) = > 0/(9 – 0) = > y = 0,
So, x =0, and y =0 are intercepts.
Step 2: Now, we will check the symmetry of function -
We put ‘-x’ as ‘x’ in above function, then,
f(-x) = (-x)^{3}/(9 – (-x)^{2}) = -x^{3}/(9 – x^{2}) = -f(x),
Means f(-x) + f(x) = 0,
If Function is odd than this function is symmetric about origin.
Step 3: Now we evaluate the Domain of given function -
(9 – x^{2}) = 0 = > x = +/-3,
Means domain of function is x = (-∞, -3)∪(-3, 3)∪(3, ∞),
Step 4: Now we calculate the asymptotes of given function, vertical asymptotes are at -3 and at 3.
Step 5: Now we sketch the curve of the above equation by using all these things, which we have calculated in above steps.
This is an example which shows what the basic steps for evaluating a curve are?
Now we will discuss applications of curve sketching:
Curve sketching technique is useful for finding out the maximum and minimum value of curve:
We use following steps for finding maximum and minimum value by curve sketching technique -
Step 1: First of all we will assume variable as a curve like y = x^{2} + 5.
Step 2: After assuming the variable, we will find derivative of that variable with respect to particular variable, like differentiation of ‘y’ with respect to ‘x’.
Step 3: Now we calculate critical points of that variable, by putting differential equation equals to 0 like (dy/dx) = 0 gives x = a and x = b.
Step 4: After finding critical points, we will calculate second order derivative for finding that critical Point is local maximum point or local minimum point.
Like we will calculate (d^{2}y/dx^{2}) and putting critical point values and if d^{2}y/dx^{2} gives positive sign, then that critical point is called as a local minima point and if after putting the value of critical point, d^{2}y/dx^{2} gives negative sign, then that critical point is called as local maxima point means -
If at x = a, (d^{2}y/dx^{2})_{x = a }< 0, then x = a is a local maxima point.
If at x = b, (d^{2}y/dx^{2})_{x = b }> 0, then x = b is a local minima point.
If at x = c, (d^{2}y/dx^{2})_{x = c }= 0, then x = c is a point of inflexion.
Step 5 : Next, we will sketch a curve of the given equation, local maxima and local minima makes this easier because at local minima, curve moves in upward direction and at local maxima, curve moves in downward direction and at point of inflexion, its center point moves between the upward and downward side of curve.
We take an example which defines the above process -
Example: Find the local maxima and local minima of following curve,
x^{2} + 2x + 1?
Solution: We use following steps to calculate local maxima and local minima -
Step 1: First of all we will assume a variable ‘y’ as a curve, means
y = x^{2} + 2x + 1
Step 2: Next, we will differentiate ‘y’ with respect to ‘x’-
dy/dx = d/dx (x^{2} + 2x + 1) = 2x + 2,
Step 3: Next, we will calculate the critical points of the given curve-
Put dy/dx = 0,
= > 2x + 2 = 0,
= > x = -1.
Step 4: Now, we will again differentiate dy/dx with respect to x means, we will calculate second order derivative -
d^{2}y/dx^{2 }= d/dx(dy/dx) = d/dx(2x + 2) = 2, which is a positive number.
So, we can say that given critical point of above curve is a local minima because at x = -1, (d^{2}y/dx^{2})_{x = -1 }> 0, then x = -1 is a local minima point.
Step 5: When we sketch a curve of above equation, it makes a sharp point at local minima point towards upside.
This is basic and best application of curve sketching.