In this page we are goimng to discuss about slope as a derivative concept . The trigonometrically tangent of the angle that a line makes with the positive direction of x – axis in anticlockwise direction is called the slope of the line.
The slope of a line in general is denoted by m. Thus m = tan Ø, where Ø is the angle which the line makes with the positive x-axis in anticlockwise direction.
The application of derivatives to plane geometry is related to slopes. Let us consider a curve whose equation is y = f(x).
On this curve we take a point P(x1, y1). Let the tangent at this point not be parallel to the co-ordinate axes, we can get the equation of tangent line at P.
The derivative of the function f(x) is the slope of the tangent.
dy /dx at P(x1,y1) is tan α = slope of the tangent at P, where α is the angle which the tangent at P(x1,y1) makes with the positive direction of x-axis.
The equation of straight line with slope m and passing through the point (x1,y1)is given as
y-y1 = m(x –x1).
For the curve y = f(x) we have dy /dx =f’(x)
For the tangent line, slope = m = f’(x1)= dy / dx at (x1,y1).
The equation line of the tangent line will be of the form
y-y1 = f’(x1)(x –x1).
If the tangent at P is parallel to the x axis then α =0, `rArr` tan α =0, `rArr` slope =dy /dx =0.
If the tangent at P is perpendicular to the x –axis, or parallel to the y-axis then,
α = `\pi/{2}`, `rArr` tan α =1, `rArr` slope =dy / dx =1.
I like to share this Derivative of a Square Root Function with you all through my article.
Slope of the tangent at P(x1,y1) on the curve y = f(x) is the value of dy/dx at (x1,y1)
Slope of the Normal at P is perpendicular to the tangent and is negative 1/ slope of the tangent.
Slope as a Derivative Examples
Below are the examples on slope as a derivative -
Example 1:
Find the slopes of tangent and normal to the curve y = x4 - 3x2 at (1, -2)
Solution:
Step 1 : Let us write the function y = x4 - 3x2
Step 2 : Then slope is dy/dx = 4x3 - 6x
Step 3 : Slope at (1,-2 ) is 4(1)3 - 6(1) = 4 - 6 = -2 Hence slope of the tangent is m= -2
Step 4 : Slope of the normal = -1/ slope of the tangent = - (-1/2) = 1/2. Hence slope of the normal = 1/2
Let us find the slope and normal to a function which involves the time element also.
Example 2 :
Find the slope and normal to the curve x = ct and y = c/t at t
Solution:-
Step 1 : Let us write the equation x = ct
Step 2 : Then the slope of the function is dx/dt = c
Step 3 : Let us find dy/dt which is equal to -c/t2
Step 4 : We have to find dy/dx. Hence here we use chain rule . That is dy/dx = dy/dt ÷ dx/dt = -c/t2 ÷ c = -c/t2 * 1/c
= -1 /t2
Step 5 : Hence slope of the tangent to the curve x =ct , y = c/t is -1/t2
Step 6 : Slope of the normal at 't'is the reciprocal of the tangent with sign change = t2
Hence slope of the normal at 't' is t2
Answer ; Slope of the tangent at 't' is -1/t2 and slope pf the normal at 't' = t2
Integrals is the opposite of differential equations. Hence if slope of an equation is given , then we can find the equation using integrals.
Example 3: Find the equation of the curve whose slope at the point (x,y) is 3x2 + 2, if the curve is passing through the point (1,-1)
Solution:-
Step 1 : We have the slope dy /dx = 3x2 + 2
Step 2 : That gives us dy = (3x2 + 2) dx
Step 3 : Let us integrate both sides ∫dy = ∫ (3x2+ 2) dx
y = 3 x3 + 2x + c = > x3 + 2x + c
3
Step 4 : This passes through (1,-1) Hence we get -1 = (1)3 + 2(1) + c
-1 = 1 +2+c
-1 = 3 + c
That is c = -4
Step 5 : Let us now write the equation as y = x3 + 2x - 4
Answer :- Equation of the curve whose slope at a point (x,y) is 3x2 + 2 and the point is (1,-1) is y = x3 + 2x - 4
We have used differential equation to find slope of the tangent and slope of the normal to the curve and we have used integrals to find the equation of the curve when slope is given.
The slope of a line in general is denoted by m. Thus m = tan Ø, where Ø is the angle which the line makes with the positive x-axis in anticlockwise direction.
The application of derivatives to plane geometry is related to slopes. Let us consider a curve whose equation is y = f(x).
On this curve we take a point P(x1, y1). Let the tangent at this point not be parallel to the co-ordinate axes, we can get the equation of tangent line at P.
The derivative of the function f(x) is the slope of the tangent.
dy /dx at P(x1,y1) is tan α = slope of the tangent at P, where α is the angle which the tangent at P(x1,y1) makes with the positive direction of x-axis.
The equation of straight line with slope m and passing through the point (x1,y1)is given as
y-y1 = m(x –x1).
For the curve y = f(x) we have dy /dx =f’(x)
For the tangent line, slope = m = f’(x1)= dy / dx at (x1,y1).
The equation line of the tangent line will be of the form
y-y1 = f’(x1)(x –x1).
If the tangent at P is parallel to the x axis then α =0, `rArr` tan α =0, `rArr` slope =dy /dx =0.
If the tangent at P is perpendicular to the x –axis, or parallel to the y-axis then,
α = `\pi/{2}`, `rArr` tan α =1, `rArr` slope =dy / dx =1.
I like to share this Derivative of a Square Root Function with you all through my article.
Slope of the tangent at P(x1,y1) on the curve y = f(x) is the value of dy/dx at (x1,y1)
Slope of the Normal at P is perpendicular to the tangent and is negative 1/ slope of the tangent.
Slope as a Derivative Examples
Below are the examples on slope as a derivative -
Example 1:
Find the slopes of tangent and normal to the curve y = x4 - 3x2 at (1, -2)
Solution:
Step 1 : Let us write the function y = x4 - 3x2
Step 2 : Then slope is dy/dx = 4x3 - 6x
Step 3 : Slope at (1,-2 ) is 4(1)3 - 6(1) = 4 - 6 = -2 Hence slope of the tangent is m= -2
Step 4 : Slope of the normal = -1/ slope of the tangent = - (-1/2) = 1/2. Hence slope of the normal = 1/2
Let us find the slope and normal to a function which involves the time element also.
Example 2 :
Find the slope and normal to the curve x = ct and y = c/t at t
Solution:-
Step 1 : Let us write the equation x = ct
Step 2 : Then the slope of the function is dx/dt = c
Step 3 : Let us find dy/dt which is equal to -c/t2
Step 4 : We have to find dy/dx. Hence here we use chain rule . That is dy/dx = dy/dt ÷ dx/dt = -c/t2 ÷ c = -c/t2 * 1/c
= -1 /t2
Step 5 : Hence slope of the tangent to the curve x =ct , y = c/t is -1/t2
Step 6 : Slope of the normal at 't'is the reciprocal of the tangent with sign change = t2
Hence slope of the normal at 't' is t2
Answer ; Slope of the tangent at 't' is -1/t2 and slope pf the normal at 't' = t2
Integrals is the opposite of differential equations. Hence if slope of an equation is given , then we can find the equation using integrals.
Example 3: Find the equation of the curve whose slope at the point (x,y) is 3x2 + 2, if the curve is passing through the point (1,-1)
Solution:-
Step 1 : We have the slope dy /dx = 3x2 + 2
Step 2 : That gives us dy = (3x2 + 2) dx
Step 3 : Let us integrate both sides ∫dy = ∫ (3x2+ 2) dx
y = 3 x3 + 2x + c = > x3 + 2x + c
3
Step 4 : This passes through (1,-1) Hence we get -1 = (1)3 + 2(1) + c
-1 = 1 +2+c
-1 = 3 + c
That is c = -4
Step 5 : Let us now write the equation as y = x3 + 2x - 4
Answer :- Equation of the curve whose slope at a point (x,y) is 3x2 + 2 and the point is (1,-1) is y = x3 + 2x - 4
We have used differential equation to find slope of the tangent and slope of the normal to the curve and we have used integrals to find the equation of the curve when slope is given.
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