Application of Differentiation

 
2.4   Application of Differentiation
 
  • Safety and maximum enjoyment of customers
  • The building of a roller coaster also take into consideration on users' maximum enjoyment out of the ride
  • The gradient of a curve at a point is also the gradient of the tangent at that point
 
Tangent and normal:
 
Gradient Function, \(f'(x)\)
     
   

The function \(f'(x)\) is used to determine the gradient of tangent to the curve at any point on the function graph \(f(x)\).

   
     
 
 
  • Based on the above diagram, the formula for the equation of straight line \(l\) with gradient \(m\) that passes through point \(P(x_1, y_1)\) can be written as:

     
   \(y-y_1=m(x-x_1)\)   
     

 

  • The equation of the tangent is:
     
   \(y-f(a)=f'(a)(x-a)\)   
     
 
  • The line which is perpendicular to the tangent is the normal to the curve  \(y=f(x) \text{ pada titik }P(a,f(a))\)  
 
  • If the gradient of the tangent, \(f'(a)\) exists and is non-zero, the gradient of the normal based on the relation of \(m_1m_2=-1\) is

     
   \(-\dfrac{1}{f'(a)}\)   
     

 

  • The equation of the normal is:
     
   \(y-f(a)=-\dfrac{1}{f'(a)}(x-a)\)   
     
 
Example
     
 

Find the equation of the tangent and normal to the curve

\( f(x) = x^3 – 2^2 + 5 \text{ at point } P(2, 5)\).

 
     
     
 

Solution:

 
     
   \(\text{Given }f(x) = x^3 – 2^2 + 5 , \\ \text{so }f'(x) = 3x^2 -4x\)  
     
   

When \(x=2,\ f'(2) \ = \ 3(2)^2 -4(2) = \ 12-8 \ = \ 4\)

 

Gradient of the tangent at point \(P(2,5) \text{ is } 4.\)

 

\(\begin{aligned} \text{Equation of the tangent is }y – 5 &= 4(x – 2) \\ y – 5 &= 4x – 8 \\&y = 4x – 3 \end{aligned}\)

 

Gradient of the normal at point \(P(2,5) \text{ is } -\dfrac{1}{4}.\)

 

\(\begin{aligned} \text{Equation of the normal is }y – 5 &= -\dfrac{1}{4}(x – 2) \\ 4y – 20 &= -x + 2 \\4y +x &= 22 \end{aligned}\)

   
     
 
 
Stationary points of curve \(y=f(x)\) :
 
 
  • There are three types of stationary points:

     
   
1)   Maximum point
 
  • Turning points
  • When \(x\) increases through \(x=a\), the value of \(\dfrac{dy}{dx}\) changes sign from positive to negative
   
2)   Minimum point
 
  • Turning points
  • When \(x\) increases through \(x=b\), the value of \(\dfrac{dy}{dx}\) changes sign from negative to positive
   
3)   Point of inflection
 
  • This stationary point which is not a maximum or a minimum point
  • A point on the curve at which the curvature of the graph changes.
   
     
 
  • In general,
     
  A turning point on a curve \(y=f(x)\)is a minimum point when  
     
    \(\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ > }0\)    
     

 

     
   

A turning point on a curve \(y=f(x)\) is a maximum point when

   
     
  \(\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ < }0\)  
     
 

Rate of change for related quantities:

  • If two variables, \(y\) and \(x\) change with time, \(t\) and are related by the equation \(y=f(x)\), then:

     
   Chain rule : \(\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}\)   
     
 
Remark
     
   

If the rate of change of y over time is negative, for example

\(\dfrac{dy}{dt} = -6\), then 

\(y\) is said to decrease at a rate of \(6 \text{ unit }s^{-1}\).

   
     
 

Small changes and approximations of certain quantities:

  • If \(y=f(x)\) and the small change in \(x\), that is \(\delta x\) causes a small change in \(y\), that is \(\delta y\), then:

     
    \(\begin{aligned} \dfrac{\delta y}{\delta x} &\approx \dfrac{dy}{dx}\\\\ \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ \text{and }f(x+ \delta x) &\approx y + \delta y\\\\ &\approx y+\dfrac{dy}{dx}(\delta x) \end{aligned}\)    
     
 
 
Example
     
 

Given \(y = x^3\), find

 
    
(a)

the approximate change in \(y\) when \(x\) increases from \(4\) to \(4.05\),

   
(b)

the approximate change in \(x\) when \(y\) decreases from \(8\) to \(7.97\).

   
   
Solution:
   
(a)

\(\begin{aligned} y&=x^3\\\\ \dfrac{dy}{dx} &=3x^2 \end{aligned}\)

 
When \(\begin{aligned} x=4, \delta x &= 4.05-4\\ &=0.05 \end{aligned}\)
   
and \(\dfrac{dy}{dx} = 3(4)^2 = 48\)
   
Then, \(\begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ &=48 \times 0.05\\\\ \delta y &= 2.4 \end{aligned}\)
   
 

Therefore, the approximate change in \(y\), that is \(\delta y\), is \(2.4\).

\(\delta y \text{ > }0\) means there is a small increase in \(y\) of \(2.4\).

   
(b)
When \(\begin{aligned} y=8, x^3&=8\\ x&=2\\ \delta y &= 7.97-8 \\ &= -0.03 \end{aligned}\)
   
and \(\dfrac{dy}{dx} = 3(2)^2 = 12\)
   
Then, \(\begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ -0.03&=12 \times \delta x\\\\ \delta x &= \dfrac{-0.03}{12}\\\\ &=-0.0025 \end{aligned}\)
   
 

Therefore, the approximate change in \(x\), that is \(\delta x\), is \(-0.0025\).

\(\delta y \text{ < }0\) means there is a small decrease in \(x\) of \(0.0025\).