Class 12 Chapter 6 (Application Of Derivative) Class Notes part I

APPLICATIONS OF DERIVATIVE

 

SOME IMPORTANT FORMULAE/KEYCONCEPTS

 

1. RATE OF CHANGE OF QUANTITIES

    

   Whenever one quantity y varies with another quantity x, satisfying some rule y = f(x), then $\dfrac{{dy}}{{dx}}$ or f'(x) represents the rate of change of y with resprect to x and ${\left[ {\dfrac{{dy}}{{dx}}} \right]_{x = {x_0}}}$ or f'(x₀) represents rate of change of y with resprect to x at x = x₀

    

   EXAMPLE :

   QUESTION :The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
   SOLUTION :Let any instant of time t, the radius of circle = r
   Then, circumference C = $2\pi r$
   Diff. Both sides w.r.t t, we get

   $\dfrac{{dC}}{{dt}} = \dfrac{{d(2\pi r)}}{{dt}} = 2\pi \dfrac{{dr}}{{dt}}$

   $\dfrac{{dr}}{{dt}} = 0.7\,\,cm/s$ {Given}

   Here, $\dfrac{{dC}}{{dt}} = 2\pi \times 0.7 = 1.4\pi \,\,\,cm/s$

2. INCREASING AND DECREASING FUNCTIONS:

   Let I be an open interval contained in the domain of a real valued function f. Then f is said to be

   1. increasing on I, if x₁ < x₂ in I, $\Rightarrow f({x_1}) \le f({x_2})$ for all ${x_1},{x_2} \in I$

   2. strictly increasing on I, if x₁ < x₂ in I, $\Rightarrow f({x_1}) < f({x_2})\,for\,\,all\,{x_1},{x_2} \in I.$

   3. increasing on I, if x₁ < x₂ in I, $\Rightarrow f({x_1}) \ge f({x_2})\,for\,\,all\,{x_1},{x_2} \in I.$

   4. increasing on I, if x₁ < x₂ in I, $\Rightarrow f({x_1}) > f({x_2})\,for\,\,all\,{x_1},{x_2} \in I.$

    

   USING THE CONCEPTS OF DERIVATIVES

   1. f is strictly increasing in (a, b) if f ′(x) > 0 for each x $\in$ (a, b)

   2. f is strictly decreasing in (a, b) if f ′(x) < 0 for each x $\in$ (a, b)

   3. A function will be increasing (decreasing) in R if it is so in every interval of R.

    

   
   
   

    

   EXAMPLE

   QUESTION : Find the interval in which the function $f(x) = 2{x^3} - 9{x^2} + 12x + 15$ is:
   (i) strictly increasing (ii) strictly decreasing

   SOLUTION: $f(x) = 2{x^3} - 9{x^2} + 12x + 15$

   $f'(x) = 6({x^2} - 3x + 2)$

   1. For strictly increasing, f’(x) > 0

      6(x² - 3x + 2) > 0

      (x - 1)(x - 2) > 0

      x< 1 or x > 2

      $x \in ( - \infty ,1) \cup (2,\infty )$

      so, f(x) is strictly increasing on $x \in ( - \infty ,1) \cup (2,\infty )$

       

   2. for strictly decreasing f’(x) < 0

   6(x² - 3x + 2) < 0

   (x - 1)(x - 2) < 0

   1 < x < 2

   so, f(x) is strictly decreasing on (1, 2).

    

3. TANGENTS AND NORMALS:

   • Slope of the tangent to the curve y = f (x) at the point (x_o, y_o) is given by

     $${\left[ {\frac{{dy}}{{dx}}} \right]_{({x_0},{y_0})}} = f'({x_0})$$

      

   • The equation of the tangent at (x_o, y_o) to the curve y = f (x) is given by y – y_o = 𝑓′(x_o)(x – x_o).

     
     
     

      

   • Slope of the normal to the curve y = f (x) at (x , y ) = $\dfrac{1}{{slope\,\,of\,\,tangent\,\,at\,({x_0},{y_0})}}$

   • Slope of the normal to the curve y = f (x) at (x , y) is given by $\dfrac{1}{{f'({x_0})}}$

   • The equation of the normal (x_o, y_o) to the curve y = f (x) is given by $y - {y_0} = \dfrac{1}{{f'({x_0})}}(x - {x_0})$

   • If slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point (x_o, y_o) is given by y=y_o.

   • If $\theta \to \dfrac{\pi }{2}$, then $\tan \theta \to \infty$, which means the tangent line is perpendicular to the y - axis. In this case, the equation of the tangent at (x_o, y_o) is given by x = x_o.

    

   EXAMPLE

   QUESTION: Find the equation of normal to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

   SOLUTION: y = x³ + 2x + 6 

   $\dfrac{{dy}}{{dx}} = 3{x^2} + 2$

   Slope of the tangent at (x₁, y₁) = $3x_1^2 + 2$

   Slope of normal at point (x₁, y₁) = $\dfrac{1}{{3x_1^2 + 2}}$

   Slope of the given line is $\dfrac{{ - 1}}{{14}}$

   According to the given condition, Slope of normal = Slope of line

   $\dfrac{{ - 1}}{{3x_1^2 + 2}} = \dfrac{{ - 1}}{{14}}$

   $x = \pm 2$

   When x₁ = 2, then y₁ = 18 and When x₁ = - 2, then y₁ = - 6

   Equation of normal at (2,18) is: $y - 18 = \dfrac{{ - 1}}{{14}}(x - 2)$

   x + 14y = 254 .....(1)

   Equation of normal at (-2, -6) is $y + 6 = \dfrac{{ - 1}}{{14}}(x + 2)$

   x + 14y + 86 = 0

    

    

4. Approximations:

   $$f(x + \Delta x) = f(x) + \Delta y$$ $$\,\,\,\,\, \approx f(x) + f'(x) \times \Delta x\,\,\,\,\,\,\,\,\,\,\,\,\,(as\,dx = \Delta x)$$

    

   • Increment $\Delta y$ in the function y = f(x) corresponding to increment $\Delta x$ in x is

                $\Delta y = \dfrac{{dy}}{{dx}}\Delta x$

   • Relative error in $y = \dfrac{{\Delta y}}{y}$.

   • Percentage error in $y = \dfrac{{\Delta y}}{y} \times 100$.

• 
  

  EXAMPLE

  QUESTION: Evaluate $\sqrt[4]{{81.5}}$

   

  SOLUTION: $\sqrt[4]{{81.5}} = \sqrt[4]{{81 + 0.5}}$

  Let x = 81 and $\Delta x$ = 0.5

  $$y = \sqrt[4]{x} = \sqrt[4]{{81}}\,\,\,\,\,\,\,\,..........(i)$$
  $$y + \Delta y = \sqrt[4]{{81.5}}\,\,\,\,..........(ii)$$
  $$\Delta y = \sqrt[4]{{81.5}} - \sqrt[4]{x}$$
  $$\sqrt[4]{{81.5}} = \Delta y + \sqrt[4]{x}$$

  using approximation $\Delta y \approx dy$

  $$\sqrt[4]{{81.5}} = \frac{{dy}}{{dx}} \times dx + \sqrt[4]{x}$$
  $$= \frac{1}{4}{x^{\frac{{ - 3}}{4}}} \times 0.5 + 3$$
  $$= \frac{1}{4} \times {81^{\frac{{ - 3}}{4}}} \times \frac{1}{2} + 3$$
  $$= \frac{1}{4} \times \frac{1}{{27}} \times \frac{1}{2} + 3 = \frac{1}{{216}} + 3 = 3.0046$$

   

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