Important Questions On Applications of Derivative

Important Board Questions

4 marks questions

1. Question:  A balloon, which always remains spherical, has a variable diameter \(\dfrac{3}{2}(2x + 1)\) . Find the rate of change of its volume with respect to x. 

   Answer:  

   \[Diameter = \frac{3}{2}(2x + 1)\] \[r = \frac{3}{4}(2x + 1)\] Volume of spherical balloon = \(\dfrac{4}{3}\pi {r^3}\)
   \[V = \frac{4}{3}\pi {\left[ {\frac{3}{4}(2x + 1)} \right]^3}\] \[\frac{{dV}}{{dx}} = \frac{{9\pi }}{{16}}\left[ {3{{(2x + 1)}^2}} \right] \times 2\] \[\frac{{dV}}{{dx}} = \frac{{27\pi }}{8}{(2x + 1)^2}\]

2. 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? 

   Answer:

   Let any instant of time t, the radius of circle = r 
   Then, circumference C = 2 π r
   differentiating both sides w.r.t. to t, we get
   \[\frac{{dC}}{{dt}} = \frac{{d(2\pi r)}}{{dt}} = 2\pi \frac{{dr}}{{dt}}\] \(Here,\,\frac{{dr}}{{dt}} = 0.7\,cm/s\)
   \[\frac{{dC}}{{dt}} = 2\pi \times 0.7 = 1.4\pi \,\,cm/s\]

3. Question:  Find the interval in which the function f(x) = (x + 1)³ × (x − 1)³ is: (i) strictly increasing (ii) strictly decreasing 

   Answer: f(x) = (x + 1)³ × (x − 1)³ 

   f(x) = (x² − 1)³

   f'(x) = 3×2x(x² − 1)²

   For strictly increasing f'(x) > 0

   f'(x) = 3×2x(x² − 1)² > 0
   x > 1
   \(x \in (1,\infty )\)
   so, f(x) is increasing in \(x \in (1,\infty )\)

   For strictly decreasing f'(x) < 0

   f'(x) = 3×2x(x² − 1)² > 0
   x > 1
   \(x \in ( - \infty ,1)\)
   so, f(x) is increasing in \(x \in ( - \infty ,1)\) .

4. Question:  Find the interval in which the function f (x) = sinx + cosx, 0  ≤  x  ≤  2π is strictly increasing or strictly decreasing. 

   Answer: f(x) = sinx + cosx
   f'(x) = cosx − sinx  \[f'(x) = \sqrt 2 \sin \left( {\frac{\pi }{4} - x} \right)\]

   For strictly increasing f'(x) > 0

   \[f'(x) = \sqrt 2 \sin \left( {\frac{\pi }{4} - x} \right) > 0\]
   \[\pi < x - \frac{\pi }{4} < 2\pi \] \[\frac{{5\pi }}{4} < x < \frac{{9\pi }}{4}\] \[\frac{{5\pi }}{4} < x < 2\pi \]
   so, f(x) is increasing in \(\left( {0,\dfrac{\pi }{4}} \right) \cup \left( {\dfrac{{5\pi }}{4},2\pi } \right)\)

   For strictly decreasing f'(x) < 0

   \[f'(x) = \sqrt 2 \sin \left( {\frac{\pi }{4} - x} \right) < 0\]
   \[0 < x - \frac{\pi }{4} < \pi \] \[\frac{\pi }{4} < x < \frac{{5\pi }}{4}\]
   so, f(x) is increasing in \(\left( {\dfrac{\pi }{4},\dfrac{{5\pi }}{4}} \right)\)

5. Question:  Find the equations of all lines having slope 0, which are tangent to the curve \(y = \dfrac{1}{{{x^2} - 2x + 3}}\) 

   Answer:  

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6. Question: Find the equation of tangent and normal to the hyperbola \(\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\) at the point (x₀, y₀) 

   Answer:  

   aksdjvn;avsdkvb

7. Question:  Find the approximate value of f(5.001), where f(x) = x³ – 7x² +15 

   Answer:  

   aksdjvn;avsdkvb

8. Question: If the radius of sphere is measured as 7 mtr with error of 0.02 m, than find the approximate error in calculating its volume. 

   Answer:  

   aksdjvn;avsdkvb

9. Question: The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x = 8 cm and y = 6 cm, find the rates of change of (i) Perimeter (ii) area of the rectangle. 

   Answer:  

   aksdjvn;avsdkvb

10. Question: A ladder 5m long is leaning against a wall. Bottom of ladder is pulled along the ground away from wall at the rate of 2m/s. How fast is the height on the wall decreasing when the foot of ladder is 4m away from the wall? 

    Answer:  

    aksdjvn;avsdkvb

6 marks questions

1. Question:  Find the area of the greatest rectangle that can be inscribed in an ellipse \(\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\)

   Answer:  

   aksdjvn;avsdkvb

2. Question: A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum sunlight through the whole opening. Explain the importance of sunlight. 

   Answer:  

   aksdjvn;avsdkvb

3. Question:  Find the point on the curve y²= 2x which is at minimum distance from the point (1, 4)

   Answer:  

   aksdjvn;avsdkvb

4. Question:  Show that the semivertical angle of a cone of maximum volume and of given slant height is \({\tan ^{ - 1}}\sqrt 2 \)

   Answer:  

   aksdjvn;avsdkvb

5. Question: Show that the height of cylinder of maximum volume that can be inscribed in a sphere of radius R is \(\dfrac{{2R}}{{\sqrt 3 }}\)

   Answer:  

   aksdjvn;avsdkvb

HOTS

1. Prove that: \(y = \dfrac{{4\sin \theta }}{{2 + \cos \theta }} - \theta \) is an incresing function in \(\left[ {0,\frac{\pi }{2}} \right]\)

2. Prove that the curves 𝑥 = 𝑦² 𝑎𝑛𝑑 𝑥𝑦 = 𝑘 are orthogonal if 8k² = 1
   ( HINT: If the curves are Orthogonal , the tangents at point of intersection to the given curves are perpendicular i.e the product of slopes of the tangents = -1) 

3. Prove that the volume of the largest cone that can be inscribed in a sphere of radius a is \(\frac{8}{{27}}\) of the volume of the sphere. 

4. Find the sub intervals of \(\left[ {0,\dfrac{\pi }{2}} \right]\) in which the function f(x) = sin⁴ + cos⁴ is (i) Strictly incresing (ii) Strictly decreasing 

5. Find the maximum area of the isosceles triangle inscribed in the ellipse \(\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\) with its vertex at one end of the major axis 

6. Show that the semi-vertical angle of the right circular cone of given total surface area and maximum volume is \({\sin ^{ - 1}}\dfrac{1}{3}\) .

7. Show that the volume of the greatest cylinder that can be inscribed in acone of height h and semi vertical angle α is \(\dfrac{4}{{27}}\pi {h^3}{\tan ^2}\alpha \) 

8. Find the point on the curve \(y = \dfrac{x}{{1 + {x^2}}}\) where the tangent to the curve has the greatest slope. 

9. Find the value of p for which the curves 𝑥² = 9𝑝(9 − 𝑦) and 𝑥² = 𝑝(𝑦 + 1) cut each other at right angle. 

10. Find the equation of the tangent to the curve \(y = \dfrac{{x - 7}}{{(x - 2)(x - 3)}}\) at the point where it cuts the x-axis 

 

 

 

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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 \(\Delta y \approx dy\)

  \[\sqrt[4]{{81.5}} = \frac{{dy}}{{dx}} \times dx + \sqrt[4]{x}\]

  \[ = \frac{1}{4}{x^{\frac{{ - 3}}{4}}} \times 0.5 + 3\]

  \[ = \frac{1}{4} \times {81^{\frac{{ - 3}}{4}}} \times \frac{1}{2} + 3\]

  \[ = \frac{1}{4} \times \frac{1}{{27}} \times \frac{1}{2} + 3 = \frac{1}{{216}} + 3 = 3.0046\]

   

  Click Here for Maxima and Minima
  

   

 

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