Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.1

The study of derivatives is a cornerstone in calculus providing the essential tools for understanding and analyzing functions. The Application of Derivatives a key topic in Class 12 Mathematics involves using the derivatives to solve real-world problems. This topic helps in understanding how rates of change affect various scenarios, optimizing functions, and determining the behavior of functions in different contexts. Exercise 6.1 from the NCERT Class 12 Mathematics textbook focuses on these applications offering practical problems to solidify the concepts learned.

What is the Application of Derivatives?

The Application of Derivatives refers to the use of derivatives to solve practical problems in various fields such as physics, engineering, economics, and more. Here are some key applications:

• Rate of Change: The Derivatives help in finding the rate at which one quantity changes with respect to another. For example, velocity is the derivative of the displacement with respect to time.
• Optimization: Derivatives are used to find the maximum and minimum values of the functions which is crucial for optimizing resources and making decisions. This includes finding the most efficient use of the resources or maximizing profit.
• Curve Sketching: The Derivatives assist in understanding the shape of graphs including identifying concavity, inflection points, and intervals where functions are increasing or decreasing.
• Motion Analysis: In physics, derivatives describe the motion of objects such as the acceleration being the derivative of velocity.

Question 1. Find the rate of change of the area of a circle with respect to its radius r when a) r=3cm b) r=4cm

Solution:

Given,

radius of circle=r=3cm

Now, we know that area=πr²=A

Rate of change of the area of a circle with respect to r=dA/dr

dA/dr=d/dr πr²=2πr

so, when r=3

dA/dr  

=2π(3)=6π

when r=4

dA/dr

=2π(4)=8π

Question 2. The volume of a cube is increasing at the rate of cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Solution:

Given,

Rate of increase of the volume =8cm³/s

length of edge of cube=12cm=s

Now,

volume(v) of a cube with side length 's'

v=s³

Now,\frac{dv}{dt}=\frac{d(s)^3}{dt}=3s^2\frac{ds}{dt}            [chain rule]

\frac{dv}{dt}=8cm^3/s=3(12)^2\frac{ds}{dt}

\frac{ds}{dt}=\frac{8cm^3}{3(12)^2cm^2}

so, the rate of change of surface Area(A)

A=6s²

\frac{dA}{dt}=\frac{d}{dt}(6s^2)=36(2s)\frac{ds}{dt}

\frac{dA}{dt}=12.12.\frac{ds}{dt}=\frac{8}{3}cm^2/s

Question 3. The radius of a circle is increasing uniformly at rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10cm.

Solution:

Given,

rate of increase of radius = 3cm/s=r

so, \frac{dr}{dt}   = 3cm\s

To find : Ratio of increase of Area(A=πr²)

\frac{dA}{dt}  = π \frac{d}{dt}(r^2)  =2πr.\frac{dr}{dt}           [chain rule]

\frac{dA}{dt}  =2π(10)³=60πr

Question 4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10cm long?

Solution:

Given: rate of increase of edge of cube, \frac{ds}{dt}         =3cm/s

To find: Rate of increase of volume (v) of the cube \frac{dv}{dt}=?

Now, \frac{dv}{dt}=\frac{d}{dt}(s^3)=3s^2.\frac{ds}{dt}           [chain rule]

So, \frac{dv}{dt}=3(10)^2.3cm^3/s=900cm^3/s

Question 5. A stone is dropped into a quiet lake and wave move in circles at the speed of 5cm/s. At the instant when the radius of the circular wave is 8cm, how fast is the enclosed area increasing?

Solution:

Given, Speed of water=rate of change of radius=5cm/s

To find: rate of increase of area=\frac{dA}{dt}=?

\frac{dr}{dt}=5cm/s

\frac{dA}{dt}   = π \frac{d}{dt}(r^2)   =2πr. \frac{dr}{dt}     [chain rule]

\frac{dA}{dt}   = 2π(8)5cm²/s

=80πcm²/s

Question 6. The radius of a circle is increasing at the rate of 0.7cm/s. What is the rate of increase of its circumstances? 

Solution:

Given: rate of increase of radius, \frac{dr}{dt}=0.7cm/s

Circumference(P)=2πr

\frac{dP}{dt}  =2π.\frac{dr}{dt}  =2π.(0.7)

\frac{dP}{dt}  =1.4π cm/s              or 4.4cm/s              [taking π=22/7]

Question 7. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x=8cm and y=6cm, find the rates of change of (a) the perimeter, and b) the area of the rectangle.  

Solution:

Given: Rate of change of length, \frac{dx}{dt}=-5cm/min

Rate of change of width, \frac{dy}{dt}=4cm/min

Now, perimeter P=2(r+y)

Area A=x.y

so, a) \frac{dP}{dt}=\frac{d}{dt}(2(x+Y))=\frac{2dx}{dt}+\frac{2dy}{dt}

           \frac{dP}{dt}=2.(-5)+2.(4)=-2cm/min

      b)\frac{dA}{dt}=\frac{d}{dx}(x.y)=x\frac{dy}{dt}+y\frac{dx}{dt}         [x is decreasing, y is increasing]

          \frac{dA}{dt}=8(4)+6(-5)=2cm/min           

Question 8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15cm.

Solution:

Given, Amount of gas pumped in per second/ Rate of change of volume \frac{dv}{dt}  =900 cm³/s

To find: Rate of change of radius, \frac{dr}{dt}          when r=15cm.

v=\frac{4}{3}  πr³

\frac{dv}{dt} = 4πr²\frac{dr}{dt}

Now, \frac{dv}{dt} =4π(15)².\frac{dr}{dt}

900=900π.\frac{dr}{dt}

\frac{dr}{dt}   = 1/π cm/s

Question 9. A balloon, in which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10cm.

Solution:

Let the radius be r & volume be v.

v=\frac{4}{3}  πr³

To find: Rate of change of volume with respect to 

i.e \frac{dv}{dr}=?

Now, \frac{dv}{dt}=\frac{4}{3}  π\frac{d}{dr}  .r³=4πr²

\frac{dv}{dr}  =400π cm²

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

Solution:

 Given: Length of ladders=5m

In ∆ ABC, AC=5m,  BC=4m,  & ∠ABC=90°,

so by Pythagoras theorem,

AB=\sqrt{(5)^2-(4)^2}  =3

Now,  let AB=x   & BC=y

so, x²,y²=5²   or  x²,y²=25               ---------1

Differentiating   both sides  of 1 by t,  we get 

2x.\frac{dx}{dt}+2y.\frac{dy}{dt}=0         

or

x.\frac{dx}{dt}=-y.\frac{dy}{dt}

Now  at BC=y=4,

\frac{dy}{dt}=2cm/s

so x.\frac{dx}{dt}=-y\frac{dy}{dt}

3\frac{dx}{dt}=-4(2)

\frac{dx}{dt}=\frac{-8}{3}cm/s           [negative sign means AB is decreasing]

Question 11. A particle moves along the curve 6y=x³ + 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x-coordinate.

Solution:

Given: curve  6y=x³+2  ------------1

and  \frac{dy}{dt}=8.\frac{dx}{dt}       ----------2

Differentially 1 with respect to it, we get,

\frac{6dy}{dt}=3x^2.\frac{dx}{dt}    

from 2          6.8.\frac{dx}{dt}=3x^2.\frac{dx}{dt}

16=x²

x=±4           -------------3

Now for y coordinates, put 3

6y=x³+2

when x = -4

6y = -64+2

y = \frac{32}{3}

and, when x = 4

6y = 66

y = 11

Question 12. The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Solution:

Given: Rate of increase of radius \frac{dr}{dt}=\frac{1}{2}     cm/s

To Find: Rate of increase of volume,\frac{dv}{dt}=?

Now, v=4/3πr³

\frac{dv}{dt}  =\frac{4}{3}  π \frac{d}{dt}(r^3)  =4πr²\frac{dr}{dt}     
\frac{dv}{dt}  =4π(1)².\frac{1}{2}   cm³/s

\frac{dv}{dt}  =2π cm³/s

Question 13. A balloon, which always remains spherical, has a variable diameter\frac{3}{2}  (2x+1). Find the rate of change of its volume with respect to x.

Solution:

Given: Diameter of sphere=3/2(2x+1)=d

So, radius of the sphere will be d/2=3/4(2x+1)=r

\frac{dr}{dx} = \frac{3}{4}   (2)

\frac{dr}{dx} = \frac{3}{2}

Now, volume =4/3πr³

Rate of change of volume with respect to radius 

\frac{dv}{dr}  =\frac{4}{3}  π\frac{d}{dr}(r^3)  =4πr²

\frac{dv}{dr} . \frac{dr}{dx}   = 4π((\frac{3(2x+1)}{4})^2 . \frac{3}{2}

\frac{dv}{dx} = \frac{27}{8}  π (2x+1)²

Question 14. Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the cone increasing when the height is 4cm?

Solution:

Given: Rate of falling sand=12cm³/s

Now this rate is basically the rate of change of the cone.

so, \frac{dv}{dt}=12cm^3/s

Now, radius =r

height =r

height is always one-sixth of the radius so,

h=r/6  or r=6h

To find : Rate of change of height =dh/dt=?

Now, volume v=1/3πr²h=1/3.π(6h)².h

v=12πh³

\frac{dv}{dt}  =12π\frac{d}{dt}(h^3)  =12π3h².\frac{dh}{dt}

12 =36π.(4)2.\frac{dh}{dt}

\frac{12}{36.16}  = π \frac{dh}{dt}

\frac{dh}{dt}  =1 / 48π cm/s

Question 15. The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x)=0.003x²+15x+4000. Find the marginal cost when 17 units are produced.

Solution:

Given: c(x)=0.007x³-0.003x²+15c+4000

Now change in total cost with respect to units is known as marginal cost i.e \frac{dc(x)}{dx}

\frac{d[(x)]}{dx}=\frac{d}{dx}(0.007x^3-0.003x^2+15x+4000)

Marginal cost=0.021x²-0.006x+15

Marginal cost when 17 units are produced

=0.221(17)²-0.006(17)+15

=6.069-0.102+15

=20.967

Question 16. The total revenue in Rupees from the scale of x units of a product is given by R(x)=13x³+26x+15. Find the marginal revenue when x=7.

Solution:

Marginal revenue is the rate of change of total revenue with respect to no. of units.

So, Marginal revenue =\frac{d}{dx}[R(x)]

Marginal revenue=\frac{d}{dx}(13x^2+26x+15)

=26x+26=26(x+1)

Marginal revenue  when (x=7)

=26 (8)

=208

Question 17. The rate of change of area of circle with respect to its radius r at r= 6cm is (A) 10π (B) 12π (C) 8π (D) 11π.

Solution:

Area, A=πr²,where r  is the radius 

Rate of change of area with respect to its radius r is,

\frac{dA}{dr}  =π\frac{d}{dr}(r^2)  =2πr

\frac{dA}{dr}  =2π(6)=12π

Question 18. The total revenue in Rupees received from the sale of x units of a product is given by R(x)=3x² +36x+5. The marginal revenue, when x=15 is (A) 116 (B) 96 (C) 90 (D) 126

Solution:

Marginal Revenue=\frac{d}{dx}(R(x))

Marginal Revenue=6x + 36

Marginal Revenue at x=15 is    6 (15)+36=126

Conclusion

The Application of Derivatives is a powerful tool that extends the basic concepts of the calculus to solve real-world problems. Exercise 6.1 in the NCERT Class 12 Mathematics textbook provides the range of problems designed to the enhance understanding and application of derivatives in the various contexts. Mastery of this topic equips students with the skills to tackle complex problems and make informed decisions based on the mathematical analysis.