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

**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=πr^{2}=ARate of change of the area of a circle with respect to r=dA/dr

dA/dr=d/dr πr

^{2}=2πrso, 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 cm**^{3}/s. How fast is the surface area **increasing when the length of an edge is 12 cm?**

^{3}/s. How fast is the surface area

**Solution:**

Given,

Rate of increase of the volume =8cm

^{3}/slength of edge of cube=12cm=s

Now,

volume(v) of a cube with side length ‘s’

v=s

^{3}Now, [chain rule]

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

A=6s

^{2}

**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,= 3cm\s

To find : Ratio of increase of Area(A=πr^{2})= π =2πr. [chain rule]

=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, =3cm/s

To find: Rate of increase of volume (v) of the cube

Now, [chain rule]

So,

**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== π =2πr. [chain rule]

= 2π(8)5cm

^{2}/s=80πcm

^{2}/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,

Circumference(P)=2πr=2π.=2π.(0.7)

=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,

Rate of change of width,

Now, perimeter P=2(r+y)

Area A=x.y

so, a)

b)[x is decreasing, y is increasing]

**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=900 cm^{3}/s

To find: Rate of change of radius,when r=15cm.v=πr

^{3}= 4πr

^{2}Now, =4π(15)

^{2}.900=900π.

= 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=πr

^{3}To find: Rate of change of volume with respect to

i.e

Now, π.r

^{3}=4πr^{2}=400π cm

^{2}

**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==3

Now, let AB=x & BC=y

so, x

^{2},y^{2}=5^{2}or x^{2},y^{2}=25 ———1Differentiating both sides of 1 by t, we get

or

Now at BC=y=4,

so

[negative sign means AB is decreasing]

**Question 11. A particle moves along the curve 6y=x**^{3} + 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x-coordinate.

^{3}+ 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=x

^{3}+2 ————1and ———-2

Differentially 1 with respect to it, we get,

from 2 6.8.

16=x

^{2}x=±4 ————-3

Now for y coordinates, put 3

6y=x

^{3}+2

when x = -46y = -64+2

y =

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 radiuscm/s

To Find: Rate of increase of volume,

Now, v=4/3πr^{3}=π =4πr

^{2}

=4π(1)^{2}. cm^{3}/s=2π cm

^{3}/s

**Question 13. A balloon, which always remains spherical, has a variable diameter****(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

(2)

Now, volume =4/3πr^{3}Rate of change of volume with respect to radius

=π=4πr

^{2}= 4π(

π (2x+1)

^{2}

**Question 14. Sand is pouring from a pipe at the rate of 12 cm**^{3}/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?

^{3}/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=12cm^{3}/s

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

so,

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=1/3πr^{2}h=1/3.π(6h)^{2}.hv=12πh

^{3}=12π=12π3h

^{2}.12 =36π.(4)2.

= π

=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)=0.003x**^{2}+15x+4000. Find the marginal cost when 17 units are produced.

^{2}+15x+4000. Find the marginal cost when 17 units are produced.

**Solution:**

Given: c(x)=0.007x^{3}-0.003x^{2}+15c+4000

Now change in total cost with respect to units is know as marginal cost i.eMarginal cost=0.021x

^{2}-0.006x+15Marginal 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)=13x**^{3}+26x+15. Find the marginal revenue when x=7.

^{3}+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 =

Marginal revenue=

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=πr

^{2},where r is the radiusRate of change of area with respect to its radius r is,

=π=2πr

=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)=3x**^{2} +36x+5. The marginal revenue, when x=15 is (A) 116 (B) 96 (C) 90 (D) 126

^{2}+36x+5. The marginal revenue, when x=15 is

**Solution:**

Marginal Revenue=

Marginal Revenue=6x + 36

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