RD Sharma Class 12 Exercise 12 FBQ Derivative as a rate measure Solutions Maths

RD Sharma Class 12 Exercise 12 FBQ Derivative as a rate measure Solutions Maths - Download PDF Free Online

Updated on Jan 20, 2022 07:14 PM IST

RD Sharma is a highly favoured book for Mathematics and is arguably one of the most reliable coursebooks for students in class 12. The book contains complete solved questions of the entire CBSE class 12 maths syllabus and is useful in finding any required information on all concepts. This great advantage of using RD Sharma solutions has enabled students to have an advanced knowledge of maths. Mathematics is a subject that is feared by many students. In class 12, the syllabus gets tougher while the stakes are raised high. In this situation, it becomes necessary for students to have a proper guide that will help them solve the book and improve their knowledge. The book is super useful in all kinds of doubt-clearing and the Class 12 RD Sharma chapter 12 exercise FBQ solution in particular is an exceptional specimen of the series.

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RD Sharma Class 12 Solutions Chapter 12 FBQ Derivative as a Rate Measure - Other Exercise
Derivative as a Rate Measurer Excercise: FBQ
Derivatives as a rate measure exercise fill in the blanks question 5
RD Sharma Chapter-Wise Solutions:

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 12 FBQ Derivative as a Rate Measure - Other Exercise

Derivative as a Rate Measurer Excercise: FBQ

Derivatives as a rate measure exercise fill in the blanks question 1

Answer: $\frac{- 12}{5}$
Hint: Here, we use the basic concept of $\frac{d y}{d x}$ and $\frac{d z}{d x}$
Given: $\sqrt{x^{2} + 16}$ with respect to $[\frac{x}{x - 1}]$ at $x = 3$
Solution: Let $y = \sqrt{x^{2} + 16}$
$z = \sqrt{\frac{x}{(x - 1)}}$
$\frac{d y}{d x} = x {(x^{2} + 16)}^{\frac{- 1}{2}}$
$\frac{d z}{d x} = [\frac{{(x - 1) - x}}{{{(x - 1)}^{2}}}]$
$∴ \frac{d y}{d x} = x {(x^{2} + 16)}^{\frac{- 1}{2}}$ and $(\frac{d z}{d x}) = \frac{- 1}{{(x - 1)}^{2}}$
Hence,
$\frac{d y}{d x} = [\frac{(\frac{d y}{d x})}{(\frac{d z}{d x})}]$
$∴ \frac{d y}{d x} = \frac{{[x (x^{2} + 16)]}^{\frac{- 1}{2}}}{- 1 / (x - 1)^{2}}$
${\frac{d y}{d z}}_{x = 3} = [3 \times \frac{1}{5} / \frac{- 1}{4}] = (\frac{3}{- 5}) \times 4$
$= \frac{- 12}{5}$

Derivatives as a rate measure exercise fill in the blanks question 2

Answer: $r$
Hint: Here, we use the of surface area $S = 4 π r^{2}$
Given: $\frac{d r}{d t} = 2 c m / s$
Solution: $\frac{d s}{d t} = 8 π r \frac{d r}{d t}$
$\frac{d s}{d t} = 8 π r \times 2$
$\frac{d s}{d t} = 16 π r$
Rate of change of surface area

\infty

radius of sphere.

Derivatives as a rate measure exercise fill in the blanks question 3

Answer: $10 \sqrt{2} o r 14.14 c m^{2} / s$
Hint: Here, we use the length of diagonal

\frac{d s}{d t} = 0.5 c m / s

and

l = \sqrt{2} a

Given:

\frac{d l}{d t} = 0.5 c m / s

,Area = 400
Solution: $\frac{d (\sqrt{2} a)}{d t} = 0.5$
$\frac{d a}{d t} = \frac{0.5}{\sqrt{2}}$
$a^{2} = 400$
So, a = 20
$\frac{d a^{2}}{d t} = 2 \times \frac{20 \times 0.5}{\sqrt{2}} => 2 \times \frac{10}{\sqrt{2}}$

= 10 \sqrt{2} c m^{2} / s e c

Derivatives as a rate measure exercise fill in the blanks question 4

Answer: $1 c m^{^{3}} / c m^{2}$
Hint: Here, we use the volume of surface area of a cone

V = \frac{4}{3} π r^{3}

Given: Radius is 2 cm
Solution:

V = \frac{4}{3} π r^{3}

\Rightarrow \frac{d V}{d r} = 3 \times \frac{4}{3} π r^{2}

S = 4 π r^{2}

\frac{d S}{d r} = 2 \times 4 π r

= 8 π r

\frac{d V}{d S} = \frac{d V}{d r} \times \frac{d r}{d S}

=> \frac{r}{2} = 1 a t r = 2

So, $1 c m^{^{3}} / c m^{2}$ is the answer.

Derivatives as a rate measure exercise fill in the blanks question 5

Answer: $θ = \frac{π}{3}$
Hint: Here, we use the concept of

θ

angle
Given: Let $θ, \frac{d θ}{d t} = \frac{d (\sin θ)}{d t}$
Solution: Let

θ

increase twice as fast as it’s sine
$\frac{d θ}{d t} = \frac{2 d (\sin θ)}{d t}$
$\frac{d θ}{d t} = 2 \cos θ \frac{d θ}{d t}$

1 = 2 \cos θ

\cos θ = \frac{1}{2}

θ = \frac{π}{3}

Derivatives as a rate measure exercise fill in the blanks question 6

Answer: $10 \sqrt{3} c m^{2} / s$
Hint: Let the side of

c m

equilateral triangle be

x c m

Given: $\frac{d x}{d t} = 2 c m^{2} / s$
Area of equilateral triangle

A = \frac{\sqrt{3}}{4} x^{2}

.......

(i)

Solution: Differentiating equation

(i)

with respect to

x

we get

\frac{d A}{d t} = \frac{\sqrt{3}}{4} \times 2 x \times \frac{d x}{d t}

= \frac{\sqrt{3} x}{2} \frac{d x}{d t}

When

x = 10

\frac{d A}{d t} = \frac{10 \sqrt{3}}{2} \times 2 = 10 \sqrt{3} c m^{2} / s

Derivatives as a rate measure exercise fill in the blanks question 7

Answer: $\frac{1}{30 π} f t / m i n$
Hint: Here we use the formula of volume,

V = \frac{4}{3} π r^{3}

Given: $\frac{d V}{d t} = 30 c m^{3} / s e c$
Solution:

V = \frac{4}{3} π r^{3}

\frac{d V}{d t} = 4 π r^{2} (\frac{d r}{d t})

\frac{d r}{d t} = [\frac{d V}{d t}] / 4 π r^{2}

\frac{d r}{d t} = [\frac{30}{[4 \times π \times 15 \times 15]}]

= \frac{1}{30 π} f t / m i n

Derivatives as a rate measure exercise fill in the blanks question 8

Answer: $s$
Hint: Here we use the concept and formula of acceleration and velocity
Given: $S = a e^{t} + (\frac{b}{e^{t}})$
Solution: Acceleration

= d^{2} s / d t^{2}

$S = a e^{t} + (\frac{b}{e^{t}})$
$\frac{d S}{d t} = a e^{t} + b e^{- t} (- 1)$
$= a e^{t} - b e^{- t}$
$\frac{d^{2} S}{d t^{2}} = a e^{t} + b e^{- t}$
$a = a e^{t} + b e^{- t} = s$
So, $a = s$

Derivatives as a rate measure exercise fill in the blanks question 9

Answer: $\frac{15}{13} c m / s$
Hint: Here we use the concept of acceleration and velocity
Given: $V = 5 x - (x^{2} / 6)$
Solution: So,
$\frac{d V}{d t} = 5 \times \frac{d x}{d t} - \frac{x}{3} \times \frac{d x}{d t}$
$\frac{d x}{d t} = \frac{\frac{d V}{d t}}{5 - (\frac{x}{3})}$
When $x = 2$ and $\frac{d V}{d t} = 5 c m^{3} / s e c$
$\frac{d x}{d t} = \frac{5}{5 - \frac{2}{3}} = \frac{15}{13} c m / s$

Derivatives as a rate measure exercise fill in the blanks question 10

Answer: $\frac{2}{π} f t / m i n u t e$
Hint: Here we use the formula of volume of vertical cylindrical tank
Given: Radius=2 ft as the r of 8 cubic ft/min
Solution: $V = π r^{2} h$
$\frac{d V}{d t} = 4 π (\frac{d h}{d t})$
$(\frac{d h}{d t}) = (\frac{1}{4 π}) (\frac{d V}{d t})$
$\frac{d V}{d t} = (\frac{1}{4 π}) \times 8$
$= \frac{2}{π} f t / m i n u t e$

Derivatives as a rate measure exercise fill in the blanks question 11

Answer: $\frac{d C}{d t} π c m / s e c$
Hint: Here we use the circumference of circle,

C = 2 π r

Given: The radius of a circle is increasing at the rate of 0.5 cm/sec
Solution:

C = 2 π r

$\frac{d C}{d t} = 2 π \frac{d r}{d t}$
$= 2 π \times 0.5$ $[∵ \frac{d r}{d t} = 0.5]$
$\frac{d C}{d t} = π c m / s e c$

Derivatives as a rate measure exercise fill in the blanks question 12

Answer: $\frac{d A}{d t} = 12 π c m^{2} / s e c$
Hint: Here we use the formula of area of circle (A) with radius r is given by
Given: $A = π r^{2}$
Solution: $\frac{d A}{d t} = \frac{d (π r^{2})}{d t} \times \frac{d r}{d t} = 2 π r \frac{d r}{d t}$ .......[by chain rule]
It is given that

\frac{d r}{d t} = 3 c m / s

∴ \frac{d A}{d t} = 2 π r (3) = 6 π r

Thus, when r=2 cm

∴ \frac{d A}{d t} = 6 π (2) = 12 π c m^{2} / s

The RD Sharma class 12th exercise FBQ is important to practice as these questions may appear in boards. All questions of the FBQ exercise will be solved in the RD Sharma solutions book and students can use these to tackle the chapter. This chapter deals with concepts such as finding the radius of a cube, calculating the volume of the hub, measuring the rate of change of area with respect to radius, etc.

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RD Sharma Class 12th exercise FBQ has 12 questions and It includes the following concepts

Definition of derivative as a rate measurer.
Rate of increase of the perimeter, circumference of any figure
Rate of Change in Marginal cost and Marginal Revenue.
Application based question on derivative as a rate measurer

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