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Ncert Solutions Maths class 12th Application of Derivatives

x and y are the sides of two squares such that y = x – $x^{2}$ . Find the rate of change of the area of the second square with respect to the area of the first square.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t a r e a o f t h e f i r s t s q u a r e A_{1} = x^{2} \\ a n d a r e a o f t h e s e c o n d s q u a r e A_{2} = y^{2} \\ N o w A_{1} = x^{2} a n d A_{2} = y^{2} = {(x - x^{2})}^{2} \\ D i f f e r e n t i a t i n g b o t h A_{1} a n d A_{2} w . r . t . t, w e g e t \\ \frac{d A_{1}}{d t} = 2 x . \frac{d x}{d t} a n d \frac{d A_{2}}{d t} = 2 (x - x^{2}) . (1 - 2 x) . \frac{d x}{d t} \\ ∴ \frac{d A_{2}}{d A_{1}} = \frac{\frac{d A_{2}}{d t}}{\frac{d A_{1}}{d t}} = \frac{2 (x - x^{2}) . (1 - 2 x) . \frac{d x}{d t}}{2 x . \frac{d x}{d t}} \\ = \frac{x (1 - x) (1 - 2 x)}{x} = (1 - x) (1 - 2 x) \\ = 1 - 2 x - x + 2 x^{2} = 2 x^{2} - 3 x + 1 \\ H e n c e, t h e r a t e o f c h a n g e o f a r e a o f t h e s e c o n d s q u a r e w i t h r e s p e c t t o f i r s t i s 2 x^{2} - 3 x + 1 . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

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Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol.

$\begin{array}{l} L e t x b e t h e l e n g t h o f t h e c u b e \\ ∴ V o l u m e o f t h e c u b e V = x^{3} \dots (i) \\ G i v e n t h a t \frac{d V}{d t} = K \\ D i f f e r e n t i a t i n g e q (i) w . r . t . t, w e g e t \\ \frac{d V}{d t} = 3 x^{2} . \frac{d x}{d t} = K (constant) \\ ∴ \frac{d x}{d t} = \frac{K}{3 x^{2}} \\ N o w s u r f a c e ? ? a r e a o f t h e c u b e, S = 6 x^{2} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . t, w e g e t \\ \frac{d S}{d t} = 6.2 . x . \frac{d x}{d t} = 12 x . \frac{K}{3 x^{2}} \\ \Rightarrow \frac{d S}{d t} = \frac{4 K}{x} \Rightarrow \frac{d S}{d t} \propto \frac{1}{x} (4 K = constant) \\ H e n c e, s u r f a c e ? ? a r e a o f t h e c u b e v a r i e s i n v e r s e l y a s t h e l e n g t h o f t h e s i d e . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

A swimming pool is to be drained for cleaning. If L represents the number of liters of water in the pool t seconds after the pool has been plugged off to drain and L = 200(10 – t)², how fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t L = 200 {(10 - t)}^{2} \\ w h e r e L r e p r e s e n t s t h e n u m b e r o f l i t r e s o f w a t e r i n t h e p o o l . \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., t, w e g e t \\ \frac{d L}{d t} = 200 \times 2 (10 - t) (- 1) = - 400 (10 - t) \\ B u t t h e r a t e a t w h i c h t h e w a t e r i s r u n n i n g o u t \\ = - \frac{d L}{d t} = 400 (10 - t) \dots (i) \\ R a t e a t w h i c h t h e w a t e r i s r u n n i n g a f t e r 5 seconds \\ = 400 (10 - 5) = 2000 L / s (f i n a l r a t e) \\ F o r i n i t i a l r a t e p u t t = 0 \\ = 400 (10 - 0) = 4000 L / s \\ T h e a v e r a g e r a t e a t w h i c h t h e w a t e r i s r u n n i n g o u t \\ = \frac{I n i t i a l r a t e + F i n a l r a t e}{2} = \frac{4000 + 2000}{2} = \frac{6000}{2} = 3000 L / s \\ H e n c e, t h e r e q u i r e d r a t e = 3000 L / s . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I n t e r n a l r a d i u s r = 3 c m \\ E x t e r n a l r a d i u s R = r + Δ r = 3.0005 c m \\ ∴ Δ r = 3.0005 - 3 = 0.0005 c m \\ L e t y = r^{3} \Rightarrow y + Δ y = {(r + Δ r)}^{3} = R^{3} = {(3.0005)}^{3} \dots (i) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., r, w e g e t \\ \frac{d y}{d x} = 3 r^{2} \\ ∴ Δ y = (\frac{d y}{d r}) . Δ r = 3 r^{2} \times 0.0005 \\ = 3 \times {(3)}^{2} \times 0.0005 = 27 \times 0.0005 = 0.0135 \\ ∴ {(3.0005)}^{3} = y + Δ y [F r o m e q n (i)] \\ = {(3)}^{3} + 0.0135 = 27 + 0.0135 = 27.0135 \\ V o l u m e o f t h e s h e l l = \frac{4}{3} π [R^{3} - r^{3}] \\ = \frac{4}{3} π [27.0135 - 27] = \frac{4}{3} π \times 0.0135 \\ = 4 π \times 0.0045 = 4 \times 3.14 \times 0.0045 = 0.018 π c m^{3} \\ H e n c e, a p p r o x i m a t e V o l u m e o f t h e m e t a l i n t h e s h e l l i s 0.018 π c m^{3} . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

Find the approximate value of (1.999) ?.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} {(1.999)}^{5} = {(2 - 0.001)}^{5} \\ L e t x = 2 a n d Δ x = - 0.001 \\ L e t y = x^{5} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., x, w e g e t \\ \frac{d y}{d x} = 5 x^{4} = 5 {(2)}^{4} = 80 \\ N o w Δ y = (\frac{d y}{d x}) . Δ x = 80 . (- 0.001) = - 0.080 \\ ∴ {(1.999)}^{5} = y + Δ y \\ = x^{5} - 0.080 = {(2)}^{5} - 0.080 = 32 - 0.080 = 31.92 \\ H e n c e, a p p r o x i m a t e v a l u e o f {(1.999)}^{5} i s 31.92 . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

Find an angle θ, 0 < θ < $\frac{π}{2}$ , which increases twice as fast as its sine.

12 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A s p e r t h e g i v e n c o n d i t i o n, \\ \frac{d θ}{d t} = 2 \frac{d}{d t} (s i n θ) \\ \Rightarrow \frac{d θ}{d t} = 2 c o s θ \frac{d θ}{d t} \Rightarrow 1 = 2 c o s θ \\ ∴ c o s θ = \frac{1}{2} \Rightarrow c o s θ = c o s \frac{π}{3} \Rightarrow θ = \frac{π}{3} \\ H e n c e, t h e r e q u i r e d a n g l e i s \frac{π}{3} . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius.

8 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t \\ A r e a o f c i r c l e, A = π r^{2}, w h e r e r = r a d i u s o f t h e c i r c l e a n d p e r i m e t e r = 2 π r \\ A s p e r Q u e s t i o n, \frac{d A}{d t} = Κ, w h e r e K = constant \\ \Rightarrow \frac{d}{d t} (π r^{2}) = K \\ \Rightarrow π . 2 r . \frac{d r}{d t} = K \\ ∴ \frac{d r}{d t} = \frac{Κ}{4 π r^{2}} \dots (1) \\ N o w P e r i m e t e r c = 2 π r \\ D i f f e r e n t i a t i n g b o t h s i d e s ? w . r . t . t, w e g e t \\ \Rightarrow \frac{d c}{d t} = \frac{d}{d t} (2 π r) \\ \Rightarrow \frac{d c}{d t} = 2 π . \frac{d r}{d t} \\ \Rightarrow \frac{d c}{d t} = 2 π . \frac{K}{2 π r} = \frac{K}{r} [F r o m e q n (1)] \\ \Rightarrow \frac{d c}{d t} \propto \frac{1}{r} \\ H e n c e, t h e p e r i m e t e r o f t h e c i r c l e varies inversely a s the r a d i u s o f t h e c i r c l e . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol.

$\begin{array}{l} B a l l o f s a l t i s s p h e r i c a l \\ ∴ V o l u m e o f b a l l, V = \frac{4}{3} π r^{3}, w h e r e r = r a d i u s o f t h e b a l l . \\ A s p e r Q u e s t i o n, \frac{d V}{d t} \propto S, w h e r e S = s u r f a c e a r e a o f t h e b a l l \\ \Rightarrow \frac{d}{d t} (\frac{4}{3} π r^{3}) \propto 4 π r^{2} [? S = 4 π r^{2}] \\ \Rightarrow \frac{4}{3} π . 3 r^{2} . \frac{d}{d t} \propto 4 π r^{2} \\ \Rightarrow 4 π r^{2} . \frac{d r}{d t} = Κ . 4 π r^{2} [K = constant o f p r o p o r t i o n a l i t y] \\ \Rightarrow \frac{d r}{d t} = Κ . \frac{4 π r^{2}}{4 π r^{2}} \\ \Rightarrow \frac{d r}{d t} = Κ . 1 = K \\ H e n c e, t h e r a d i u s o f t h e b a l l i s decreasing a t constant r a t e . \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

**The sum of the surface areas of a rectangular parallelepiped with sides x, 2x, and and a sphere is given to be constant. Prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also, find the minimum value of the sum of their volumes.**

10 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t' r' b e t h e r a d i u s o f t h e s p h e r e . \\ ∴ S u r f a c e a r e a o f t h e s p h e r e = 4 π r^{2} \\ V o l u m e o f t h e s p h e r e = \frac{4}{3} π r^{3} \\ T h e s i d e s o f t h e p a r a l l e l o p i p e d a r e x, 2 x a n d \frac{x}{3} \\ ∴ I t s s u r f a c e a r e a = 2 [x \times 2 x + 2 x \times \frac{x}{3} + x \times \frac{x}{3}] \\ = 2 [2 x^{2} + \frac{2 x^{2}}{3} + \frac{x^{2}}{3}] = 2 [2 x^{2} + x^{2}] \\ = 2 [3 x^{2}] = 6 x^{2} \\ V o l u m e o f t h e p a r a l l e l o p i p e d = x \times 2 x \times \frac{x}{3} = \frac{2}{3} x^{3} \\ A s p e r t h e c o n d i t i o n s o f t h e q u e s t i o n, \\ s u r f a c e a r e a o f t h e p a r a l l e l o p i p e d + s u r f a c e a r e a o f t h e s p h e r e = constant \\ \Rightarrow 6 x^{2} + 4 π r^{2} = K (constant) \Rightarrow 4 π r^{2} = K - 6 x^{2} \\ ∴ r^{2} = \frac{K - 6 x^{2}}{4 π} \dots (i) \\ N o w l e t V = V o l u m e o f t h e p a r a l l e l o p i p e d + V o l u m e o f t h e s p h e r e \\ \Rightarrow V = \frac{2}{3} x^{3} + \frac{4}{3} π r^{3} \\ \Rightarrow V = \frac{2}{3} x^{3} + \frac{4}{3} π {[\frac{K - 6 x^{2}}{4 π}]}^{3 / 2} [F r o m e q . (i)] \\ \Rightarrow &thins \end{array}$

$\begin{array}{l} S q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow 4 π x^{2} = 9 (K - 6 x^{2}) \Rightarrow 4 π x^{2} = 9 K - 54 x^{2} \\ \Rightarrow 4 π x^{2} + 54 x^{2} = 9 K \\ \Rightarrow K = \frac{4 π x^{2} + 54 x^{2}}{9} (i i) \\ \Rightarrow 2 x^{2} (2 π + 27) = 9 K \\ ∴ x^{2} = \frac{9 K}{2 (2 π + 27)} = 3 \\ N o w f r o m e q . (i) w e h a v e \\ r^{2} = \frac{K - 6 x^{2}}{4 π} \\ \Rightarrow r^{2} = \frac{\frac{4 π x^{2} + 54 x^{2}}{9} - 6 x^{2}}{4 π} \\ \Rightarrow r^{2} = \frac{4 π x^{2} + 54 x^{2} - 54 x^{2}}{9 \times 4 π} = \frac{4 π x^{2}}{9 \times 4 π} = \frac{x^{2}}{9} \Rightarrow r = \frac{x}{3} ∴ x = 3 r \\ N o w w e h a v e \frac{d V}{d x} = 2 x^{2} - \frac{3 x}{} {(K - 6 x^{2})}^{1 / 2} \end{array}$

$\begin{array}{l} = 12 - \frac{3}{} \frac{\frac{4 K π + 54 K - 108 K}{4 π + 54}}{} \\ = 12 - \frac{3}{} [\frac{\frac{4 K π - 54 K}{4 π + 54}}{}] \\ = 12 - \frac{3}{} [\frac{4 K π - 54 K}{.}] \\ = 12 - \frac{6}{} [\frac{2 π - 27}{.}] \\ = 12 + \frac{6 K}{} [\frac{27 - 2 π}{.}] > 0 [? 27 - 2 π > 0] \\ ∴ \frac{d^{2} V}{d x^{2}} > 0 S o, i t i s minima. \\ H e n c e, t h e s u m o f v o l u m e i s min \end{array}$

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Ncert Solutions Maths class 12th Application of Derivatives

A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs Rs 5/cm² and the material for the sides costs Rs 2.50/cm². Find the least cost of the box

10 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

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Ncert Solutions Maths class 12th

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x and y are the sides of two squares such that y = x – $x^{2}$ . Find the rate of change of the area of the second square with respect to the area of the first square.

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.

Find the approximate value of (1.999) ?.

Find an angle θ, 0 < θ < $\frac{π}{2}$ , which increases twice as fast as its sine.

If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius.

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

**The sum of the surface areas of a rectangular parallelepiped with sides x, 2x, and and a sphere is given to be constant. Prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also, find the minimum value of the sum of their volumes.**

A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs Rs 5/cm² and the material for the sides costs Rs 2.50/cm². Find the least cost of the box