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Class 12th Application of Derivatives

An open box with a square base is to be made of a given quantity of cardboard of area c². Show that the maximum volume of the box is $\frac{c ³}{6 \sqrt 3}$ cubic units.

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alok kumar singh

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

Class 12th Application of Derivatives

If the straight-line x cosα + y sinα = p touches the curve, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then prove that a² cos²α + b² sin²α =p².

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} W e k n o w t h a t y = m x + c w i l l t o u c h t h e e l l i p s e \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 i f c^{2} = a^{2} m^{2} + b^{2} \\ H e r e e q u a t i o n o f s t r a i g h t l i n e i s x c o s α + y s i n α = p a n d t h a t o f e l l i p s e i s \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ x c o s α + y s i n α = p \\ \Rightarrow y s i n α = - x c o s α + p \\ \Rightarrow y = - x \frac{c o s α}{s i n α} + \frac{p}{s i n α} \Rightarrow y = - x c o t α + \frac{p}{s i n α} \\ C o m p a i n g w i t h y = m x + c, w e g e t \\ m = - c o t α a n d c = \frac{p}{s i n α} \\ S o, a c c o r d i n g t o t h e c o n d i t i o n, w e g e t c^{2} = a^{2} m^{2} + b^{2} \\ \frac{p^{2}}{s i n^{2} α} = a^{2} {(- c o t α)}^{2} + b^{2} \\ \Rightarrow \frac{p^{2}}{s i n^{2} α} = a^{2} \frac{c o s^{2} α}{s i n^{2} α} + b^{2} \Rightarrow p^{2} = a^{2} c o s^{2} α + b^{2} s i n^{2} α \\ H e n c e, a^{2} c o s^{2} α + b^{2} s i n^{2} α = p^{2} H e n c e p r o v e d . \end{array}$

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Class 12th Application of Derivatives

A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription, and it is believed that for every increase of Re 1/-, one subscriber will discontinue the service. Find what increase will bring maximum profit.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t u s c o n s i d e r t h a t t h e c o m p a n y i n c r e a s e s t h e a n n u a l s u b s c r i p t i o n b y R s . x \\ S o, x i s t h e n u m b e r o f s u b s c r i b e r s w h o d i s c o n t i n u e t h e s e r v i c e s . \\ ∴ T o t a l r e v e n u e, R (x) = (500 - x) (300 + x) \\ = 150000 + 500 x - 300 x - x^{2} \\ = - x^{2} + 200 x + 150000 \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t R^{'} (x) = - 2 x + 200 \\ F o r l o c a l maxima a n d l o c a l minima, R^{'} (x) = 0 \\ - 2 x + 200 = 0 \Rightarrow x = 100 \\ R^{''} (x) = - 2 < 0 M a x i m a \\ S o, R (x) i s maximum a t x = 100 \\ H e n c e, i n o r d e r t o g e t maximum profit, the company should increase its a n n u a l s u b s c r i p t i o n \\ b y R s . 100 . \end{array}$

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Class 12th NEET

Consider a thin target (10^–2m square, 10^–3m thickness) of sodium, which produces a photocurrent of 100μA when a light of intensity 100W/m²(λ = 660nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m³]

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New answer posted

a year ago

Class 12th Application of Derivatives

Find the points of local maxima, local minima, and the points of inflection of the function f(x) = $x^{5} - 5 x ? + 5 x ³ - 1$ Also, find the corresponding local maximum and local minimum values.

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{5} - 5 x^{4} + 5 x^{3} - 1 \\ f^{'} (x) = 5 x^{4} - 20 x^{3} + 15 x^{2} \\ F o r l o c a l maxima a n d l o c a l minima, f^{'} (x) = 0 \\ ∴ 5 x^{4} - 20 x^{3} + 15 x^{2} = 0 \Rightarrow 5 x^{2} (x^{2} - 4 x + 3) = 0 \\ \Rightarrow 5 x^{2} (x^{2} - 3 x - x + 3) = 0 \Rightarrow x^{2} (x - 3) (x - 1) = 0 \\ ∴ x = 0, x = 1 a n d x = 3 \\ N o w f^{''} (x) = 20 x^{3} - 60 x^{2} + 30 x \\ \Rightarrow f^{''} {(x)}_{a t x = 0} = 20 {(0)}^{3} - 60 {(0)}^{2} + 30 (0) = 0 w h i c h i s n e i t h e r M a x i m a n o r M i n i m a . \\ ∴ f (x) h a s t h e point o f inflextion a t x = 0 \\ \Rightarrow f^{''} {(x)}_{a t x = 1} = 20 {(1)}^{3} - 60 {(1)}^{2} + 30 (1) \\ = 20 - 60 + 30 = - 10 < 0 M a x i m a \\ \Rightarrow f^{''} {(x)}_{a t x = 3} = 20 {(3)}^{3} - 60 {(3)}^{2} + 30 (3) \\ = 540 - 540 + 90 = 90 > 0 M i n i m a \\ T h e m a x i m u m v a l u e o f t h e f u n c t i o n a t x = 1 \\ f (x) = {(1)}^{5} - 5 {(1)}^{4} + 5 {(1)}^{3} - 1 \\ = 1 - 5 + 5 - 1 = 0 \\ T h e minimum v a l u e o f t h e f u n c t i o n a t x = 3 \\ f (x) = {(3)}^{5} - 5 {(3)}^{4} + 5 {(3)}^{3} - 1 \\ &thins \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\frac{π}{3}$ .

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alok kumar singh

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

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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alok kumar singh

Contributor-Level 10

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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New question posted

a year ago

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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New answer posted

a year ago

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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alok kumar singh

Contributor-Level 10

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

Sol.

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Six

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

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alok kumar singh

Contributor-Level 10

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

Sol:

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

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