s = sin t 1 - cos t Find d s d t .

Given, s = sin t 1 - cos t . So, d s d t = d ( sin t 1 - cos t ) d t , Using quotient rule i.e; if y = u v then, d y d x = v d u d x - u d v d x v 2 , Then, d s d t = ( 1 - cos t ) d ( sin t ) d t - sin t d ( 1 - c o s t ) d t ( 1 - c o s t ) 2 = ( 1 - cos t ) cos t - sin 2 t ( 1 - cos t ) 2 = cos t - cos 2 t - sin 2 t ( 1 - cos t ) 2 = cos t - 1 ( 1 - cos t ) 2 ( As cos 2 θ + sin 2 θ = 1 ) = - ( 1 - cos t ) ( 1 - cos t ) 2 = - 1 1 - cos t = 1 cos t - 1 .

Let ∫ 4 sin 2 y d y = I , So, I = ∫ 4 sin 2 y d y = 4 ∫ 1 - cos 2 y 2 d y ( As cos 2 θ = 1 - 2 sin 2 θ ) ⇒ I = 4 2 [ ∫ d y - ∫ cos 2 y d y ] ⇒ I = 2 [ y - ( sin 2 y 2 ) ] + C ⇒ I = 2 y - sin 2 y + C which is the required value.

Let ∫ cos 2 y 7 d y = I As we know, cos 2 θ = 2 cos 2 θ - 1 ⇒ cos 2 θ = 1 + cos 2 θ 2 , So, I = ∫ cos 2 y 7 d y = 1 7 ∫ 1 + cos 2 y 2 d y ⇒ I = 1 14 [ ∫ d y + ∫ cos 2 y d y ] ⇒ I = y 14 + sin 2 y 28 + C ( where C is some constant )

Find d r d θ , r = cosec θ + cot θ - 1

r = ( cosec θ + cot θ ) - 1 The differentiation by the chain rule holds good to find d r d θ , Let cosec θ + cot θ = x ⇒ r = x - 1 ⇒ d r d x = - x - 2 From chain rule d r d θ = d r d x × d x d θ ⇒ d x d θ = - cosec θ cot θ - cosec 2 θ , ⇒ d r d θ = - x - 2 × ( - cosec θ cot θ - cosec 2 θ ) ⇒ d x d θ = cosec θ cosec θ + cot θ

(i) The magnitude of sum of three vectors A → , B → and C → representing the sides of a cube of length A is equal to (ii) If A → = 3 i ^ + 4 j ^ and B → = 7 i ^ + 24 j ^ then, the vector having the same magnitude as B → and parallel to A → is (iii) If A → ‖ B → then, A → × B → = ? (iv) The magnitude of area of the parallelogram formed by the adjacent sides of vectors A → = 3 i ^ + 2 j ^ and B → = 2 i ^ - 4 k ^ is .. (v) Sum of two opposite vector to each other is a.... vector. (vi) The unit vector along vector i ^ + j ^ + k ^ is ...... (vii) If A → is ...... to B → then, A → · B → = 0 (viii). The vector A → = i ^ + j ^ , where i ^ and j ^ are unit vectors along x -axis and y -axis respectively, makes an angle of ..... degree with x -axis. (ix) If A → + B → + C → = 0 → then, A → · B → × C → = ?

(i) Here A → = B → = B → = A So, the magnitude of the sum of these vectors will be R = A → + B → + C → = A 2 + A 2 + A 2 = 3 A . (ii) Let the required vector be S → whose magnitude is same as B → i.e; S → = B → = 7 2 + 24 2 = 49 + 576 = 625 = 25 , Now, S → = B → . A → A → = 25 × ( 3 i ^ + 4 j ^ ) 3 2 + 4 2 = 25 5 × ( 3 i ^ + 4 j ^ ) = 15 i ^ + 20 j ^ . (iii) As the two vectors are parallel i.e; the angle between them is 0 ∘ , Now, A → × B → = A B sin θ = A B sin 0 ∘ = 0 i.e; a null vector. (iv) Area of the parallelogram will be given as A → × B → , So, A → × B → = i ^ j ^ k ^ 3 2 0 2 0 - 4 = ( 2 × - 4 - 0 ) i ^ - ( 3 × - 4 - 2 × 0 ) j ^ + ( 3 × 0 - 2 × 2 ) k ^ = - 8 i ^ + 12 j ^ - 4 k ^ . Now, A → × B → = ( - 8 ) 2 + 12 2 + ( - 4 ) 2 = 64 + 144 + 16 = 224 units . (v) Negative, Positive or Zero. (vi) Let the given vector be V → = i ^ + j ^ + k , ^ So, the required unit vector will be given as; V ^ = V → V → = i ^ + j ^ + k ^ 1 2 + 1 2 + 1 2 = i ^ + j ^ + k ^ 3 = i ^ 3 + j ^ 3 + k ^ 3 . (vii) A → . B → = A B C o s θ = 0 i.e; the angle between the two vectors is 90 ∘ as C o s 90 ∘ = 0 which means means A → ⊥ B → . (viii) From the plot shown above for the given vector we have; slope = θ = tan - ( 1 1 ) = tan - 1 1 = 45 ° . (ix) As A → + B → + C → = 0 , So, R → = A → + B → + C → = 0 i.e; components of resultant vector are 0 , So A → . ( B → × C → ) = 0 .

E M B I B E

Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4

Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 4: Exercise-4

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Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 4: Exercise-4

Attempt the practice questions on Chapter 2: Mathematical Tools in Physics, Exercise 4: Exercise-4 with hints and solutions to strengthen your understanding. Alpha Question Bank for Medical: Physics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Mathematical Tools in Physics, Exercise 4: Exercise-4 with Hints & Solutions

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 4

$s = \frac{\sin t}{1 - \cos t}$ Find $\frac{d s}{d t}$ .

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 5

Find the differentiation of $r = - {(\sec θ + \tan θ)}^{- 1}$

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 6

Evaluate the Integrals of ${(1 + 2 \cos x)}^{2}$ and check your answers by differentiation.

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 7

$\int 4 \sin^{2} y d y$

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 8

$\int \frac{\cos^{2} y}{7} d y$

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 10

Find $\frac{d r}{d θ}$ , $r = {(cosec θ + \cot θ)}^{- 1}$

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 12

The resultant of two equal forces is double of either of the forces. The angle between them is:

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-4>Q 14

(i) The magnitude of sum of three vectors $\vec{A}, \vec{B}$ and $\vec{C}$ representing the sides of a cube of length A is equal to

Question Image

(ii) If $\vec{A} = 3 \hat{i} + 4 \hat{j}$ and $\vec{B} = 7 \hat{i} + 24 \hat{j}$ then, the vector having the same magnitude as $\vec{B}$ and parallel to
$\vec{A}$ is
(iii) If $\vec{A} ‖ \vec{B}$ then, $\vec{A} \times \vec{B} = ?$
(iv) The magnitude of area of the parallelogram formed by the adjacent sides of vectors $\vec{A} = 3 \hat{i} + 2 \hat{j}$ and $\vec{B} = 2 \hat{i} - 4 \hat{k}$ is ..
(v) Sum of two opposite vector to each other is a.... vector.
(vi) The unit vector along vector $\hat{i} + \hat{j} + \hat{k}$ is ......
(vii) If $\vec{A}$ is ...... to $\vec{B}$ then, $\vec{A} \cdot \vec{B} = 0$
(viii). The vector $\vec{A} = \hat{i} + \hat{j},$ where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ -axis and $y$ -axis respectively, makes an angle of ..... degree with $x$ -axis.
(ix) If $\vec{A} + \vec{B} + \vec{C} = \vec{0}$ then, $\vec{A} \cdot (\vec{B} \times \vec{C}) = ?$

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