s = 1 + cosec t 1 - cosec t . Find d s d t .

Given, s = 1 + cos e c t 1 - cos e c t , As we know by quotient rule, if p = v u then, d p d t = u d v d t - v d u d t u 2 . So, we have; d s d t = d ( ( 1 + cos e c t ) 1 - cos e c t ) d t = ( 1 - cos e c t ) d ( 1 + cos e c t ) d t - ( 1 + cos e c t ) d ( 1 - cos e c t ) d t ( 1 - cos e c t ) 2 = ( 1 - cos e c t ) ( - cos e c t c o t t ) - ( 1 + cos e c t ) ( cos e c t c o t t ) ( 1 - cos e c t ) 2 = - cos e c t c o t t + cos e c 2 t c o t t - cos e c t c o t t - cos e c 2 t c o t t ( 1 - cos e c t ) 2 = - 2 cos e c t c o t t ( 1 - cos e c t ) 2 = - 2 cos e c t c o t t 1 + cos e c 2 t - 2 cos e c t = - 2 cos t sin 2 t × sin 2 t 1 - 2 sin t + sin 2 t = - 2 cos t ( 1 - sin t ) 2 .

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.

E M B I B E

Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-3>Q 6

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If the magnitude of sum of two vectors is equal to magnitude of difference of two vectors, the angle between these vectors is

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Important Questions on Mathematical Tools in Physics

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-3>Q 7

In the cube of side $' a'$ shown in the figure, the vector from the central point of the face $A B O D$ to the central point of the face $B E F O$ will be :

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Alpha Question Bank for Medical: Physics>Chapter 2 - Mathematical Tools in Physics>Exercise-3>Q 8

Two vectors

\vec{A}

and

\vec{B}

have equal magnitudes. The magnitude of

(\vec{A} + \vec{B})

is

n

times the magnitude of

(\vec{A} - \vec{B})

. The angle between

\vec{A}

and

\vec{B}

is:

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

Suppose $u$ and $v$ are differentiable functions of $x$ and $u (1) = 2, u^{'} (1) = 0, v (1) = 5, v^{'} (1) = - 1 .$ Find the values of the following derivatives at $x = 1 .$