Question

The efficiency $E$ of a screw jack is given by $E = \dfrac{{\tan \theta }}{{\tan (\theta + \phi )}}$ where $\theta$ is variable and $\phi$ is some constant angle lying in $(0,\dfrac{\pi }{2})$. The maximum efficiency is given by,
A. $\dfrac{{1 - \cos \phi }}{{1 + \cos \phi }}$
B. $\dfrac{{1 - \sin \phi }}{{1 + \sin \phi }}$
C. $\dfrac{{\cos \phi }}{{3(1 + \sin \phi )}}$
D. $\dfrac{{\cos \phi }}{{1 + \cos \phi }}$

Answer 1

The efficiency of the screw jack will be maximum if the second order derivative of the efficiency is a negative quantity. To solve this question from the maxima or minima condition, find the maximum value of the angle $\theta$ in terms of the angle $\phi$ to get the maximum efficiency.

Formula used: The value of any function will be a maxima or minima if $\dfrac{{dy}}{{dx}} = 0$ where y is a function of x.
Relation of sine of compound angles, $2\sin a\cos b = \sin (a + b) + \sin (a - b)$]

Complete step by step answer:
We have given here that the efficiency of a screw jack is $E = \dfrac{{\tan \theta }}{{\tan (\theta + \phi )}}$ where $\theta$is variable and $\phi$ is some constant angle lying in $(0,\dfrac{\pi }{2})$. Now, we have to find the maximum value of the efficiency.
We know that the value of any function will be a maxima or minima if $\dfrac{{dy}}{{dx}} = 0$ where y is a function of x. So, if we find the first order derivative of the efficiency w.r.t $\theta$ we can get the minima or maxima condition.

Let’s first rewrite the expression of the efficiency using the formulas of compound angles.
$E = \dfrac{{\tan \theta }}{{\tan (\theta + \phi )}}$
$\Rightarrow E = \dfrac{{\sin \theta \cos (\theta + \phi )}}{{\sin (\theta + \phi )\cos \theta }}$
$\Rightarrow E = \dfrac{{2\sin \theta \cos (\theta + \phi )}}{{2\sin (\theta + \phi )\cos \theta }}$
$\Rightarrow E = \dfrac{{\sin (2\theta + \phi ) + \sin (\theta - \theta - \phi )}}{{\sin (2\theta + \phi ) + \sin (\theta + \phi - \theta )}}$ [ using the formula, $2\sin a\cos b = \sin (a + b) + \sin (a - b)$]
$\Rightarrow E = \dfrac{{\sin (2\theta + \phi ) - \sin (\phi )}}{{\sin (2\theta + \phi ) + \sin (\phi )}}$…..(i)
Now, differentiating with respect to $\theta$ we get,
$\dfrac{{dE}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {\dfrac{{\sin (2\theta + \phi ) - \sin (\phi )}}{{\sin (2\theta + \phi ) + \sin (\phi )}}} \right)$
$\Rightarrow \dfrac{{dE}}{{d\theta }} = \dfrac{{[\sin (2\theta + \phi ) + \sin (\phi )]\dfrac{d}{{d\theta }}\left[ {\sin (2\theta + \phi ) - \sin (\phi )} \right] - [\sin (2\theta + \phi ) - \sin (\phi )]\dfrac{d}{{d\theta }}\left[ {\sin (2\theta + \phi ) + \sin (\phi )} \right]}}{{{{[\sin (2\theta + \phi ) + \sin (\phi )]}^2}}}$
$\Rightarrow \dfrac{{dE}}{{d\theta }} = \dfrac{{[\sin (2\theta + \phi ) + \sin (\phi )] \times 2\cos (2\theta + \phi ) - [\sin (2\theta + \phi ) - \sin (\phi )] \times 2\cos (2\theta + \phi )}}{{{{[\sin (2\theta + \phi ) + \sin (\phi )]}^2}}}$

Now, for maxima or minima conditions, this will be equal to zero. Hence,
$\dfrac{{dE}}{{d\theta }} = \dfrac{{[\sin (2\theta + \phi ) + \sin (\phi )] \times 2\cos (2\theta + \phi ) - [\sin (2\theta + \phi ) - \sin (\phi )] \times 2\cos (2\theta + \phi )}}{{{{[\sin (2\theta + \phi ) + \sin (\phi )]}^2}}} = 0$
$\Rightarrow 2\cos (2\theta + \phi )\dfrac{{[\sin (2\theta + \phi ) + \sin (\phi )] - [\sin (2\theta + \phi ) - \sin (\phi )]}}{{{{[\sin (2\theta + \phi ) + \sin (\phi )]}^2}}} = 0$
$\Rightarrow 2\cos (2\theta + \phi )[\sin (2\theta + \phi ) + \sin (\phi ) - \sin (2\theta + \phi ) + \sin (\phi )] = 0$
$\Rightarrow 4\cos (2\theta + \phi )\sin \phi = 0$
Hence, from this we can write,
$\sin \phi = 0$or $\cos (2\theta + \phi ) = 0$
For, $\sin \phi = 0$
$\phi = 0$
But $\phi$ is lying in $(0,\dfrac{\pi }{2})$or $0 < \phi < \dfrac{\pi }{2}$
Hence, $\cos (2\theta + \phi ) = 0$
$\theta = \dfrac{\pi }{4} - \dfrac{\phi }{2}$

From the term $\dfrac{{dE}}{{d\theta }} = 2\cos (2\theta + \phi )\dfrac{{2\sin (\phi )}}{{{{[\sin (2\theta + \phi ) + \sin (\phi )]}^2}}}$we can see that the if again differentiated $\dfrac{{{d^2}E}}{{d{\theta ^2}}}$will be negative at $\theta = \dfrac{\pi }{4} - \dfrac{\phi }{2}$ since differentiation of cosine is $\dfrac{d}{{d\theta }}\cos \theta = - \sin \theta$ and the square term will always be positive. Hence, $E$will have maxima at $\theta = \dfrac{\pi }{4} - \dfrac{\phi }{2}$since $\dfrac{{{d^2}E}}{{d{\theta ^2}}}$is negative.
Hence, putting the value of $\theta$ in equation(i) we will have,
$E = \dfrac{{\sin (\dfrac{\pi }{2} - \phi + \phi ) - \sin (\phi )}}{{\sin (\dfrac{\pi }{2} - \phi + \phi ) + \sin (\phi )}}$
$\Rightarrow E = \dfrac{{\sin (\dfrac{\pi }{2}) - \sin (\phi )}}{{\sin (\dfrac{\pi }{2}) + \sin (\phi )}}$
$\therefore E = \dfrac{{1 - \sin \phi }}{{1 + \sin \phi }}$
Hence, the value of maximum efficiency is $E = \dfrac{{1 - \sin \phi }}{{1 + \sin \phi }}$.

Hence, option B is the correct answer.

Note: When solving this type of problem, deduce the differentiation of the function carefully. Any missing sign or value will lead to a different or wrong result. We can also solve this problem by directly differentiating the given function without using the relation of compound angle at first, but when equating the relation to zero we have to use the relation of compound angles or else we will miss the require solution $\theta = \dfrac{\pi }{4} - \dfrac{\phi }{2}$ and will get only the solution $\phi = 0$.