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Question: If \(F = \dfrac{{\mu mg}}{{\cos \theta + \mu \sin \theta }}\) , then \(F\) is minimum for A. \(\co...

If F=μmgcosθ+μsinθF = \dfrac{{\mu mg}}{{\cos \theta + \mu \sin \theta }} , then FF is minimum for
A. cosθ=μ\cos \theta = \mu
B. sinθ=μ\sin \theta = \mu
C. tanθ=μ\tan \theta = \mu
D. cotθ=μ\cot \theta = \mu

Explanation

Solution

We can find the minimum value of the given function by using the concept of maxima and minima. We first derivate the given function with respect to θ\theta and then equate it to zero, to obtain the condition for minimum value of the function. Also, we calculate the minimum value of the function.

Complete step by step answer:
The given function is
F=μmgcosθ+μsinθF = \dfrac{{\mu mg}}{{\cos \theta + \mu \sin \theta }}
Here, μ,m,g\mu ,m,g are constants.
Differentiating the given equation with respect to θ\theta , we get
dFdθ=μmg(1cosθ+μsinθ)2(sinθ+μcosθ)\dfrac{{dF}}{{d\theta }} = - \mu mg{\left( {\dfrac{1}{{\cos \theta + \mu \sin \theta }}} \right)^2}\left( { - \sin \theta + \mu \cos \theta } \right)
Now, For minima, we have to put dFdθ=0\dfrac{{dF}}{{d\theta }} = 0, then
As μmg(1cosθ+μsinθ)20\mu mg{\left( {\dfrac{1}{{\cos \theta + \mu \sin \theta }}} \right)^2} \ne 0
So,
0=sinθμcosθ0 = \sin \theta - \mu \cos \theta
μcosθ=sinθ\Rightarrow \mu \cos \theta = \sin \theta
We get, tanθ=μ\tan \theta = \mu
The value of function for tanθ=μ\tan \theta = \mu is
F=tanθmgcosθ+tanθsinθF = \dfrac{{\tan \theta mg}}{{\cos \theta + \tan \theta \sin \theta }}
F=sinθmg\therefore F = \sin \theta mg. This is the minimum value of the given function for tanθ=μ\tan \theta = \mu .

Hence, option C is correct.

Note: We should know the rules of differentiation i.e. ddx[1f(x)]=[1(f(x))2]f(x)\dfrac{d}{{dx}}\left[ {\dfrac{1}{{f(x)}}} \right] = \left[ {\dfrac{{ - 1}}{{{{\left( {f(x)} \right)}^2}}}} \right]f'(x). We can also solve this question by putting the various options into the value of μ\mu , but it will generate false results for every option, so we make use of differentiation.