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Question: A force \(F\) depends on \(\theta \) as \(F = \dfrac{K}{{\sin \theta + n\cos \theta }}\), where \(K\...

A force FF depends on θ\theta as F=Ksin⁡θ+ncos⁡θF = \dfrac{K}{{\sin \theta + n\cos \theta }}, where KK and nn are positive constants. Find the minimum value of FF given that θ\theta is acute.

Explanation

Solution

In the given question, we are provided with the expression of a force dependent on angle θ\theta . We are provided with the expression for force as F=Ksin⁡θ+ncos⁡θF = \dfrac{K}{{\sin \theta + n\cos \theta }}. We have to find the value of force given that K and n are positive constants. So, we have to find the maximum value of the denominator to evaluate the minimum value of the rational function of force. To solve the problem, we must know the range of the trigonometric expression (asin⁡x+bcos⁡x)\left( {a\sin x + b\cos x} \right).

Complete step by step answer:
So, the given expression for force is F=Ksin⁡θ+ncos⁡θF = \dfrac{K}{{\sin \theta + n\cos \theta }}, where KK and nn are positive constants. We have to find the minimum value of FF given that KK and nn are positive constants. So, we will find the maximum value of the denominator (sin⁡θ+ncos⁡θ)\left( {\sin \theta + n\cos \theta } \right) such that the overall rational function of force acquires minimum value.

We know that the range of the trigonometric expression (asin⁡x+bcos⁡x)\left( {a\sin x + b\cos x} \right) is [−a2+b2,a2+b2]\left[ { - \sqrt {{a^2} + {b^2}} ,\sqrt {{a^2} + {b^2}} } \right].
Now, the range of the trigonometric expression sin⁡θ+ncos⁡θ\sin \theta + n\cos \theta is [−12+n2,12+n2]\left[ { - \sqrt {{1^2} + {n^2}} ,\sqrt {{1^2} + {n^2}} } \right].
So, the minimum and maximum values for the left side of the equation are −1+n2- \sqrt {1 + {n^2}} and 1+n2\sqrt {1 + {n^2}} respectively.
Hence, we get, −1+n2⩽(sin⁡θ+ncos⁡θ)⩽1+n2- \sqrt {1 + {n^2}} \leqslant \left( {\sin \theta + n\cos \theta } \right) \leqslant \sqrt {1 + {n^2}}
Now, we know that the maximum value of sin⁡θ+ncos⁡θ\sin \theta + n\cos \theta is 1+n2\sqrt {1 + {n^2}} .
So, the expression for force acquires a minimum value of sin⁡θ+ncos⁡θ\sin \theta + n\cos \theta as 1+n2\sqrt {1 + {n^2}} .

Hence, we get the minimum value of force as F=K1+n2F = \dfrac{K}{{\sqrt {1 + {n^2}} }}.

Note: Such questions require grip over the concepts of trigonometry and inequalities. One must know the methodology to calculate the range of trigonometric expressions of the form (asin⁡x+bcos⁡x)\left( {a\sin x + b\cos x} \right) in order to solve the given problem.We must also know that a rational function acquires minimum value for maximum value of its denominator given that the numerator is constant.