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Question: If \(x=a\left( \cos \theta +\theta \sin \theta \right)\) and \(y=a\left( \sin \theta -\theta \cos \t...

If x=a(cos⁡θ+θsin⁡θ)x=a\left( \cos \theta +\theta \sin \theta \right) and y=a(sin⁡θ−θcos⁡θ)y=a\left( \sin \theta -\theta \cos \theta \right) , then dydx\dfrac{dy}{dx} is equal to.
(a) cos⁡θ\cos \theta
(b) tan⁡θ\tan \theta
(c) sec⁡θ\sec \theta
(d) csc⁡θ\csc \theta

Explanation

Solution

Hint:For solving this question we will use some standard results differentiation of y=sin⁡xy=\sin x , y=cos⁡xy=\cos x and then apply the product rule of differentiation to differentiate the given term with respect to θ\theta and get the expression of dydθ&dxdθ\dfrac{dy}{d\theta }\And \dfrac{dx}{d\theta } as a function of θ\theta . After that, we will use the formula dydx=dydθ×1(dxdθ)\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)} and solve further to get the value of dydx\dfrac{dy}{dx} easily.

Complete step-by-step answer:
Given:
It is given that, if x=a(cos⁡θ+θsin⁡θ)x=a\left( \cos \theta +\theta \sin \theta \right) and y=a(sin⁡θ−θcos⁡θ)y=a\left( \sin \theta -\theta \cos \theta \right) , then we have to find the value of dydx\dfrac{dy}{dx} .
Now, before we proceed we should know the following formulas and concepts of trigonometry and differential calculus:
1. If y=f(x)⋅g(x)y=f\left( x \right)\cdot g\left( x \right) , then dydx=d(f(x)⋅g(x))dx=f′(x)⋅g(x)+f(x)⋅g′(x)\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right)\cdot g\left( x \right) \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right) . This is also known as the product rule of differentiation.
2. If y=sin⁡xy=\sin x , then dydx=d(sin⁡x)dx=cos⁡x\dfrac{dy}{dx}=\dfrac{d\left( \sin x \right)}{dx}=\cos x .
3. If y=cos⁡xy=\cos x , then dydx=d(cos⁡x)dx=−sin⁡x\dfrac{dy}{dx}=\dfrac{d\left( \cos x \right)}{dx}=-\sin x .
4. If x=f(θ)x=f\left( \theta \right) and y=g(θ)y=g\left( \theta \right) , then dydx=dydθ×1(dxdθ)=g′(θ)f′(θ)\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)}=\dfrac{{g}'\left( \theta \right)}{{f}'\left( \theta \right)} . This is also called differentiation of a parametric function.
Now, we will use the above-mentioned formulas and concepts to differentiate x=a(cos⁡θ+θsin⁡θ)x=a\left( \cos \theta +\theta \sin \theta \right) and y=a(sin⁡θ−θcos⁡θ)y=a\left( \sin \theta -\theta \cos \theta \right) separately with respect to θ\theta .
Calculation of dydθ\dfrac{dy}{d\theta } :
Now, we will differentiate y=a(sin⁡θ−θcos⁡θ)y=a\left( \sin \theta -\theta \cos \theta \right) with respect to θ\theta . Then,
y=a(sin⁡θ−θcos⁡θ) ⇒dydθ=d(a(sin⁡θ−θcos⁡θ))dθ \begin{aligned} & y=a\left( \sin \theta -\theta \cos \theta \right) \\\ & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{d\left( a\left( \sin \theta -\theta \cos \theta \right) \right)}{d\theta } \\\ \end{aligned}
Now, as aa is a constant term so, we can write d(a(sin⁡θ−θcos⁡θ))dθ=ad(sin⁡θ−θcos⁡θ)dθ\dfrac{d\left( a\left( \sin \theta -\theta \cos \theta \right) \right)}{d\theta }=a\dfrac{d\left( \sin \theta -\theta \cos \theta \right)}{d\theta } in the above equation. Then,
dydθ=d(a(sin⁡θ−θcos⁡θ))dθ ⇒dydθ=ad(sin⁡θ−θcos⁡θ)dθ ⇒dydθ=a(d(sin⁡θ)dθ−d(θcos⁡θ)dθ) \begin{aligned} & \dfrac{dy}{d\theta }=\dfrac{d\left( a\left( \sin \theta -\theta \cos \theta \right) \right)}{d\theta } \\\ & \Rightarrow \dfrac{dy}{d\theta }=a\dfrac{d\left( \sin \theta -\theta \cos \theta \right)}{d\theta } \\\ & \Rightarrow \dfrac{dy}{d\theta }=a\left( \dfrac{d\left( \sin \theta \right)}{d\theta }-\dfrac{d\left( \theta \cos \theta \right)}{d\theta } \right) \\\ \end{aligned}
Now, we will use the formula d(sin⁡x)dx=cos⁡x\dfrac{d\left( \sin x \right)}{dx}=\cos x to write d(sin⁡θ)dθ=cos⁡θ\dfrac{d\left( \sin \theta \right)}{d\theta }=\cos \theta and product rule of differential calculus to write d(θcos⁡θ)dθ=cos⁡θdθdθ+θd(cos⁡θ)dθ\dfrac{d\left( \theta \cos \theta \right)}{d\theta }=\cos \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \cos \theta \right)}{d\theta } in the above equation. Then,
dydθ=a(d(sin⁡θ)dθ−d(θcos⁡θ)dθ) ⇒dydθ=a(cos⁡θ−(cos⁡θdθdθ+θd(cos⁡θ)dθ)) \begin{aligned} & \dfrac{dy}{d\theta }=a\left( \dfrac{d\left( \sin \theta \right)}{d\theta }-\dfrac{d\left( \theta \cos \theta \right)}{d\theta } \right) \\\ & \Rightarrow \dfrac{dy}{d\theta }=a\left( \cos \theta -\left( \cos \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \cos \theta \right)}{d\theta } \right) \right) \\\ \end{aligned}
Now, we will write dθdθ=1\dfrac{d\theta }{d\theta }=1 and use the formula d(cos⁡x)dx=−sin⁡x\dfrac{d\left( \cos x \right)}{dx}=-\sin x to write d(cos⁡θ)dθ=−sin⁡θ\dfrac{d\left( \cos \theta \right)}{d\theta }=-\sin \theta in the above equation. Then,
dydθ=a(cos⁡θ−(cos⁡θdθdθ+θd(cos⁡θ)dθ)) ⇒dydθ=a(cos⁡θ−(cos⁡θ−θsin⁡θ)) ⇒dydθ=a(cos⁡θ−cos⁡θ+θsin⁡θ) ⇒dydθ=aθsin⁡θ..............(1) \begin{aligned} & \dfrac{dy}{d\theta }=a\left( \cos \theta -\left( \cos \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \cos \theta \right)}{d\theta } \right) \right) \\\ & \Rightarrow \dfrac{dy}{d\theta }=a\left( \cos \theta -\left( \cos \theta -\theta \sin \theta \right) \right) \\\ & \Rightarrow \dfrac{dy}{d\theta }=a\left( \cos \theta -\cos \theta +\theta \sin \theta \right) \\\ & \Rightarrow \dfrac{dy}{d\theta }=a\theta \sin \theta ..............\left( 1 \right) \\\ \end{aligned}
Calculation of dxdθ\dfrac{dx}{d\theta } :
Now, we will differentiate x=a(cos⁡θ+θsin⁡θ)x=a\left( \cos \theta +\theta \sin \theta \right) with respect to θ\theta . Then,
x=a(cos⁡θ+θsin⁡θ) ⇒dxdθ=d(a(cos⁡θ+θsin⁡θ))dθ \begin{aligned} & x=a\left( \cos \theta +\theta \sin \theta \right) \\\ & \Rightarrow \dfrac{dx}{d\theta }=\dfrac{d\left( a\left( \cos \theta +\theta \sin \theta \right) \right)}{d\theta } \\\ \end{aligned}
Now, as aa is a constant term so, we can write d(a(cos⁡θ+θsin⁡θ))dθ=ad(cos⁡θ+θsin⁡θ)dθ\dfrac{d\left( a\left( \cos \theta +\theta \sin \theta \right) \right)}{d\theta }=a\dfrac{d\left( \cos \theta +\theta \sin \theta \right)}{d\theta } in the above equation. Then,
dxdθ=d(a(cos⁡θ+θsin⁡θ))dθ ⇒dxdθ=ad(cos⁡θ+θsin⁡θ)dθ ⇒dxdθ=a(d(cos⁡θ)dθ+d(θsin⁡θ)dθ) \begin{aligned} & \dfrac{dx}{d\theta }=\dfrac{d\left( a\left( \cos \theta +\theta \sin \theta \right) \right)}{d\theta } \\\ & \Rightarrow \dfrac{dx}{d\theta }=a\dfrac{d\left( \cos \theta +\theta \sin \theta \right)}{d\theta } \\\ & \Rightarrow \dfrac{dx}{d\theta }=a\left( \dfrac{d\left( \cos \theta \right)}{d\theta }+\dfrac{d\left( \theta \sin \theta \right)}{d\theta } \right) \\\ \end{aligned}
Now, we will use the formula d(cos⁡x)dx=−sin⁡x\dfrac{d\left( \cos x \right)}{dx}=-\sin x to write d(cos⁡θ)dθ=−sin⁡θ\dfrac{d\left( \cos \theta \right)}{d\theta }=-\sin \theta and product rule of differential calculus to write d(θsin⁡θ)dθ=sin⁡θdθdθ+θd(sin⁡θ)dθ\dfrac{d\left( \theta \sin \theta \right)}{d\theta }=\sin \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \sin \theta \right)}{d\theta } in the above equation. Then,
dxdθ=a(d(cos⁡θ)dθ+d(θsin⁡θ)dθ) ⇒dxdθ=a(−sin⁡θ+(sin⁡θdθdθ+θd(sin⁡θ)dθ)) \begin{aligned} & \dfrac{dx}{d\theta }=a\left( \dfrac{d\left( \cos \theta \right)}{d\theta }+\dfrac{d\left( \theta \sin \theta \right)}{d\theta } \right) \\\ & \Rightarrow \dfrac{dx}{d\theta }=a\left( -\sin \theta +\left( \sin \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \sin \theta \right)}{d\theta } \right) \right) \\\ \end{aligned}
Now, we will write dθdθ=1\dfrac{d\theta }{d\theta }=1 and use the formula d(sin⁡x)dx=cos⁡x\dfrac{d\left( \sin x \right)}{dx}=\cos x to write d(sin⁡θ)dθ=cos⁡θ\dfrac{d\left( \sin \theta \right)}{d\theta }=\cos \theta in the above equation. Then,
dxdθ=a(−sin⁡θ+(sin⁡θdθdθ+θd(sin⁡θ)dθ)) ⇒dxdθ=a(−sin⁡θ+(sin⁡θ+θcos⁡θ)) ⇒dxdθ=a(−sin⁡θ+sin⁡θ+θcos⁡θ) ⇒dxdθ=aθcos⁡θ..............(2) \begin{aligned} & \dfrac{dx}{d\theta }=a\left( -\sin \theta +\left( \sin \theta \dfrac{d\theta }{d\theta }+\theta \dfrac{d\left( \sin \theta \right)}{d\theta } \right) \right) \\\ & \Rightarrow \dfrac{dx}{d\theta }=a\left( -\sin \theta +\left( \sin \theta +\theta \cos \theta \right) \right) \\\ & \Rightarrow \dfrac{dx}{d\theta }=a\left( -\sin \theta +\sin \theta +\theta \cos \theta \right) \\\ & \Rightarrow \dfrac{dx}{d\theta }=a\theta \cos \theta ..............\left( 2 \right) \\\ \end{aligned}
Now, we will use the formula for the differentiation of a parametric function to write dydx=dydθ×1(dxdθ)\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)} . So, we will divide the equation (1) by equation (2). Then,
dydx=dydθ×1(dxdθ) ⇒dydx=aθsin⁡θaθcos⁡θ ⇒dydx=sin⁡θcos⁡θ \begin{aligned} & \dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{1}{\left( \dfrac{dx}{d\theta } \right)} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{a\theta \sin \theta }{a\theta \cos \theta } \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{\sin \theta }{\cos \theta } \\\ \end{aligned}
Now, as we know that, sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin \theta }{\cos \theta }=\tan \theta . Then,
dydx=sin⁡θcos⁡θ ⇒dydx=tan⁡θ \begin{aligned} & \dfrac{dy}{dx}=\dfrac{\sin \theta }{\cos \theta } \\\ & \Rightarrow \dfrac{dy}{dx}=\tan \theta \\\ \end{aligned}
Now, from the above result, we conclude that if x=a(cos⁡θ+θsin⁡θ)x=a\left( \cos \theta +\theta \sin \theta \right) and y=a(sin⁡θ−θcos⁡θ)y=a\left( \sin \theta -\theta \cos \theta \right) , then dydx=tan⁡θ\dfrac{dy}{dx}=\tan \theta .
Hence, (b) will be the correct option.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct result quickly. And we should not try to eliminate θ\theta from the given equations as in every option dydx\dfrac{dy}{dx} is a function of θ\theta so, we should apply the formula for the differentiation for the parametric function accurately, and use every formula and product rule of differentiation to solve further. Moreover, we should avoid making calculation mistakes while solving to get the correct result.