Question

show that the evolute of cycloid
x=a(t+ sint),y=a(1-cost)
is given by x = a(t-sint)
y-2a = a(t cost)​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given, $x=a(t+sint),y=a(1-cost)$

Step 1. Find $y_{1}$ and $y_{2}$

Now, $\dfrac{dx}{dt}=a(1+cost)$

and $\dfrac{dy}{dt}=asint$

$\therefore y_{1}=\dfrac{dy}{dx}$

$=\dfrac{dy}{dt}\times\dfrac{dt}{dx}$

$=\dfrac{asint}{a(1+cost)}$

$=\dfrac{2sin\dfrac{t}{2}cos\dfrac{t}{2}}{2cos^{2}\dfrac{t}{2}}$

$=\dfrac{sin\dfrac{t}{2}}{cos\dfrac{t}{2}}$

$=tan\dfrac{t}{2}$

Then $y_{2}=\dfrac{d^{2}y}{dx^{2}}$

$=\dfrac{d}{dt}(\dfrac{dy}{dx})\times\dfrac{dt}{dx}$

$=\dfrac{d}{dt}(tan\dfrac{t}{2})\times\dfrac{1}{a(1+cost)}$

$=\dfrac{1}{2}sec^{2}\dfrac{t}{2}\times\dfrac{1}{2acos^{2}\dfrac{t}{2}}$

$=\dfrac{1}{4acos^{4}\dfrac{t}{2}}$

Step 2. Find $\bar{x}$ and $\bar{y}$

Therefore, $\bar{x}=x-\dfrac{y_{1}(1+y_{1}^{2})}{y_{2}}$

$=a(t+sint)-\dfrac{tan\dfrac{t}{2}(1+tan^{2}\dfrac{t}{2})}{\dfrac{1}{4acos^{4}\dfrac{t}{2}}}$

$=a(t+sint)-4atan\dfrac{t}{2}sec^{2}\dfrac{t}{2}cos^{4}\dfrac{t}{2}$

$=a(t+sint)-4atan\dfrac{t}{2}cos^{2}\dfrac{t}{2}$

$=a(t+sint)-4asin\dfrac{t}{2}cos\dfrac{t}{2}$

$=a(t+sint)-2a(2sin\dfrac{t}{2}cos\dfrac{t}{2})$

$=a(t+sint)-2asint$

$=a(t-sint)$

$\Rightarrow \bar{x}=a(t-sint)$

and $\bar{y}=y+\dfrac{1+y_{1}^{2}}{y_{2}}$

$=a(1-cost)+\dfrac{1+tan^{2}\dfrac{t}{2}}{\dfrac{1}{4acos^{4}\dfrac{t}{2}}}$

$=a(1-cost)+4asec^{2}\dfrac{t}{2}cos^{4}\dfrac{t}{2}$

$=a(1-cost)+4acos^{2}\dfrac{t}{2}$

$=a(1-cost)+2a(2cos^{2}\dfrac{t}{2})$

$=a(1-cost)+2a(1+cost)$

$=a-acost+2a+2acost$

$=a+acost+2a$

$\Rightarrow \bar{y}-2a=a(1+cost)$

Answer:

Locus of $(\bar{x},\bar{y})$ is the evolute. Hence the evolute is given by the parametric equation

$x=a(t-sint),y-2a=a(1+cost)$ which is another cycloid.

Hence proved.

Note:

The question has a mistake which has been corrected in the proof.

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