Question

A partical moves in the x-y plane according to the scheme x = -8 sinπt and y = -2 cos²πt, where t is time. Find the equation of path of the particle.

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Answer 1

$\blue{\Large{\boxed{\tt{ \red{Solution:- }}}}}$

$\dashrightarrow$$\tt{y = -2\cos^{2}\pi t}$

$\dashrightarrow$$\tt{y = -2(\cos^{2}\pi t-\sin^{2}\pi t)}$

$\dashrightarrow$$\tt{y = 1-2\sin^{2}\pi t ---(1)}$

Now,

$\dashrightarrow$$\tt{x = -8\sin\pi t}$

$\dashrightarrow$$\tt{\sin\pi t = \frac{-x}{8}---(2)}$

Putting the value of $\sf{eq^{n}}$(2)in $\sf{eq^{n}}$ (1)

we'll get,

$\green\rightarrowtail$$\sf{y = - 2 \{1 - 2( \frac{ - x}{8})^{2} \}}$

$\green\rightarrowtail$$\sf{y = - 2(1 - \frac{2x ^{2} }{64})}$

$\green\rightarrowtail$$\sf{y = - 2 + \frac{4x^{2} }{64}}$

$\green\rightarrowtail$$\sf{y = \frac{x^2}{16} -2}$

$\green\rightarrowtail$$\sf{\frac{x^2}{16}-2 = 0}$

$\green\rightarrowtail$$\sf{x^2 = 2×16}$

$\green\rightarrowtail$$\sf{x =\sqrt{32}}$

$\green\rightarrowtail$$\sf\boxed{x =\pm4\sqrt{2}}$

Identities used:-

$\purple\mapsto$$\rm{\cos2\theta= \cos^2- \sin^2\theta}$

$\purple\mapsto$$\rm{\sin^2\theta+\cos^2\theta= 1}$

NOTE:- for the figure you may refer to the attachment.

Hope it helps :)