Question

The parametric equation of a curve is given by $x=t-t^3$ & $y=1-t^4$ form a loop for all values of $tϵ[-1,1]$, then its area equals,

• A. $\frac{2}{7}$
• B. $\frac{3}{7}$
• C. $\frac{16}{35}$
• D. $\frac{11}{35}$

Answer 1

Answer: (C)

$\frac{16}{35}$

$x=t-t^3$
$\therefore$ $\frac{dx}{dt}=1-3t^2$, $y=1-t^4$
$\therefore$ $\frac{dy}{dt}=-4t^3$
$\therefore$ $x\frac{dy}{dt}-y\frac{dx}{dt}=(1-t^3)(-4t^3)-(1-t^4)(1-3t^2)$
$=t^6-3t^4+3t^2-1$
$\therefore$ Area of the loop $=\frac{1}{2}{\int }_{-1}^1(x\frac{dy}{dt}-y\frac{dx}{dt})dt$

$=\frac{1}{2}{\int }_{-1}^1(t^6-3t^4+3t^2-1)dt=\frac{1}{2}{\int }_0^1(t^6-3t^4+3t^2-1)dt$
$={(\frac{t^7}{7}-\frac{3}{5}{t}^5+t^3-t)}_0^1=\frac{1}{7}-\frac{3}{5}+1-1$ $=\frac{16}{35}$