The equation of the normal at t=frac{pi}{2} t | vedclass

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(B) Given the parametric equations of the curve: $x=2 \sin t$ and $y=2 \cos t$. Squaring and adding, we get $x^2+y^2=4 \sin^2 t + 4 \cos^2 t = 4$. Thi

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$y=0$

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(B) Given the parametric equations of the curve: $x=2 \sin t$ and $y=2 \cos t$. Squaring and adding, we get $x^2+y^2=4 \sin^2 t + 4 \cos^2 t = 4$. This represents a circle with radius $2$. Differentiating $x^2+y^2=4$ with respect to $x$: $2x + 2y \frac{dy}{dx} = 0$, which gives $\frac{dy}{dx} = -\frac{x}{y}$. The slope of the tangent at any point $(x, y)$ is $m_T = -\frac{x}{y}$. The slope of the normal $m_N$ is $-\frac{1}{m_T} = \frac{y}{x}$. At $t = \frac{\pi}{2}$, the point is $x = 2 \sin(\frac{\pi}{2}) = 2$ and $y = 2 \cos(\frac{\pi}{2}) = 0$. The slope of the normal at this point is $m_N = \frac{0}{2} = 0$. The equation of the normal is $y - y_1 = m_N(x - x_1)$. Substituting the values: $y - 0 = 0(x - 2)$, which simplifies to $y = 0$.

The equation of the normal at $t=\frac{\pi}{2}$ to the curve $x=2 \sin t, y=2 \cos t$ is

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