# Parametric equations and a value for the parameter t are given x = (60 cos 30^{circ})t, y = 5 + (60 sin 30^{circ})t - 16t2, t = 2. Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t. Question
Parametric equations, polar coordinates, and vector-valued functions Parametric equations and a value for the parameter t are given $$x = (60 cos 30^{\circ})t, y = 5 + (60 sin 30^{\circ})t - 16t2, t = 2$$. Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t. 2021-03-02
Step 1 Given Parametric equations and a value for the parameter t $$x = (60 cos 30^{\circ})t, y = 5 + (60 sin 30^{\circ})t − 16t^2$$ and $$t=2$$ We have to find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t. Step 2 Given parametric equation $$x = (60 cos 30^{\circ})t, y = 5 + (60 sin 30^{\circ})t − 16t^2$$ As cos $$30^{\circ} = \frac{\sqrt3}{2} and sin 30^{\circ} = \frac{1}{2}$$ So, $$x = (30 \sqrt3)t, y = 5 + 30t − 16t^2$$ (1) Now we have to find the coordinates of the point on the plane curve corresponding to the value$$t=2$$. On plugging in $$t = 2 in (1) we get: x = 60 sqrt3, y = 1$$ So, $$(60 \sqrt3, 1)$$ is the point.

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