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.