calc
posted by kelly .
for the parametric curve defined by x=32t^2 and y=52t ...sketch the curve using the parametric equation to plot of the point. use an arrow to indicate the direction of the curve for o<t<1. Find an equation for the line tangent to the curve at the point where t=1.
x=3  2t^2 , so dx/dt = 4t
y= 5  2t , so dy/dt = 2
it follows that (dy/dt)÷(dx/dt)
=dy/dt
=2/4t
= 1/2 when t=1
by assigning different values to t, you can generate x and y values, thus a few points.
It should be obvious quickly that your would get a parabola with vertex at (3,5), axis of symmetry y = 5, opening to the left
when t=1, the point is (1,7) and slope =1/2
use that information and y = mx + b to find the tangent equation
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