We are given a curve in parametric form:

`x=sin(pi t), y=t^2+t`

We are asked to graph the curve and then find and graph the tangent line to the curve at (0,2).

To graph the curve we can substitute in values for t. Here are a few points found by starting at t=0 and incrementing t by .25:

(0,0), (.707,.3125), (1,.75), (.707, 1.3125), (0,2), (-.707, 2.8125), (-1,3.75), (-.707, 4.8125), (0,6)

(See attachment for curve.)

To find the tangent line we need a point (here (0,2)), and the slope of the tangent line at that point. To find the slope we need the derivative and we use `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))` .

`(dy)/(dt)=2t+1; (dx)/(dt)=pi cos(pi t)`

So `(dy)/(dx)=(2t+1)/(pi cos(pi t))`

Now at (0,2) we have t=1 so `(2(1)+1)/(pi cos(pi))=-3/pi`

Using the point-slope formula for the tangent line we get:

`y-2=-3/pi(x-0) ==> y=-3/pi x+2`

(See the attachment for the graph of the tangent line.)

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