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how do i find the equaiton of the line tangent to the curve at the point defined by the given value of t.

1. x=t, y= the square root of t, t=1/4

2. x=sec^2t-1, y= tan t, t= -pi/4

  • math - ,

    I assume you are in calculus.

    y= t^.5

    at t= .25 y'= .5/(.25)^5= .5/.5=1

    y= mx + b
    y= 1 x + b
    put in x,y and solve for b.

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