Posted by eleanor on .
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
I assume you are in calculus.
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.