Question

How do you find the equation of the line tangent to `y=sqrt(t-1)` at (5,2)?

Answer 1

`y-2=1/4 (t-5)`

Explanation:

Slope of the given function is `y' =1/(2sqrt (t-1)`. At (5,2), this would be `1/4` (put t=5)

The tangent line in point-slope form would be,

`y-2=1/4 (t-5)`