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

1 answer

I assume you are in calculus.

y= t^.5
y'=.5/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.
Similar Questions
  1. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  2. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  3. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 0 answers
  4. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 3 answers
more similar questions