Question

The problem asks:

A curve i nthe plane is defined parametrically by the equation
x = t^3 + t and y = t^4 + 2t^2.
An equation of the line tangent to the cerve at t = 1 is...?

please help. thanks!

Answer 1

dx/dt = 3t^2 + 1
dy/dt = 4t^3 + 4t

then (dy/dt) / (dx/dt) = (3t^2 + 1)/(4t^3 + 4t)

dy/dt = (3t^2 + 1)/(4t^3 + 4t)
when t = 1, dy/dt = 4/8 = 1/2

when t=1, x = 2 , y = 3
so we have the point (2,3) and slope 1/2

equation
y-3 = 1/2(x-2)
2y-6 = x-2

x - 2y = -4 or y = (1/2)x + 2