Calc 3
posted by Rebecca .
Find the parametric equations for the tangent line to the curve with the given parametric equations at specified point.
x= e^t
y=te^t
z=te^(t^2)
(1,0,0)

Try a circle with parameter t=θ
x=cos(t)
y=sin(t)
dx/dt=sin(t)
dy/dt=cos(t)
at t=0, (dx/dt,dy/dt)=(0,1)
at t=π/4, (dx/dt,dy/dt)=(√2/2,√2/2)
...etc
Here:
x= e^t
y=te^t
z=te^(t^2)
(1,0,0)
Solve for t at the point (1,0,0) gives t=0 (the only way to get y=z=0).
dx/dt=e^t
dy/dt=(1+t)e^t
dz/dt=t(2+t)e^t
Substitute t=0 and find the equation of the line passing through (1,0,0) with the given slopes.
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