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April 28, 2016
Posted by **Rebecca ** on Wednesday, November 10, 2010 at 11:24pm.

x= e^t

y=te^t

z=te^(t^2)

(1,0,0)

- Calc 3 -
**MathMate**, Thursday, November 11, 2010 at 9:16amTry 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.