Question

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)

Answer 1

To find the parametric equations for the tangent line, we will first find the vectors representing the tangent lines at the given point in each direction, which are obtained by the derivative of the position vector with respect to the parameter.

dx/dt = d(e^t)/dt = e^t
dy/dt = d(te^t)/dt = e^t + te^t
dz/dt = d(te^(t^2))/dt = e^(t^2) + 2t^2 * e^(t^2)

Now, we plug in the given point (1,0,0), which corresponds to t = 0:

dx/dt|_(t=0) = 1
dy/dt|_(t=0) = 1
dz/dt|_(t=0) = 1

So the tangent vector is (1,1,1).

The parametric equations for the tangent line are:

x = 1 + t
y = 0 + t
z = 0 + t