A particle following the curve traced out by the path $c(t) = \left(\sin\left[e^{t}\right], t, 4-t^2\right)$ flies off on a tangent at time $t = 1$. Find the position of particle at time $t = 2$.
Solution:
$$C(t) = \left(\sin\left[e^t\right], t, 4-t^2\right),\qquad C'(t) = \left(e^t\cos\left[e^t\right], 1, -2t\right)$$
$$C(1) = \big(\sin[e], 1, 3\big),\qquad C'(t) = \big(e\cos[e], 1, -2\big)$$
At $t = 2$ the parametric is
$$[x, y, z] = \big(\sin[e], 1, 3\big) + \big(e\cos[e], 1, -2\big)(2-1)$$
so
$$(x, y, z) = \big(\sin[e] + e\cos[e], 2, 1\big)$$
Is it correct?