Solving ODE to find the unit tangent vector 
A parametric curve is described by the following equations
  $$\frac{dx}{dt} = x, y = \cos(t), z = \sin(t)$$ and passes through
  $(1,1,0)$ when $t = 0$. By solving the ODE for $x(t)$, or otherwise,
  find an expression for $x$ in terms of $t$ and use this to write the
  space curve as a vector function. Hence, find the unit tangent to the
  curve $T(t)$ at the point $(1,1,0)$.

First I started by solving the differential equation and obtained $x=Ae^t$, where $A$ is an arbitrary constant (so $A=e^c$ or $A=0$ or $A=-e^c$). And then I found that $T(0)$ is $$T(0)=\Bigl(\frac{A}{\sqrt{A^2+1}},0,\frac{1}{\sqrt {A^2+1}}\Bigl)$$ 
But I'm not sure if this is correct. I can't seem to find what the constant's value would be. I thought of putting the initial condition to be $x(0)=1$ but that doesn't seem right. Please somebody help!
 A: It is correct to say that the general solution to
$\dfrac{dx}{dt} = x \tag 1$
takes the form
$x(t) = Ae^t; \tag 2$
it remains to determine the value of the constant $A$; this can be done using the given datum that
$x(0) = 1; \tag 3$
then (2) becomes
$A = Ae^0 = x(0) = 1; \tag 4$
thus,
$x(t) = e^t; \tag 5$
the space curve is thus
$(x(t), y(t), z(t)) = (e^t, \cos t, \sin t); \tag 6$
the tangent vector is
$(x'(t), y'(t), z'(t)) = (e^t, -\sin t, \cos t); \tag 7$
at $t = 0$ we find
$(x'(0), y'(0), z'(0)) = (1,  0,  1); \tag 8$
the norm of this vector is thus
$\sqrt{(x'(0))^2 + (y'(0))^2 + (z'(0))^2)} = \sqrt 2, \tag 9$
so the unit tangent vector at $t = 0$, where the curve passes through $(1, 1, 0)$, is
$T(0) = \left ( \dfrac{1}{\sqrt 2}, 0, \dfrac{1}{\sqrt 2} \right ) = \left ( \dfrac{\sqrt 2}{2}, 0, \dfrac{\sqrt 2}{2} \right ). \tag{10}$
In fact, we don't need the solution to (1) to solve this problem, since when $t = 0$ we are given that $x = 1$, and (1) directly asserts that
$x'(0) = \dfrac{dx}{dt}(0) = x(0) = 1. \tag{11}$
