A parametric curve is described by the following equations $\dfrac{\text{d}x}{\text{d}t} =x$, $y=\cos(t)$, $z=\sin(t)$,
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⟩.
I am very unsure how to tackle this question and feel embarrassed about it, can anyone help me out please?
We first use an Anzats (educated guess) that $x(t)$ can be of the form $$x(t) = e^{a t + b},$$ And observe that $$\frac{\partial x}{\partial t} = e^{a t + b} \frac{\partial}{\partial t} (a t + b) = e^{t + b} = a x(t) $$ So we see that for the condition $x(t) = \frac{\partial x}{\partial t}$, we require that $a = 1$. This means that the parameter $b$ remains and for all values of $b$ the condition $x(t) = \frac{\partial x}{\partial t}$ will be satisfied. However to satisfy the conditon that the line passes through $(1, 1, 0)$ for $t=0$ we require that $$ \begin{align} x(0) = e^{0 + b} = 1 \\ \ln [e^{0+b}] = \ln 0 \\ 0 + b = 0 \\ b = 0 \end{align}$$ Thus for $a = 1$ and $b=0$ the function $x(t) = e^{t}$ solves our problem.
Solving the ODE for $x(t)$, $\frac{dx}{dt}=x$ is equivalent to $x^{-1}dx=dt$. Integrate both sides and we get $\ln(x)+C_0=t$. Taking $\exp$ value of both sides, $C_1x=e^t$. We get $x(t)=Ce^t$.
At the point $\langle1,1,0\rangle$, we have $t=2k\pi$ where $k\in\mathbb{N}$ and $x=Ce^{2k\pi}=1$. Therefore we get $C=e^{-2k\pi}$.
We get the curve's function to be $$T(t)=\langle e^{t-2k\pi},\cos(t),\sin(t)\rangle.$$
Let $T'(t)$ to be the tangent vector at $T(t)$, we have $$T'(t)=\langle e^{t-2k\pi},-\sin(t),\cos(t)\rangle.$$
When $t=2k\pi$, $T'$ is the tangent vector at $\langle 1,1,0\rangle$. $$T'(2k\pi)=\langle e^0,-\sin(2k\pi),\cos(2k\pi)\rangle=\langle 1,0,1\rangle.$$
The unit vector of $\langle 1,0,1\rangle$ is $\frac1{\sqrt2}\langle 1,0,1\rangle=\langle\frac{\sqrt2}2,0,\frac{\sqrt2}2\rangle$
