Q: A particle moves on the curve $ y = ln(x) $ so that the x-component has velocity $x'(t)= t+1$ for t=0, the particle is at the point (1,0). At time t=1, the particle is at the point?
My work thus far: I recognize that x(t) -> v(t)->a(t) by differentials in a linear path. I believe this extend to all further dimensions as well. Since we are provided with dx/dt, and I can easily integrate the x-component velocity to arrive at $$ x(t)=t^2/2+t+C $$. C= -3/2 when t=0, so $$ x(t)=t^2/2+t-3/2.$$
Here is where I am stuck: I know that since $y = ln(x) $ increases in y-component as the x-component goes to the right just by knowledge of the graph shape, and I believe the answer I want should be the magnitude of both components x & y. But I am at a loss on how to find the y-component vector since it should be a function of time. How do I finish this conceptually (I don't even really care about the answer want to know what I'm missing)?
First of all, you solved for $C$ wrongly. $x = 1$ when $t = 0$, so $C = 1$, giving $\displaystyle x(t) = \frac 12 t^2 + t + 1$.
To get $y(t)$, proceed as follows:
By chain rule, $\displaystyle \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac 1x \cdot (t+1)$.
Putting the expression for $x(t)$ we've derived into that you get:
$\displaystyle \frac{dy}{dt} = \frac{t+1}{\frac 12 t^2 + t + 1}$
Solve that by separating variables,
$\displaystyle \int dy = \int \frac{t+1}{\frac 12 t^2 + t + 1}dt$
Note that on the RHS, the numerator is the exact derivative of the denominator, so:
$\displaystyle y(t) = \ln|\frac 12 t^2 + t + 1| + D$
where $D$ is another constant to solve for. When $t = 0, y = 0$ so $D = 0$.
So you get $\displaystyle y(t) = \ln|\frac 12 t^2 + t + 1|$
When $t = 1$, therefore $\displaystyle y = \ln\frac 52$ and $\displaystyle x = \frac 52$ so at $t=1$, the particle is at $\displaystyle (\frac 52, \ln\frac 52)$.
EDIT: As you correctly observed, you can just write $y(t) = \ln x(t)$ and immediately get the answer that way. So we can really shorten the working. I'm leaving the full working here as an example of how one can overthink a simple problem!