Finding Curvature

Question

A curve is described by the vector $f(t)=(3e^t, 4e^t)$. I found that the curvature is equal to $0$. I am confused on how to explain geometrically why the curvature is zero.

Answer 1

Simple: because $f(t)$ describes a line, in this case, a line of slope $4/3$, that moves outward from the origin ($t \to -\infty$) to positive $x$ and $y$ values.

Answer 2

I think the comment by David was a great hint that you seem to have missed and, perhaps, this is what is confusing you: if you have the curve

$$f(t)=\left(3e^t\,,\,4e^t\right)$$

Then, calling the first coordinate $\,x\,$ and the second one $\,y\,$ , you get

$$y=4e^t=\frac43\cdot3e^t=\frac43x\implies y=\frac43x\ldots\leftarrow\;\text{this is a straight line!}$$

Now, just as Ron answered and commented, the curvature of a straight line is zero...