Q:

Finding Curvature

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.

A:

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...

