Question

calculate the curvature of curve:
r(t)=(e^t)+(e^-1)+(t)

Answer 1

differentiate twice

Answer 2

here is formula

http://en.wikipedia.org/wiki/Curvature#Curvature_of_a_graph

Answer 3

i still dont get it

Answer 4

can you find the 1st and 2nd derivatives ?

Answer 5

its the mathematical constant that is the problem .

Answer 6

which constant?
e?

Answer 7

differentiate a constant u get zero

Answer 8

diff t u get 1

Answer 9

diff e^t u get back the same

Answer 10

\[y' = e^{t}+1\]
\[y'' = e^{t}\]
\[k = \frac{e^{t}}{(1+(e^{t}+1)^{2})^{3/2}}\]

Answer 11

actuallywhat you have write defines a function and in its natüre curves andfunctions are two different concepts but one might still think that we are supposed to interpret its graph as a curve (which I can actually object since a curve is not just a trace but a trace with a parametrization on an open subset of R )So anyway let us proceed further as this is the curve as lying in the xy plane as (t,e^t+e^-1+t,0) defined on some open subset of IR then oone can note that k can be found with a formula containing 1st and 2nd derivatives differentiating once one can obtain (1,e^t+1,0) and further differentiating (0,e^t,0) so we need to take cross product of (1,e^t+1,0) * (0,e^t,0)=(0,0,-1) then since curvature is given by magnitude of this cross product over3rd power of magnitude of 1st derivative the answer becomes 1/((1+e^(2t)+2*e^t+1)^(0.5)) however i think you might ask the question wrong that is the question might be as find curvature of the following curve x(t)=(e^t,e^-1,t)