Arc length of 3D curve from (8,4,0) to (24,36,4log(3))

[mknut389]

Homework Statement

Consider the path f{r}(t) = (8t, 4t^2, 4log(t) ) defined for t > 0.
Find the length of the curve between the points (8, 4, 0) and (24, 36, 4log(3)).

Homework Equations

$$\int$$|r' (t)|dt

The Attempt at a Solution

r(t)=(8t, 4t^2, 4log(t))
r'(t)=(8, 8t, 4/(ln(10)t))
|r' (t)|=$$\sqrt{8^2+(8t)^2+(1.737177927/t)^2}$$
|r' (t)|=$$\sqrt{64+64t^2+3.01778715219/t^2}$$

At Point (8,4,0) t=1 and at Point (24, 36, 4log(3)) t=3
Therefore the integral is from 1 to 3

from here, the integral of $$\int$$$$\sqrt{64+64t^2+3.01778715219/t^2}$$
is to complex to do by hand, so with MATLAB and TI-89 I am getting an answer of 36.106527, which according to the assignment is wrong. Am I going about this problem wrong? what should I do?

 

[Office_Shredder]

Are you sure that by log(t) it's not referring to base e logarithm? Some books use log and ln as the same thing. Also, you have a 1/x when you should have a 1/t in the third coordinate

 

[mknut389]

I am sure it is log(t) and not ln(t). That would make it a bit easier... Thanks for the variable mistake catch... I fixed it...

 

[Dick]

I get 36.10652942031572. I don't think you are doing anything fundamentally wrong. Why are you sure it's not ln(x)?