# Math Help - Arclength?

1. ## Arclength?

The problem states, r(t)=(6t,3t^2,3log(t)), find the length from the points (6,3,0) and (24,48,3log(4)). I thought it would be the integral from 1-4 of the square root of the derivatives of each squared, added together. Seems that this isn't correct.

My derivatives are r'(t)=(6,6t,3/(tln10))

My final answer is 48.928761 but this was NOT correct. Can someone offer some help here?

2. Originally Posted by Bracketology
The problem states, r(t)=(6t,3t^2,3log(t)), find the length from the points (6,3,0) and (24,48,3log(4)). I thought it would be the integral from 1-4 of the square root of the derivatives of each squared, added together. Seems that this isn't correct.

My derivatives are r'(t)=(6,6t,3/(tln10))

My final answer is 48.928761 but this was NOT correct. Can someone offer some help here?

I get $\displaystyle{\int\limits^4_1\sqrt{6^2+6^2t^2+\fra c{3^2}{t^2}}\,dt=3\int\limits^4_1\frac{2t^2+1}{t}\ dt=45+3\ln 4$ , which

is slightly more that what you got.

Tonio

3. Where did you get the 3^2/t^2? I thought it was 3/tln10, that's where Mine was different. Your answer is correct.

4. Originally Posted by Bracketology
Where did you get the 3^2/t^2? I thought it was 3/tln10, that's where Mine was different. Your answer is correct.
the derivative of $3\log{t}$ is $\dfrac{3}{t}$ ... usually in more advanced math texts, "log" is understood to be base e, not base 10.

5. Originally Posted by Bracketology
Where did you get the 3^2/t^2? I thought it was 3/tln10, that's where Mine was different. Your answer is correct.

This may be confussing: though we're taught in high school, and thus we're used, to

take $\log=\log_{10}\,,\,\,and\,\,\ln=\log_e$ (as it's usual in conventional hand calculators), in

fact many mathematicians see both log and ln as one and the same.

In fact, when I saw your question that's the first thing I thought, namely: that we have

here log = the natural logarithm (i.e., ln) and not logarithm to base 10.

This was the whole problem...

Tonio