# Homework Help: Arc Length

1. Sep 3, 2016

### Turbodog66

1. The problem statement, all variables and given/known data

Consider the path r(t) = <10t,5t2,5ln(t) defined for t >0. Find the length of the curve between (10,5,0) and (20,20,5ln(2))

2. Relevant equations

L= ∫ab |r'(t)|dt

3. The attempt at a solution
r'(t) = <10, 10t, 5/t>

t values are 1 and 2 based on the x values for the points listed

√(102 + (10t)2 + (5/t)2)
√(100 + 100t2 + 25/t2)
√(25) ⋅ √(4 + 4t2 + 1/t2)
5√(4 + 4t2 + 1/t2)

This is where I am getting stuck. Is there an additional step that is missing?

2. Sep 3, 2016

### Ray Vickson

Yes. You need to figure out the initial $t_1$ final value $t_2$ of $t$, then do the integral
$$\text{Length} = \int_{t_1}^{t_2} 5 \sqrt{ \displaystyle 4 + 4 t^2 + \frac{1}{t^2}} \, dt$$

3. Sep 3, 2016

### LCKurtz

Hint: If you re-arrange the expression under the square root, you may recognize it as a perfect square.

4. Sep 3, 2016

### Turbodog66

Based on the two points listed, I found that t1 = 1 and t2=2. So next I would then substitute in the t values and solve? I attempted that previously, but I did not get the correct answer.

$$\text{Length} = \int_{1}^{2} 5 \sqrt{ \displaystyle 4 + 4 (2^2 -1^2) + \frac{1}{2^2-1^2}} \, dt$$

5. Sep 3, 2016

### Ray Vickson

If $f(t) = 5 \sqrt{4 + 4t^2 + 1/t^2}$, then $\int_{t_1}^{t_2} f(t) \, dt = F(t_2) - F(t_1)$, where $F(t) = \int f(t) \, dt$ is the indefinite integral of $f(t)$. This is not anything like what you wrote!