Length of Curve Between Points: Exploring r(t)

[Turbodog66]

Homework Statement

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))

Homework Equations

L= ∫_a^b |r'(t)|dt

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?

 

[Ray Vickson]

Homework Statement

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))

Homework Equations

L= ∫_a^b |r'(t)|dt

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?

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

 

[LCKurtz]

Homework Statement

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))

Homework Equations

L= ∫_a^b |r'(t)|dt

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?

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

 

[Turbodog66]

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

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.

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

 

[Ray Vickson]

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.

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

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!