Find the length of the curve for the given expression

– $r(t) \space = \space 8i \space + \space t^2 j \space t^3k, \space 0 \leq \space t \leq \space 1$

The main objective of this question is to find the length of the curve for the given expression.

This question uses the concept of the length of the curve. The length of an arc is how far apart two points are along a curve. It is calculated as:

$\space ||r(t)|| \space = \space \int_{a}^{b} \sqrt{(x’)^2 \space + \space (y’)^ 2 \space + \space (z’)^2 } \,dt$

We have to find the arc length. We know that it is calculated as:

$\space ||r(t)|| \space = \space \int_{a}^{b} \sqrt{(x’)^2 \space + \space (y’)^ 2 \space + \space (z’)^2 } \,dt$

Now:

$\space x’ \space = \space \frac{d}{dt}8 \space = \space 0$

$\space y’ \space = \space \frac{d}{dt}t^2 \space = \space 2t$

$\space z’ \space = \space \frac{d}{dt}t^3 \space = \space 3t$

Now substituting the values in the formula results in:

$\space ||r(t)|| \space = \space \int_{0}^{1} \sqrt{(0)^2 \space + \space (2t)^ 2 \space + \space (3t)^2 } \,dt$

By simplifying, we get:

$\space ||r(t)|| \space = \space \int_{0}^{1} t \sqrt{4 \space + \space 9t^2 } \,dt$

Let $s$ is equal to $4 \space + \space 9t^2$.

Thus:

$\space tdt \space = \space \frac{1}{18} ds$

Now $t$ equal to $0$ results in $4$ and  $t$ equal to $1$ results in $13$. \

Substituting the values, we get:

$\space ||r(t)|| \space = \space \frac{1}{18}\int_{4}^{13} \sqrt{s} \,ds$

By simplifying, we get:

$\space = \space \frac{1}{27} ( 13 ^{\frac{3}{2}} \space – \space 4 ^{\frac{3}{2}} )$

Numerical Results

The length of the curve for the given expression is:

$\space = \space \frac{1}{27} ( 13 ^{\frac{3}{2}} \space – \space 4 ^{\frac{3}{2}} )$

Example

Find the length of the curve for the given expression.

$r(t) \space = \space 10i \space + \space t^2 j \space t^3k, \space 0 \leq \space t \leq \space 1$

We have to find the arc length and calculated  as:

$\space ||r(t)|| \space = \space \int_{a}^{b} \sqrt{(x’)^2 \space + \space (y’)^ 2 \space + \space (z’)^2 } \,dt$

Now:

$\space x’ \space = \space \frac{d}{dt}10 \space = \space 0$

$\space y’ \space = \space \frac{d}{dt}t^2 \space = \space 2t$

$\space z’ \space = \space \frac{d}{dt}t^3 \space = \space 3t$

Now substituting the values in the formula results in:

$\space ||r(t)|| \space = \space \int_{0}^{1} \sqrt{(0)^2 \space + \space (2t)^ 2 \space + \space (3t)^2 } \,dt$

By simplifying, we get:

$\space ||r(t)|| \space = \space \int_{0}^{1} t \sqrt{4 \space + \space 9t^2 } \,dt$

Let $s$ is equal to $4 \space + \space 9t^2$.

$\space tdt \space = \space \frac{1}{18} ds$

Now $t$ equal to $0$ results in $4$ and  $t$ equal to $1$ results in $13$. \

Substituting the values, we get:

$\space ||r(t)|| \space = \space \frac{1}{18}\int_{4}^{13} \sqrt{s} \,ds$

By simplifying, we get:

$\space = \space \frac{1}{27} ( 13 ^{\frac{3}{2}} \space – \space 4 ^{\frac{3}{2}} )$