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Arc Length of Curves in ThreeDimensional Space Examples 2
Recall from the Arc Length of Curves in ThreeDimensional Space that the arc length of a curve given by the vector equation $\vec{r}(t) = (x(t), y(t), z(t))$ that traces out a continuous and bounded curve $C$ once for $a ≤ t ≤ b$ can be calculated with the following formula:
(1)We will now look at some examples of computing arc lengths of curves. For examples can be found on the Arc Length of Curves in ThreeDimensional Space Examples 1 page.
Example 1
Find the length of the curve defined by the vectorvalued function $\vec{r}(t) = t^3 \vec{i} + \frac{\sqrt{6}}{2}t^2 \vec{j} + t \vec{k}$ for $0 ≤ t ≤ 1$.
If we differentiate component by component we get that $\frac{dx}{dt} = 3t^2$, $\frac{dy}{dt} = \sqrt{6}t$, and $\frac{dz}{dt} = 1$. Applying the formula above for arc length and we get that:
(2)Example 2
Compute the arc length of the intersection between the surfaces $x^2 = 2y$ and $3z = xy$ from the origin to $(6, 18, 36)$.
We first need to find parametric equations to describe the intersection. Let $x = t$. Then from the equations above, we have that $y = \frac{t^2}{2}$ and $z = \frac{t^3}{6}$, and so let $\vec{r}(t) = \left ( t, \frac{t^2}{2}, \frac{t^3}{6} \right )$. This vectorvalued function represents the intersection between the surfaces $x^2 = 2y$ and $3z = xy$.
Now notice that the point $(0, 0, 0)$ corresponds to $t = 0$, and the point $(6, 18, 36)$ corresponds to $t = 6$.
Now we need to differentiate each component of our vectorvalued function to get $\frac{dx}{dt} = 1$, $\frac{dy}{dt} = t$, and $\frac{dz}{dt} = \frac{t^2}{2}$. Applying the formula for arc length and we get that:
(3)