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Chapter 11: Problem 4

Suppose an object moves in space with the position function \(\mathbf{r}(t)=\langle x(t), y(t), z(t)\rangle .\) Write the integral that gives the distance it travels between \(t=a\) and \(t=b\)

Short Answer

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Question: Find the integral that represents the distance an object moves in space between time \(t=a\) and \(t=b\) with a given position function \(\mathbf{r}(t)=\langle x(t), y(t), z(t)\rangle\). Answer: The integral representing the distance the object travels between time \(t=a\) and \(t=b\) can be found using the following formula: $$ L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} dt, $$ where \(x'(t)\), \(y'(t)\), and \(z'(t)\) represent the derivatives of the position function components \(x(t)\), \(y(t)\), and \(z(t)\), respectively.

Step by step solution

01

Calculate the derivative of the position function

To find the derivative of the position function with respect to time \(t\), we differentiate each component separately: $$ \frac{d\mathbf{r}}{dt} = \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle. $$ Now we can plug in the given components \(x(t)\), \(y(t)\), and \(z(t)\) and compute their derivatives: $$ \frac{d\mathbf{r}}{dt} = \langle x'(t), y'(t), z'(t) \rangle. $$
02

Calculate the magnitude of the derivative of the position function

Next, we need to find the magnitude of the derivative vector \(\frac{d\mathbf{r}}{dt}\). The magnitude of a vector \(\langle a, b, c \rangle \) is given by \(\sqrt{a^2+b^2+c^2}\). So, the magnitude of the derivative vector is: $$ \left\lVert\frac{d\mathbf{r}}{dt}\right\rVert = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}. $$
03

Set up the integral for the distance traveled

Now that we have found the magnitude of the derivative vector, we can use it to set up the integral that represents the distance the object travels between \(t=a\) and \(t=b\). The arc length of a curve in space is given by: $$ L = \int_a^b \left\lVert\frac{d\mathbf{r}}{dt}\right\rVert dt. $$ Using the magnitude of the derivative vector calculated in Step 2, we can set up the integral: $$ L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} dt. $$ This integral is the distance the object travels between time \(t=a\) and \(t=b\).

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