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Chapter 10: Problem 58

The \(z\) -component of the derivative of the vector-valued function \(\mathbf{u}\) is 0 for \(t\) in the domain of the function. What does this information imply about the graph of \(\mathbf{u}\) ?

Short Answer

Expert verified
The information that the \(z\)-component of the derivative of the vector-valued function \(\mathbf{u}\) is always zero implies that the graph of \(\mathbf{u}\) is entirely in a plane parallel to the \(xy\)-plane, with no change, or 'height' variation, in the \(z\)-direction.

Step by step solution

01

Understanding the \(z\)-component of a derivative

When a derivative of a function at a point is zero, it means that the function is not changing at that point. Alternatively said, the function has a local maximum, minimum, or a constant function in its neighborhood. Now, for a vector-valued function, if the derivative of its \(z\)-component is zero, it means that the function does not change in the \(z\)-direction. This is true not for the whole function, but only for the \(z\)-component of the function.
02

Implication for the graph

Since the \(z\)-component of \(\mathbf{u}'\) is always zero, this means that the function \(\mathbf{u}\) does not change in the \(z\)-direction. Therefore, when we draw the graph of \(\mathbf{u}\), we will not observe any movement up or down at any point in the graph. Instead, the function will only change in the \(x\)- and \(y\)-directions, creating a curve or a line on the \(xy\)-plane. That is, the graph of \(\mathbf{u}\) will be a curve or line lying entirely in a plane parallel to the \(xy\)-plane.
03

Formulate the conclusion

In conclusion, if the \(z\) -component of the derivative of a vector-valued function is always zero, then the graph of the function will lie entirely in a plane parallel to the \(xy\)-plane. The function does not change, or have no 'height' variations, in the \(z\)-direction at any point of the graph, meaning it does not move above or below the defined plane.

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Most popular questions from this chapter

In Exercises 41 and \(42,\) use the definition of the derivative to find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=(3 t+2) \mathbf{i}+\left(1-t^{2}\right) \mathbf{j} $$

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