Find the open interval(s) on which the curve given by the vector-valued function is smooth. (Enter your answer using interval notation.) r(t) = 7t'i – t6j
Angles in Circles
Angles within a circle are feasible to create with the help of different properties of the circle such as radii, tangents, and chords. The radius is the distance from the center of the circle to the circumference of the circle. A tangent is a line made perpendicular to the radius through its endpoint placed on the circle as well as the line drawn at right angles to a tangent across the point of contact when the circle passes through the center of the circle. The chord is a line segment with its endpoints on the circle. A secant line or secant is the infinite extension of the chord.
Arcs in Circles
A circular arc is the arc of a circle formed by two distinct points. It is a section or segment of the circumference of a circle. A straight line passing through the center connecting the two distinct ends of the arc is termed a semi-circular arc.
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![### Smoothness of a Curve Defined by a Vector-Valued Function
**Problem Statement:**
Find the open interval(s) on which the curve given by the vector-valued function is smooth. (Enter your answer using interval notation.)
\[ \mathbf{r}(t) = 7t^7 \mathbf{i} - t^5 \mathbf{j} \]
**Explanation:**
The smoothness of a curve for a vector-valued function \(\mathbf{r}(t)\) can be determined by analyzing its components. A curve is smooth if \(\mathbf{r}(t)\) is differentiable and its derivative \(\mathbf{r}'(t)\) is continuous and nowhere equal to the zero vector within the interval.
For the given vector-valued function \(\mathbf{r}(t) = 7t^7 \mathbf{i} - t^5 \mathbf{j}\):
- \(\mathbf{i}\) and \(\mathbf{j}\) are the unit vectors in the x and y directions respectively.
- \(7t^7\) and \(t^5\) are the scalar components of the function.
First, compute the derivative of \(\mathbf{r}(t)\):
\[ \mathbf{r}'(t) = \frac{d}{dt}(7t^7 \mathbf{i} - t^5 \mathbf{j}) \]
This becomes:
\[ \mathbf{r}'(t) = \frac{d}{dt}(7t^7) \mathbf{i} - \frac{d}{dt}(t^5) \mathbf{j} \]
\[ \mathbf{r}'(t) = 49t^6 \mathbf{i} - 5t^4 \mathbf{j} \]
For the curve to be smooth, \(\mathbf{r}'(t)\) must not be equal to the zero vector at any point in the interval. This implies that both components must not simultaneously be zero.
\[ 49t^6 \neq 0 \quad \text{or} \quad 5t^4 \neq 0 \]
Since the only value of \(t\) that makes both \(49t^6\) and \(5t^4\) zero is \(t = 0\), the curve is not smooth at \(t = 0\).
Therefore, the smoothness interval excludes](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafc8379c-b6a2-4db8-9fce-fab4471db5da%2F9ea6eabb-deb8-48d3-a2ec-58d295b2e049%2F9ab7tjj_processed.png&w=3840&q=75)
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