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.
12.2q3
![### Problem: Finding the Derivative of a Vector-Valued Function
Given the vector-valued function:
\[ \mathbf{r}(t) = 8 \sqrt{t} \mathbf{i} + t^2 \sqrt{t} \mathbf{j} + \ln(t^2) \mathbf{k} \]
We are tasked with finding the derivative \(\mathbf{r}'(t)\).
### Steps to Find the Derivative \(\mathbf{r}'(t)\):
1. **Differentiate each component of the vector function with respect to \(t\):**
- \[\mathbf{r}(t) = 8\sqrt{t} \mathbf{i}\]
- \[\mathbf{r}(t) = t^2 \sqrt{t} \mathbf{j}\]
- \[\mathbf{r}(t) = \ln(t^2) \mathbf{k}\]
2. **Apply the differentiation rules for each component:**
- For **\( \mathbf{i} \)-component**:
- The derivative of \( 8\sqrt{t} \):
- \( \frac{d}{dt}\left(8\sqrt{t}\right) = 8 \cdot \frac{1}{2}t^{-\frac{1}{2}} = 4t^{-\frac{1}{2}} = \frac{4}{\sqrt{t}} \)
- For **\( \mathbf{j} \)-component**:
- The derivative of \( t^2 \sqrt{t} \):
- \( t^2 \sqrt{t} = t^2 \cdot t^{\frac{1}{2}} = t^{2 + \frac{1}{2}} = t^{\frac{5}{2}} \)
- The derivative of \( t^{\frac{5}{2}} \):
- \( \frac{d}{dt}\left(t^{\frac{5}{2}}\right) = \frac{5}{2} t^{\frac{3}{2}} \)
- For **\( \mathbf{k} \)-component**:
- The derivative of \( \ln(t^2) \):
- Using the property of logarithms, \( \ln(t^2) = 2 \ln(t) \)
- The derivative of \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafc8379c-b6a2-4db8-9fce-fab4471db5da%2F8ccfa0e1-667e-4497-b546-e39b51cd8dbf%2Fydfnke_processed.png&w=3840&q=75)

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