. Find the vector ~r
0
(t); then sketch the graph of ~r(t) in 2-space and draw the tangent vector
~r
0
(t0).
~r(t) = 2 sin(t)
ˆi + 3 cos(t)
ˆj; t0 = π/6

Let's solve for the vector \(\vec{r}_0(t)\) (which is the derivative of \(\vec{r}(t)\)) and sketch the graph of \(\vec{r}(t)\) in 2-space along with the tangent vector \(\vec{r}_0(t_0)\).

Given:
\[
\vec{r}(t) = 2 \sin(t) \, \hat{i} + 3 \cos(t) \, \hat{j}
\]
and \(t_0 = \frac{\pi}{6}\).

### Step 1: Find \(\vec{r}_0(t)\)

To find \(\vec{r}_0(t)\), we need to take the derivative of \(\vec{r}(t)\) with respect to \(t\):

\[
\vec{r}_0(t) = \frac{d}{dt} \left( 2 \sin(t) \, \hat{i} + 3 \cos(t) \, \hat{j} \right)
\]

- The derivative of \(2 \sin(t)\) is \(2 \cos(t)\).
- The derivative of \(3 \cos(t)\) is \(-3 \sin(t)\).

Thus, the derivative \(\vec{r}_0(t)\) is:
\[
\vec{r}_0(t) = 2 \cos(t) \, \hat{i} - 3 \sin(t) \, \hat{j}
\]

### Step 2: Evaluate \(\vec{r}_0(t_0)\) at \(t_0 = \frac{\pi}{6}\)

Now, let's substitute \(t_0 = \frac{\pi}{6}\) into \(\vec{r}_0(t)\):

\[
\vec{r}_0\left( \frac{\pi}{6} \right) = 2 \cos\left( \frac{\pi}{6} \right) \, \hat{i} - 3 \sin\left( \frac{\pi}{6} \right) \, \hat{j}
\]

Using known trigonometric values:
- \(\cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}\)
- \(\sin\left( \frac{\pi}{6} \right) = \frac{1}{2}\)

Substituting these values gives:
\[
\vec{r}_0\left( \frac{\pi}{6} \right) = 2 \times \frac{\sqrt{3}}{2} \, \hat{i} - 3 \times \frac{1}{2} \, \hat{j}
\]
\[
\vec{r}_0\left( \frac{\pi}{6} \right) = \sqrt{3} \, \hat{i} - \frac{3}{2} \, \hat{j}
\]

### Step 3: Sketch the graph

To sketch the graph of \(\vec{r}(t)\):
- The equation describes a parametric curve in 2-space.
- \(\vec{r}(t) = 2 \sin(t) \, \hat{i} + 3 \cos(t) \, \hat{j}\) traces an ellipse with:
  - Semi-major axis of 3 along the \(y\)-axis (\(\hat{j}\)-direction).
  - Semi-minor axis of 2 along the \(x\)-axis (\(\hat{i}\)-direction).

The tangent vector \(\vec{r}_0(t_0)\) at \(t_0 = \frac{\pi}{6}\) can be drawn as an arrow starting at the point \(\vec{r}\left( \frac{\pi}{6} \right)\), pointing in the direction of \(\vec{r}_0\left( \frac{\pi}{6} \right)\).

If you'd like, I can generate the graph of this curve and the tangent vector for you.

Would you like a graphical representation, further details, or have any questions?

### 5 Related Questions:
1. How do parametric equations define curves in space?
2. What is the geometric interpretation of a tangent vector?
3. How can we generalize this method to find tangent lines for 3D parametric curves?
4. What are the applications of parametric equations in physics or engineering?
5. How can we calculate the curvature of a parametric curve?

### Tip:
When working with parametric curves, always check the range of \(t\) to ensure you understand the full shape and behavior of the curve.

Find the vector ~r'(t); then sketch the graph of ~r(t) in 2-space and draw the tangent vector ~r'(t0). ~r(t) = 2 sin(t) i + 3 cos(t) j; t0 = π/6

This page explains how to find the derivative of the parametric vector function ~r(t) = 2 sin(t) i + 3 cos(t) j and evaluates the tangent vector at t0 = π/6. The problem involves concepts from vector calculus and trigonometry. Ideal for students in Grades 11-12 studying pre-calculus or calculus.

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