Math Problem Statement

. 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

Solution

Let's solve for the vector r0(t)\vec{r}_0(t) (which is the derivative of r(t)\vec{r}(t)) and sketch the graph of r(t)\vec{r}(t) in 2-space along with the tangent vector r0(t0)\vec{r}_0(t_0).

Given: r(t)=2sin(t)i^+3cos(t)j^\vec{r}(t) = 2 \sin(t) \, \hat{i} + 3 \cos(t) \, \hat{j} and t0=π6t_0 = \frac{\pi}{6}.

Step 1: Find r0(t)\vec{r}_0(t)

To find r0(t)\vec{r}_0(t), we need to take the derivative of r(t)\vec{r}(t) with respect to tt:

r0(t)=ddt(2sin(t)i^+3cos(t)j^)\vec{r}_0(t) = \frac{d}{dt} \left( 2 \sin(t) \, \hat{i} + 3 \cos(t) \, \hat{j} \right)

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

Thus, the derivative r0(t)\vec{r}_0(t) is: r0(t)=2cos(t)i^3sin(t)j^\vec{r}_0(t) = 2 \cos(t) \, \hat{i} - 3 \sin(t) \, \hat{j}

Step 2: Evaluate r0(t0)\vec{r}_0(t_0) at t0=π6t_0 = \frac{\pi}{6}

Now, let's substitute t0=π6t_0 = \frac{\pi}{6} into r0(t)\vec{r}_0(t):

r0(π6)=2cos(π6)i^3sin(π6)j^\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(π6)=32\cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}
  • sin(π6)=12\sin\left( \frac{\pi}{6} \right) = \frac{1}{2}

Substituting these values gives: r0(π6)=2×32i^3×12j^\vec{r}_0\left( \frac{\pi}{6} \right) = 2 \times \frac{\sqrt{3}}{2} \, \hat{i} - 3 \times \frac{1}{2} \, \hat{j} r0(π6)=3i^32j^\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 r(t)\vec{r}(t):

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

The tangent vector r0(t0)\vec{r}_0(t_0) at t0=π6t_0 = \frac{\pi}{6} can be drawn as an arrow starting at the point r(π6)\vec{r}\left( \frac{\pi}{6} \right), pointing in the direction of r0(π6)\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 tt to ensure you understand the full shape and behavior of the curve.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Parametric Equations
Vector Calculus
Trigonometry

Formulas

Derivative of parametric vector function: \(\vec{r}'(t) = \frac{d}{dt} \left( 2 \sin(t) \, \hat{i} + 3 \cos(t) \, \hat{j} \right)\)
Trigonometric identities: \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)

Theorems

Basic differentiation rules for parametric curves
Trigonometric values at specific angles

Suitable Grade Level

Grades 11-12 (Pre-Calculus, Calculus)