Find the​ curve's unit tangent vector.
Bold r left parenthesis t right parenthesis equals left parenthesis 2 cosine cubed t right parenthesis Bold j plus left parenthesis 2 sine cubed t right parenthesis Bold k​,    
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Part 1
Find the​ curve's unit tangent vector.
Bold Upper T​(t)equalsleft parenthesis nothing right parenthesis Bold j plus left parenthesis nothing right parenthesis Bold k
  
  
​(Type exact​ answers, using radicals as​ needed.)

To find the unit tangent vector \( \mathbf{T}(t) \) of the given curve \( \mathbf{r}(t) \), follow these steps:

Given:
\[
\mathbf{r}(t) = \left( 2 \cos^3(t) \right) \mathbf{j} + \left( 2 \sin^3(t) \right) \mathbf{k}
\]

### Step 1: Find the derivative of \( \mathbf{r}(t) \)
The unit tangent vector \( \mathbf{T}(t) \) is found by first determining the velocity vector \( \mathbf{r}'(t) \), which is the derivative of \( \mathbf{r}(t) \) with respect to \( t \):

\[
\mathbf{r}'(t) = \frac{d}{dt} \left( 2 \cos^3(t) \right) \mathbf{j} + \frac{d}{dt} \left( 2 \sin^3(t) \right) \mathbf{k}
\]

Using the chain rule:
\[
\frac{d}{dt} \left( 2 \cos^3(t) \right) = 2 \cdot 3 \cos^2(t) \cdot (-\sin(t)) = -6 \cos^2(t) \sin(t)
\]
\[
\frac{d}{dt} \left( 2 \sin^3(t) \right) = 2 \cdot 3 \sin^2(t) \cdot \cos(t) = 6 \sin^2(t) \cos(t)
\]

Thus, the velocity vector \( \mathbf{r}'(t) \) is:
\[
\mathbf{r}'(t) = -6 \cos^2(t) \sin(t) \mathbf{j} + 6 \sin^2(t) \cos(t) \mathbf{k}
\]

### Step 2: Find the magnitude of \( \mathbf{r}'(t) \)
The magnitude \( |\mathbf{r}'(t)| \) is:
\[
|\mathbf{r}'(t)| = \sqrt{\left( -6 \cos^2(t) \sin(t) \right)^2 + \left( 6 \sin^2(t) \cos(t) \right)^2}
\]

Simplifying:
\[
|\mathbf{r}'(t)| = \sqrt{36 \cos^4(t) \sin^2(t) + 36 \sin^4(t) \cos^2(t)}
\]
\[
|\mathbf{r}'(t)| = \sqrt{36 \cos^2(t) \sin^2(t) \left( \cos^2(t) + \sin^2(t) \right)}
\]
Since \( \cos^2(t) + \sin^2(t) = 1 \), we have:
\[
|\mathbf{r}'(t)| = \sqrt{36 \cos^2(t) \sin^2(t)}
\]
\[
|\mathbf{r}'(t)| = 6 |\cos(t) \sin(t)|
\]

### Step 3: Find the unit tangent vector
The unit tangent vector \( \mathbf{T}(t) \) is given by:
\[
\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}
\]
Thus:
\[
\mathbf{T}(t) = \frac{-6 \cos^2(t) \sin(t)}{6 |\cos(t) \sin(t)|} \mathbf{j} + \frac{6 \sin^2(t) \cos(t)}{6 |\cos(t) \sin(t)|} \mathbf{k}
\]
\[
\mathbf{T}(t) = -\frac{\cos(t)}{|\cos(t)|} \mathbf{j} + \frac{\sin(t)}{|\sin(t)|} \mathbf{k}
\]

Hence, the unit tangent vector is:
\[
\mathbf{T}(t) = -\text{sgn}(\cos(t)) \mathbf{j} + \text{sgn}(\sin(t)) \mathbf{k}
\]

Would you like any clarifications or further details on any part of this solution?

Here are 5 related questions to explore:
1. How would you find the curvature of a given curve?
2. Can the unit tangent vector ever be zero?
3. What is the significance of the unit tangent vector in physics?
4. How is the principal normal vector related to the tangent vector?
5. What is the geometric interpretation of the velocity vector?

**Tip:** Always remember that the unit tangent vector represents the direction of motion at a given point on the curve.

Find the curve's unit tangent vector. Bold r(t) equals (2 cosine cubed t) Bold j plus (2 sine cubed t) Bold k. Part 1: Find the curve's unit tangent vector. Bold T(t) equals (nothing) Bold j plus (nothing) Bold k (Type exact answers, using radicals as needed).

Learn how to find the unit tangent vector for the parametric curve r(t) = (2 cos^3(t)) j + (2 sin^3(t)) k. This detailed solution covers key calculus concepts, including finding derivatives, computing magnitudes, and applying vector calculus techniques. Suitable for students in Grades 11-12.

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