Question

In Exercises 7 and \(8,\) (a) sketch the space curve represented by the vector- valued function, and (b) sketch the vectors \(\mathbf{r}\left(t_{0}\right)\) and \(\mathbf{r}^{\prime}\left(t_{0}\right)\) for the given value of \(t_{0}\). $$ \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}+t \mathbf{k}, \quad t_{0}=\frac{3 \pi}{2} $$

Short answer

The vector at time \( t_0 \) is \(\mathbf{r}(t_{0}) = -2j+\frac{3 \pi}{2}k \) and the derivative of the vector at \( t_0 \) is \(\mathbf{r}'(t_{0}) = 2i\).

Step-by-step solution

Breaking down the vector-valued function

The function is given as \( \mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}+t \mathbf{k} \). This is a vector function of a parameter t. Each of its components is given as a time dependent function.

Calculate \(\mathbf{r}(t_{0})\)

First, substitute the given \(t_{0}=\frac{3 \pi}{2}\) into the equation to get the vector \(\mathbf{r}(t_{0})\). We obtain:\[ \mathbf{r}(t_0) = 2*cos(\frac{3 \pi}{2})i+2*sin(\frac{3 \pi}{2})j+\frac{3 \pi}{2}k = 2*0*i+2*-1*j+\frac{3 \pi}{2}k = -2j+\frac{3 \pi}{2}k \]

Find the derivative of \(\mathbf{r}(t)\) and evaluate at \(t_{0}\)

The derivative of the vector-valued function \( \mathbf{r}(t) \) is:\[ \mathbf{r}'(t) = -2 \sin t \mathbf{i}+2 \cos t \mathbf{j}+\mathbf{k} \]Replace \( t \) with \( t_{0} = \frac{3 \pi}{2} \), we get:\[ \mathbf{r}'(t_0) = -2sin(\frac{3 \pi}{2})i+2cos(\frac{3 \pi}{2})j+k = 2i \]

Sketch the function

Reconstruct the geometric representation of the vector-valued function and also include the vectors \(\mathbf{r}(t_{0})\) and \(\mathbf{r}'(t_{0})\). Remember the vector \(\mathbf{r}(t_{0})\) points to the position on the curve at \(t = t_{0}\), and \(\mathbf{r}'(t_{0})\) is the tangent vector to the curve at this point.