In this question you may round each entry to the nearest hundredth where necessary.
In parts (b) and (c), note that we can measure the curvature of a curve with $k(t)=\frac{\left|\vec{T}^{\prime}(t)\right|}{\left|\vec{r}^{\prime}(t)\right|}$. The curve shown above in the xy plane is given by $\vec{r}(t)=< t, e^{0.35 t}> $ a) Compute $\vec{T}(3)$ and $\vec{N}(3)$.
\[
\begin{array}{l}
\vec{T}(3)= \\
\vec{N}(3)=
\end{array}
\]
b) What is the curvature of the above curve when $t=3$ ?
\[
k(3)=
\]
c) What is the maximum curvature for this curve?
\[
k_{\max }=
\]

Step 1 ：First, we need to find the derivative of the given vector function. The derivative of \(\vec{r}(t)=\) is \(\vec{r}^{\prime}(t)=<1, 0.35e^{0.35 t}>\).

Step 2 ：Next, we calculate the magnitude of \(\vec{r}^{\prime}(t)\) at \(t=3\). This is given by \(\left|\vec{r}^{\prime}(3)\right| = \sqrt{1^2 + (0.35e^{0.35 \cdot 3})^2}\).

Step 3 ：Then, we find the unit tangent vector \(\vec{T}(t)\) by dividing \(\vec{r}^{\prime}(t)\) by its magnitude. So, \(\vec{T}(3) = \frac{\vec{r}^{\prime}(3)}{\left|\vec{r}^{\prime}(3)\right|}\).

Step 4 ：To find the unit normal vector \(\vec{N}(t)\), we first find the derivative of \(\vec{T}(t)\), and then divide it by its magnitude. So, \(\vec{N}(3) = \frac{\vec{T}^{\prime}(3)}{\left|\vec{T}^{\prime}(3)\right|}\).

Step 5 ：For the curvature \(k(t)\), we use the formula \(k(t)=\frac{\left|\vec{T}^{\prime}(t)\right|}{\left|\vec{r}^{\prime}(t)\right|}\). So, \(k(3) = \frac{\left|\vec{T}^{\prime}(3)\right|}{\left|\vec{r}^{\prime}(3)\right|}\).

Step 6 ：To find the maximum curvature, we need to find the maximum value of \(k(t)\). This can be done by finding the derivative of \(k(t)\), setting it equal to zero, and solving for \(t\). Then substitute these \(t\) values into \(k(t)\) to find the maximum curvature \(k_{\max}\).