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 $k(t)=\frac{|{\overrightarrow{T}}^{\prime }(t)|}{|{\overrightarrow{r}}^{\prime }(t)|}$. The curve shown above in the xy plane is given by $\overrightarrow{r}(t)=<t,e^{0.35t}>$ a) Compute $\overrightarrow{T}(3)$ and $\overrightarrow{N}(3)$.
$$\begin{array}{l}\overrightarrow{T}(3)=\\ \overrightarrow{N}(3)=\end{array}$$
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 ${\overrightarrow{r}}^{\prime }(t)=<1,0.35e^{0.35t}>$.

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

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

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

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

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 $k_{max}$.