Derivatives And Integrals Of Vector Functions

(requires JavaScript)

1. The figure shows a curve $C$ given by a vector function $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)$.

   1. Draw the vectors $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4.5\right)-\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4\right)$, $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4.2\right)-\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4\right)$, ${\displaystyle \frac{\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4.5\right)-\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4\right)}{0.5}}$, and ${\displaystyle \frac{\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4.2\right)-\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(4\right)}{0.2}}$.
   2. Write expressions for $\mathbf{r}\prime <!--FUNCTION\; APPLICATION-->\left(4\right)$ and the unit tangent vector $\mathbf{T}<!--FUNCTION\; APPLICATION-->\left(4\right)$. Draw the vector $\mathbf{T}<!--FUNCTION\; APPLICATION-->\left(4\right)$.

   

2. Make a large sketch of the curve described by the vector function $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)=⟨t^2,t⟩,0\le t\le 2$, and draw the vectors $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(1\right)$, $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(1.1\right)$, and $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(1.1\right)-\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(1\right)$. Moreover, draw the vector $\mathbf{r}\prime <!--FUNCTION\; APPLICATION-->\left(1\right)$ starting at $\left(1,1\right)$ and compare it with the vector

   ${\displaystyle \frac{\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(1.1\right)-\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(1\right)}{0.1}}$

   Explain why these vectors are similar.

3. Let $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)=\left(1+\cos <!--FUNCTION\; APPLICATION-->\left(t\right)\right)\mathbf{i}+\left(2+\sin <!--FUNCTION\; APPLICATION-->\left(t\right)\right)\mathbf{j}$ and ${\displaystyle t=\frac{\pi }{6}}$, sketch the plane curve with the given vector equation, find $\mathbf{r}\prime <!--FUNCTION\; APPLICATION-->\left(t\right)$, and sketch the position vector $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)$ and the tangent vector $\mathbf{r}\prime <!--FUNCTION\; APPLICATION-->\left(t\right)$ for the given value of $t$.

4. Find the derivative of the vector function

   $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)=e^{t^2}\mathbf{i}-\mathbf{j}+\ln <!--FUNCTION\; APPLICATION-->\left(1+3t\right)\mathbf{k}$.

   $\mathbf{r}\prime <!--FUNCTION\; APPLICATION-->\left(t\right)=2t{e}^{t^2}\mathbf{i}+\frac{3}{1+3t}\mathbf{k}$

5. Find the unit tangent vector $\mathbf{T}<!--FUNCTION\; APPLICATION-->\left(t\right)$ at the point where $t=0$ for the curve given by the equation

   $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)=\cos <!--FUNCTION\; APPLICATION-->\left(t\right)\mathbf{i}+3t\mathbf{j}+2\sin <!--FUNCTION\; APPLICATION-->\left(2t\right)\mathbf{k}$.

   ${\displaystyle \frac{3}{5}\mathbf{j}+\frac{4}{5}\mathbf{k}}$

6. Find parametric equations for the tangent line to the curve

   $x=\ln <!--FUNCTION\; APPLICATION-->\left(t\right),y=2\sqrt{t},z=t^2$

   at the point $\left(0,2,1\right)$.

   $x=t$, $y=2+t$, $z=1+2t$

7. At what point do the curves

   ${\mathbf{r}}_1<!--FUNCTION\; APPLICATION-->\left(t\right)=⟨t,1-t,3+t^2⟩$

   and

   ${\mathbf{r}}_2<!--FUNCTION\; APPLICATION-->\left(s\right)=⟨3-s,s-2,s^2⟩$

   intersect? Find the angle of intersection to the nearest degree.

   $⟨1,0,4⟩$, the angle is approx. $54.74°$

8. Evaluate the integral

   ${\displaystyle {\int }_0^{\pi /2}\left(3{\sin }^2<!--FUNCTION\; APPLICATION-->\left(t\right)\cos <!--FUNCTION\; APPLICATION-->\left(t\right)\mathbf{i}+3\sin <!--FUNCTION\; APPLICATION-->\left(t\right){\cos }^2<!--FUNCTION\; APPLICATION-->\left(t\right)\mathbf{j}+2\sin <!--FUNCTION\; APPLICATION-->\left(t\right)\cos <!--FUNCTION\; APPLICATION-->\left(t\right)\mathbf{k}\right)dt}$

   $\mathbf{i}+\mathbf{j}+\mathbf{k}$
9. Find $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)$ if $\mathbf{r}\prime <!--FUNCTION\; APPLICATION-->\left(t\right)=t\mathbf{i}+e^t\mathbf{j}+t{e}^t\mathbf{k}$ and $\mathbf{r}<!--FUNCTION\; APPLICATION-->\left(0\right)=\mathbf{i}+\mathbf{j}+\mathbf{k}$.
   ${\displaystyle \mathbf{r}<!--FUNCTION\; APPLICATION-->\left(t\right)=⟨\frac{t^2}{2}+1,e^t,t{e}^t-e^t+2⟩}$