W14 - Examples

Speed, distance, displacement

Speed, distance, displacement

The parametric curve $\left(t,\frac{2}{3}{t}^{3/2}\right)$ describes the position of a moving particle ($t$ measuring seconds).
(a) What is the speed function?

Suppose the particle travels for $8$ seconds starting at $t=0$.
(b) What is the total distance traveled?
(c) What is the total displacement?

Solution

(a)

Compute derivatives:

$$\left(x^{\prime },y^{\prime }\right)=\left(1,t^{1/2}\right)$$

Now compute the speed:

$$\begin{gathered} (x^{\prime }{)}^2+(y^{\prime }{)}^2=1+(t^{1/2}{)}^2=1+t \\ \gg \gg v(t)=\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}=\sqrt{1+t} \end{gathered}$$

---

(b)

Distance traveled by using speed.

Compute total distance traveled function:

$$s(t)={\int }_{u=0}^t\sqrt{1+u}du$$

Substitute $w=1+u$ and $dw=du$. New bounds are $1$ and $1+t$. Calculate:

$$\begin{array}{c}\gg \gg {\int }_1^{1+t}\sqrt{w}dw\\ \\ \gg \gg \frac{2}{3}{w}^{3/2}{|}_1^{1+t}\gg \gg \frac{2}{3}\left((1+t{)}^{3/2}-1\right)\end{array}$$

The distance traveled up to $t=8$ is:

$$s(8)=\frac{2}{3}\left(9^{3/2}-1\right)\gg \gg \frac{2}{3}(27-1)\gg \gg \frac{52}{3}$$

---

(c)

Displacement formula: $d=\sqrt{(x_1-x_0{)}^2+(y_1-y_0{)}^2}$

Now compute starting and ending points.

For starting point, insert $t=0$:

$$(x(t),y(t)){|}_{t=0}\gg \gg \left(t,\frac{2}{3}{t}^{3/2}\right){|}_{t=0}\gg \gg (0,0)$$

For ending point, insert $t=8$:

$$\begin{array}{c}(x(t),y(t)){|}_{t=8}\gg \gg \left(t,\frac{2}{3}{t}^{3/2}\right){|}_{t=8}\\ \\ \gg \gg \left(8,\frac{2}{3}{8}^{3/2}\right)\gg \gg \left(8,\frac{32\sqrt{2}}{3}\right)\end{array}$$

Insert $(0,0)$ and $\left(8,32\sqrt{2}/3\right)$:

$$\begin{array}{c}\gg \gg \sqrt{8^2+{\left(\frac{32\sqrt{2}}{3}\right)}^2}\gg \gg \sqrt{64+\frac{2048}{9}}\\ \\ \gg \gg \frac{\sqrt{2624}}{3}\end{array}$$ Link to original

Surface of revolution - parametric circle

Surface of revolution - parametric circle

By revolving the unit upper semicircle about the $x$-axis, we can compute the surface area of the unit sphere.

Parametrization of the unit upper semicircle:

$$\begin{array}{c}(x,y)=(\cos t,\sin t)\\ \\ \gg \gg (x^{\prime },y^{\prime })=(-\sin t,\cos t)\end{array}$$

---

Therefore, the arc element:

$$\begin{array}{c}ds=\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}dt\\ \\ \gg \gg \sqrt{(-\sin t{)}^2+(\cos t{)}^2}dt\gg \gg dt\end{array}$$

Now for $R(t)$ we choose $R(t)=y(t)=\sin t$ because we are revolving about the $x$-axis.

---

Plugging all this into the integral formula and evaluating gives:

$$A={\int }_0^{\pi }2\pi \sin tdt\gg \gg -2\pi \cos t{|}_0^{\pi }\gg \gg 4\pi$$

Notice: This method is a little easier than the method using the graph $y=\sqrt{1-x^2}$.

Link to original

Surface of revolution - parametric curve

Surface of revolution - parametric curve

Set up the integral which computes the surface area of the surface generated by revolving about the $x$-axis the curve $(t^3,t^2-1)$ for $0\le t\le 1$.

Solution

For revolution about the $x$-axis, we set $R=y(t)=t^2-1$.

Then compute $ds$:

$$\begin{array}{c}ds=\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}\gg \gg \sqrt{(3t^2{)}^2+(2t{)}^2}\gg \gg \sqrt{9t^4+4t^2}\\ \\ \gg \gg \sqrt{t^2(9t^2+4)}\gg \gg t\sqrt{9t^2+4}\end{array}$$

Therefore the desired integral is:

$$A={\int }_0^{1}2\pi Rds\gg \gg {\int }_0^{1}2\pi (t^2-1)t\sqrt{9t^2+4}dt$$Link to original

Converting to polar pi-correction

Converting to polar: pi-correction

Compute the polar coordinates of $\left(-\frac{1}{2},+\frac{\sqrt{3}}{2}\right)$ and of $\left(+\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$.

Solution

For $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ we observe first that it lies in Quadrant II.

Next compute:

$${\tan }^{-1}\left(\frac{\sqrt{3}/2}{-1/2}\right)\gg \gg {\tan }^{-1}\left(-\sqrt{3}\right)\gg \gg -\pi /3$$

This angle is in Quadrant IV. We add $\pi$ to get the polar angle in Quadrant II:

$$\theta =\pi -\pi /3\gg \gg 2\pi /3$$

The radius is of course $1$ since this point lies on the unit circle. Therefore polar coordinates are $(r,\theta )=(1,2\pi /3)$.

For $\left(+\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ we observe first that it lies in Quadrant IV. (No extra $\pi$ needed.)

Next compute:

$${\tan }^{-1}\left(\frac{-\sqrt{2}/2}{+\sqrt{2}/2}\right)\gg \gg {\tan }^{-1}(-1)\gg \gg -\pi /4$$

So the point in polar is $(1,-\pi /4)$.

Link to original

Shifted circle in polar

Shifted circle in polar

For example, let’s convert a shifted circle to polar. Say we have the Cartesian equation:

$$x^2+(y-3{)}^2=9$$

Then to find the polar we substitute $x=r\cos \theta$ and $y=r\sin \theta$ and simplify:

$$\begin{array}{c}{x}^2+(y-3{)}^2=9\\ \\ \gg \gg r^2{\cos }^2\theta +(r\sin \theta -3{)}^2=9\\ \\ \gg \gg r^2{\cos }^2\theta +r^2{\sin }^2\theta -6r\sin \theta +9=9\\ \\ \gg \gg r^2({\sin }^2\theta +{\cos }^2\theta )-6r\sin \theta =0\\ \\ \gg \gg r^2-6r\sin \theta =0\gg \gg r=6\sin \theta \end{array}$$

So this shifted circle is the polar graph of the polar function $r(\theta )=6\sin \theta$.

Link to original