Examples

Tangent to a cycloid

Find the tangent line (described parametrically) to the cycloid $(4t-4\sin t,4-4\cos t)$ when $t=\pi /4$.

Solution

(1) Compute $x^{\prime }$ and $y^{\prime }$.

Find $x^{\prime }(t)$ and $y^{\prime }(t)$:

$$\begin{array}{cc}x(t)& =4t-4\sin t\gg \gg x^{\prime }(t)=4-4\cos t\\ & \\ y(t)& =4-4\cos t\gg \gg y^{\prime }(t)=4\sin t\end{array}$$

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(2) Plug in $t=\pi /4$:

$$\begin{array}{c}{x}^{\prime }(\pi /4)\gg \gg 4-4\cos (\pi /4)\gg \gg 4-2\sqrt{2}\\ \\ y^{\prime }(\pi /4)\gg \gg 4\sin (\pi /4)\gg \gg 2\sqrt{2}\end{array}$$

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(3) Apply formula: ${\displaystyle \frac{dy}{dx}=\frac{y^{\prime }}{x^{\prime }}}$:

Calculate $\frac{dy}{dx}$ at $t=\pi /4$:

$$\begin{array}{c}\frac{dy}{dx}(\pi /4)=\frac{y^{\prime }(\pi /4)}{x^{\prime }(\pi /4)}\gg \gg \frac{2\sqrt{2}}{4-2\sqrt{2}}\\ \\ \gg \gg \frac{2\sqrt{2}}{4-2\sqrt{2}}\cdot \frac{4+2\sqrt{2}}{4+2\sqrt{2}}\\ \\ \gg \gg \frac{8\sqrt{2}+8}{16-8}\gg \gg \sqrt{2}+1\end{array}$$

So:

$${\frac{dy}{dx}|}_{t=\pi /4}=\sqrt{2}+1$$

This is the slope $m$ for our line.

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(4) Need the point $P$ for our line. Find $(x,y)$ at $t=\pi /4$.

Plug $t=\pi /4$ into parametric formulas:

$$\begin{array}{c}(x(t),y(t)){|}_{t=\pi /4}\gg \gg (4\frac{\pi }{4}-4\sin (\pi /4),4-4\cos (\pi /4))\\ \\ \gg \gg (\pi -2\sqrt{2},4-2\sqrt{2})\end{array}$$

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(5) Point-slope formulation of tangent line:

$$x=a+t,y=b+mt$$

Inserting data:

$$x=(\pi -2\sqrt{2})+t,y=(4-2\sqrt{2})+(\sqrt{2}+1)t$$

Vertical and horizontal tangents of the circle

Consider the circle parametrized by $x=\cos t$ and $y=\sin t$. Find the points where the tangent lines are vertical or horizontal.

Solution

(1) For the points with vertical tangent line, we find where the moving point has $x^{\prime }(t)=0$ (purely vertical motion):

$$\begin{array}{c}{x}^{\prime }(t)=-\sin t,\\ \\ x^{\prime }(t)=0\gg \gg -\sin t=0\\ \\ \gg \gg t=0,\pi \end{array}$$

The moving point is at $(\mathrm{1,0})$ when $t=0$, and at $(-\mathrm{1,0})$ when $t=\pi$.

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(2) For the points with horizontal tangent line, we find where the moving point has $y^{\prime }(t)=0$ (purely horizontal motion):

$$\begin{array}{c}{y}^{\prime }(t)=\cos t,\\ \\ y^{\prime }(t)=0\gg \gg \cos t=0\\ \\ \gg \gg t=\frac{\pi }{2},\frac{3\pi }{2}\end{array}$$

The moving point is at $(\mathrm{0,1})$ when $t=\pi /2$, and at $(0,-1)$ when $t=3\pi /2$.

Finding the point with specified slope

Consider the parametric curve given by $(x,y)=(t^2,t^3)$. Find the point where the slope of the tangent line to this curve equals 5.

Solution

(1) Compute the derivatives:

$$x^{\prime }(t)=2t,y^{\prime }(t)=3t^2$$

Therefore the slope of the tangent line, in terms of $t$:

$$\begin{array}{c}m=\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}\\ \\ \gg \gg \frac{3t^2}{2t}\gg \gg \frac{3}{2}t\end{array}$$

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(2) Set up equation:

$$\begin{array}{c}m=5\\ \\ \frac{3}{2}t=5\\ \\ \gg \gg t=\frac{10}{3}\end{array}$$

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(3) Find the point:

$$(x,y){|}_{t=10/3}\gg \gg (\frac{100}{9},\frac{1000}{27})$$

Perimeter of a circle

(1) The perimeter of the circle $(R\cos t,R\sin t)$ is easily found. We have $(x^{\prime },y^{\prime })=(-R\sin t,R\cos t)$, and therefore:

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

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(2) Integrate around the circle:

$$\begin{array}{c}\text{Perimeter}={\int }_0^{2\pi }ds\gg \gg {\int }_0^{2\pi }Rdt\\ \\ \gg \gg Rt{|}_0^{2\pi }=2\pi R\end{array}$$

Perimeter of an asteroid

Find the perimeter length of the ‘asteroid’ given parametrically by $(x,y)=(a{\cos }^3\theta ,a{\sin }^3\theta )$ for $a=2$.

Solution

(1) Notice: Throughout this problem we use the parameter $\theta$ instead of $t$. This does not mean we are using polar coordinates!

Compute the derivatives in $\theta$:

$$(x^{\prime },y^{\prime })=(3a{\cos }^2\theta \sin \theta ,3a{\sin }^2\theta \cos \theta )$$

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(2) Compute the infinitesimal arc element.

$$\begin{array}{c}{(x^{\prime })}^2+{(y^{\prime })}^2\gg \gg 9a^2{\cos }^4\theta {\sin }^2\theta +9a^2{\sin }^4\theta {\cos }^2\theta \\ \\ \gg \gg 9a^2{\sin }^2\theta {\cos }^2\theta ({\cos }^2\theta +{\sin }^2\theta )\\ \\ \gg \gg 9a^2{\sin }^2\theta {\cos }^2\theta \end{array}$$

Plug into the arc element, simplify:

$$\begin{array}{c}ds=\sqrt{{(x^{\prime })}^2+y^{\prime }{)}^2}d\theta \\ \\ \gg \gg \sqrt{9a^2{\sin }^2\theta {\cos }^2\theta }d\theta \\ \\ \gg \gg ds=3a\left|\sin \theta \cos \theta \right|d\theta \end{array}$$

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(3) Bounds of integration?

Easiest to use $\theta \in [0,\pi /2]$. This covers one edge of the asteroid. Then multiply by 4 for the final answer.

On the interval $\theta \in [0,\pi /2]$, the factor $3a\sin \theta \cos \theta$ is positive. So we can drop the absolute value and integrate directly.

Absolute values matter!

If we tried to integrate on the whole range $\theta \in [\mathrm{0,2}\pi ]$, then $3a\sin \theta \cos \theta$ really does change sign.

To perform integration properly with these absolute values, we’d need to convert to a piecewise function by adding appropriate minus signs.

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(4) Integrate the arc element:

$$\begin{array}{ccc}& {\int }_0^{\pi /2}ds\gg \gg {\int }_0^{\pi /2}3a\sin \theta \cos \theta d\theta & \\ & & \\ & \gg \gg 3a{\int }_{u=0}^{1}udu& \text{(}u=\sin \theta \text{)}\\ & & \\ & \gg \gg 3a{\frac{u^2}{2}|}_0^1\gg \gg \frac{3a}{2}& \end{array}$$

Finally, multiply by 4 to get the total perimeter: $L=6a$

Speed, distance, displacement

The parametric curve $(t,\frac{2}{3}{t}^{3/2})$ 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:

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

Now compute the speed:

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

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(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:

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

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

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

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(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 $t=0$:

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

For ending point, insert $t=8$:

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

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

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

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:

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

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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.

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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 $y=\sqrt{1-x^2}$.

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:

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