View source for Calculus II 10.03 Parametric Equations and Calculus

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=10.3 Parametric Equations and Calculus=
* Find the slope for a tangent line to a curve given by parametric equations.
* Find the arc length for a curve given by parametric equations.
* Find the area for a rotated surface in parametric form.
==Slope and Tangent Lines==
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[[File:Calculus_II_10.03.01.PNG | frame | At time \(t\), the elevation angle is \(\theta\).<br />Figure 10.3.1]]<br clear=all/>
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Consider the projectile represented by the parametric equations
::\(x=24\sqrt{2}t \text{ and } y=-16t^{2}+ 24\sqrt{2}t\)
as shown in Figure 10.3.1. The initial angle is \(45^{\circ}\) with slope \(m=\tan 45^{\circ} = 1 \). Can the slope be found at any time \(t\)? Theorem 10.3.1 provides a formula for finding the slope as a function for \(t\).
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===Theorem 10.3.1 The Derivative for the Parametric Form===
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[[File:Calculus_II_10.03.02.PNG | frame |The slope for the secant line through the points \((f(t),g(t))\) and \((f(t+\Delta t),g(t+\Delta t))\) is \(\Delta y/ \Delta x\). Figure 10.3.2]]<br clear=all/>
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If a smooth curve \(C\) is given by the equations
::\(x=f(t) \text{ and } y=g(t)\)
then the slope for \(C\) at a point is
::$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt},\: \frac{dx}{dt} \ne 0.$$
'''Proof''' Consider \(\Delta t >0\) and let
::\(  \Delta x=f(t+\Delta t)-f(t) \text{ and } \Delta y=g(t+ \Delta t)-g(t) \)
as shown in Figure 10.3.2.<br />
Because \(\Delta x \to 0\) as \(\Delta t \to 0\), the limit can be written as
{| style="margin-left: 40px"
| $$ \frac{dy}{dx} $$ || $$= \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} $$
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| || $$= \lim_{\Delta x \to 0} \frac{g(t+ \Delta t)-g(t)}{f(t+\Delta t)-f(t)}  $$.
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Divide the numerator and denominator by \(\Delta t\) and differentiate for \(f\) and \(g\) to conclude.
{| style="margin-left: 40px"
| $$ \frac{dy}{dx} $$ || $$=\lim_{\Delta x \to 0} \frac{[g(t+ \Delta t)-g(t)]\Delta t}{[f(t+\Delta t)-f(t)]/\Delta t}  $$
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| || $$= \frac{ \lim_{\Delta x \to 0}\frac{g(t+ \Delta t)-g(t)}{\Delta t} }{ \lim_{\Delta x \to 0} \frac{f(t+\Delta t)-f(t)}{\Delta t}}  $$.
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| || $$= \frac{g{}^{\prime}(t)}{f{}^{\prime}(t)} = \frac{dy/dt}{dx/dt}.$$
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===Example 10.3.1 Differentiation and Parametric Form===
Find \(dy/dx\) for the curve given by \(x=\sin t\) and \(y=\cos t\).<br />
'''Solution'''
{| style="margin-left: 40px"
| $$ \frac{dy}{dx} $$ || $$= \frac{dy/dt}{dx/dt}$$
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| || $$= \frac{-\sin t}{\cos t} $$
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| || \(= - \tan t \)
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Because \(dy/dx\) is a function for \(t\), Theorem 10.3.1 can be used repeatedly to find ''higher-order'' derivatives. For example,
{| style="margin-left: 40px"
| $$ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left [ \frac{d}{dx} \right ] = \frac{  \frac{d}{dt} \left [ \frac{dy}{dx} \right ] }{dx/dt}$$ ||<span style="color:red">Second derivative</span>
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| $$ \frac{d^{3}y}{dx^{3}} = \frac{d}{dx} \left [ \frac{d^{2}}{dx^{2}} \right ] = \frac{  \frac{d}{dt} \left [ \frac{d^{2}y}{dx^{2}} \right ] }{dx/dt} $$ ||<span style="color:red">Third derivative</span>
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===Example 10.3.2 Finding Slope and Concavity===
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[[File:Calculus_II_10.03.03.PNG | frame | The graph is concave upward at \((2,3)\) when \(t=4\).<br />Figure 10.3.3]]<br clear=all/>
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For the curve given by
::$$ x=\sqrt{t} \text{ and } y= \frac{1}{4}(t^{2}-4), \: t \geqslant 0 $$
find the slope and concavity at the \((2,3)\).<br />
'''Solution''' Because
::$$ \frac{dy}{dx}= \frac{dy/dt}{dx/dt} = \frac{(1/2)t}{(1/2)t^{-1/2}}=t^{3/2} \:\:\:\: \color{red}{\text{Parametric form for the first derivative }}$$
the second derivative is
::$$ \frac{d^{2}y}{dx^{2}}= \frac{\frac{d}{dt}[dy/dx]}{dx/dt} = \frac{\frac{d}{dt}[t^{3/2}]}{dx/dt} =\frac{(3/2)t^{1/2}}{(1/2)t^{-1/2}}=3t \:\:\:\: \color{red}{\text{Parametric form for the second derivative }}$$
At \((2,3)\) \(t=4\) and the slope is
::$$ \frac{dy}{dx}= (4)^{3/2}=8.$$
When \(t=4\) the second derivative is
::$$ \frac{d^{2}y}{dx^{2}}= 3(4)=12 >0.$$
The graph is concave upward at \((2,3)\) as shown in Figure 10.3.3.
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===Example 10.3.3 A Curve with Two Tangent Lines at a Point===
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[[File:Calculus_II_10.03.04.PNG | frame |This prolate cycloid has two tangent
lines at the point \((0,2)\).<br />Figure 10.3.4]]<br clear=all/>
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Because the parametric equations \(x=f(t)\) and \(y=g(t)\) need not define \(y\) as a function for \(x\), it is possible for a plane curve to loop around and cross itself. At these intersecting points, the curve may have more than one tangent line.

The '''prolate cycloid'''[http://mathworld.wolfram.com/ProlateCycloid.html] given by
::\( x=2t-\pi \sin t \text{ and } y=2 - \pi \cos t \)
crosses itself at \((0,2)\), as shown in Figure 10.3.4. Find the equations for both tangent lines at this point.<br />
'''Solution''' Because \(x=0\) and \(y=2\) when \(t= \pm \pi /2\), and 
::$$  \frac{dy}{dx}= \frac{dy/dt}{dx/dt} = \frac{\pi \sin t}{2-\pi \cos t}$$
This reduces to, \(dy/dx= - \pi /2 \) when \(t= - \pi /2 \) and \(dy/dx= \pi /2 \) when \(t= \pi /2 \). Therefore the tangent lines are
::$$ y-2= - \left (\frac{\pi}{2} \right )x \:\:\:\: \color{red}{\text{Tangent line when } t=-\frac{\pi}{2}} $$
::$$ y-2= \left (\frac{\pi}{2} \right )x \:\:\:\: \color{red}{\text{Tangent line when } t=\frac{\pi}{2}}$$

The curve has a horizontal tangent at \((0,2-\pi)\) when \(t=0\).  A vertical tangent occurs at \((f(t_{0}),g(t_{0}))\) whenever \(dx/dt=0\) and \(dy/dt \ne0\), when \(t=t_{0}\). A horizontal tangent occurs at \((f(t_{0}),g(t_{0}))\) whenever \(dy/dt=0\) and \(dx/dt \ne 0\), when \(t=t_{0}\).
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==Arc Length==
How can the ''distance'' traveled by an object along parametric curve be determined? Recall from [[Calculus_II_07.04_Arc_Length_and_Surfaces_for_Revolution | Section 7.4]] that for a given curve \(C\) the formula for its arc length is given by \(y=h(x)\) over the interval \([x_{0},x_{1}]\) is
{| style="margin-left: 40px"
| $$S$$ || $$= \int_{x_{0}}^{x_{1}} \sqrt{ 1+ [h{}^{\prime}(x)]^{2} }\: dx $$
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|  || $$= \int_{x_{0}}^{x_{1}} \sqrt{ 1+ \left ( \frac{dy}{dx} \right)^{2} }\: dx $$
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|colspan="2"|If \(C\) is represented by the parametric equations \(x=f(t)\) and \(y=g(t)\),<br /> \(a \leqslant t \leqslant b\), and if \(dx/dt={f{}^{\prime}(t)} >0\), then the equation continues
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|  || $$= \int_{x_{0}}^{x_{1}} \sqrt{ 1+ \left ( \frac{dy/dt}{dx/dt} \right)^{2} }\: dx $$
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| || $$= \int_{a}^{b} \sqrt{ \frac{(dx/dt)^{2}+(dy/dt)^{2}}{(dx/dt)^{2}}  }\: \frac{dx}{dt}\: dt $$
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| || $$= \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt}\right)^{2}+\left( \frac{dy}{dt}\right)^{2}}\: dt $$
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| || $$= \int_{a}^{b} \sqrt{[ f{}^{\prime}(t)]^{2}+[ g{}^{\prime}(t)]^{2}}\: dt. $$
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===Theorem 10.3.2 Arc Length in Parametric Form===
If a smooth curve \(C\) is represented by the parametric equations \(x=f(t)\) and \(y=g(t)\) such that \(C\) doe snot intersect itself over the interval \(a \leqslant t \leqslant b\), except possibly at the endpoints, then the arc length for \(C\) over that interval is given by
::$$S = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt}\right)^{2}+\left( \frac{dy}{dt}\right)^{2}}\: dt = \int_{a}^{b} \sqrt{[ f{}^{\prime}(t)]^{2}+[ g{}^{\prime}(t)]^{2}}\: dt. $$
===Example 10.3.4 Finding the Arc Length for a Epicycloid===
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[[File:Calculus_II_10.03.05.PNG | frame |An epicycloid is traced by a point on the smaller circle as it rolls around the larger circle.<br />Figure 10.3.5]]<br clear=all/>
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An '''epicycloid'''[http://mathworld.wolfram.com/Epicycloid.html] is traced by a point on the smaller circle as it rolls around the larger circle.<br />
A circle, with radius 1, rolls around a circle, with radius 4, as shown in Figure 10.3.5. The epicycloid is given by
::\( x= 5 \cos t-\cos 5t  \text{ and } y=5 \sin t - \sin 5t. \)
Find the distance traveled by the point in one complete trip about the larger circle.<br />
'''Solution''' Before applying Theorem 10.3.2, note in Figure 10.3.5 that the curve has sharp points when \(t=0\) and \(t=\pi/2\). Between these two points, \(dx/dt\) and \(dy/dt\) are not simultaneously 0. The curve segment generated from \(t=0\) to \(t=\pi/2\) is smooth. To find the total distance traveled by the point,find the arc length for the segment in the first quadrant and multiply by 4.
{| style="margin-left: 40px;"
|style="vertical-align:middle;"| $$S$$ || $$= 4\int_{0}^{\pi/2} \sqrt{ \left( \frac{dx}{dt}\right)^{2}+\left( \frac{dy}{dt}\right)^{2}}\: dt \:\:\:\: \color{red}{\text{Parametric form for arc length }} $$
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| || $$= 4\int_{0}^{\pi/2} \sqrt{ \left( -5 \sin t + 5 \sin 5t \right)^{2}+\left( 5 \cos t-5 \cos 5t\right)^{2}}\: dt $$
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| || $$= 20\int_{0}^{\pi/2} \sqrt{ 2-2 \sin t \sin 5t - 2 \cos t \cos 5t}\: dt $$
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| || $$= 20\int_{0}^{\pi/2} \sqrt{ 2- \cos 4t}\: dt \:\:\:\: \color{red}{\text{Difference formula for cosine }}$$
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| || $$= 20\int_{0}^{\pi/2} \sqrt{ 4 \sin^{2} 2t}\: dt \:\:\:\: \color{red}{\text{Double-angle formula }}$$
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| || $$= 40\int_{0}^{\pi/2} \sin 2t\: dt $$
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| || $$= -20 \bigg[ \cos 2t \bigg]_{0}^{\pi/2} = 40 $$
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For the epicycloid shown in Figure 10.3.5, an arc length 40 long seems about right because the circle's circumference with radius 6 is
::\(2 \pi r = 12 \pi \approx 37.7\).
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==Area for a Rotated Surface==
The formula for a rotated surface area in rectangular form to develop a formula for a rotated surface area in parametric form.
===Theorem 10.3.3 Area for a Rotated Surface in Parametric Form===
If a smooth curve \(C\) given by \(x=f(t)\) and \(y=g(t)\) does not cross itself on an interval \(a \leqslant t \leqslant b\), then the  area \(S\) for the rotated surface formed by revolving \(C\) about the coordinate axis is given by the following.
::$$1. \:S= 2 \pi\int_{a}^{b} g(t) \sqrt{ \left( \frac{dx}{dt}\right)^{2}+\left( \frac{dy}{dt}\right)^{2}}\: dt  \:\:\:\: \color{red}{\text{Revolution about the }x\text{-axis: }g(t) \geqslant 0} $$
::$$2. \:S= 2 \pi\int_{a}^{b} f(t) \sqrt{ \left( \frac{dx}{dt}\right)^{2}+\left( \frac{dy}{dt}\right)^{2}}\: dt  \:\:\:\: \color{red}{\text{Revolution about the }y\text{-axis: }f(t) \geqslant 0} $$
The differential form for the arc length is
::$$ds= \sqrt{ \left( \frac{dx}{dt}\right)^{2}+\left( \frac{dy}{dt}\right)^{2}}\: dt$$
Then the formulas are written as 
::$$1. \:S= 2 \pi \int_{a}^{b} g(t)\: ds \:\:\:\: 2.\: S= 2 \pi \int_{a}^{b} f(t)\:ds $$
===Example 10.3.5 Finding the Area for a Rotated Surface===
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[[File:Calculus_II_10.03.06.PNG | frame |An epicycloid is traced by a point on the smaller circle as it rolls around the larger circle.<br />Figure 10.3.6]]<br clear=all/>
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Let \(C\) be the arc for the circle \(x^{2}+y^{2}=9\) between
::$$(3,0) \text{ and } \left ( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right ) $$
as shown in Figure 10.3.6. Find the area for the surface formed by rotating \(C\) about the \(x\)-axis.<br />
'''Solution''' The curve \(C\) can be represented parametrically by the equations
::\(x= 3 \cos t \text{ and } y = 3 \sin t, \: 0 \leqslant t \leqslant \pi /3. \)
The interval for \(t\) was determined by observing that \(t=0\) when \(x=3\) and \(t=\pi /3 \) when \(x=3/2\). On this interval, \(C\) is smooth and \(y\) is nonnegative, and Theorem 10.3.3 can be applied.
{| style="margin-left: 40px;"
|style="vertical-align:middle;"| $$S$$ || $$=2 \pi \int_{0}^{\pi/3} (3 \sin t) \sqrt{(-3 \sin t)^{2}+(3 \cos t)^{2}} \:dt $$||&nbsp;&nbsp;&nbsp;&nbsp;<span style="color:red">Theorem 10.3.3</span>
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| || $$=6 \pi \int_{0}^{\pi/3} \sin t \sqrt{9(-3 \sin t)^{2}+(3 \cos t)^{2}} \:dt $$|| 
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| || $$=6 \pi \int_{0}^{\pi/3} 3 \sin t \:dt $$||&nbsp;&nbsp;&nbsp;&nbsp;<span style="color:red">Trigonometric identity</span>
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| || $$=-18 \pi \bigg[ \cos t \bigg]_{0}^{\pi/3} $$|| 
|-
| || $$=-18 \pi \left( \frac{1}{2}-1 \right)= 9\pi $$|| 
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[[File:Square_X.jpg | right ]]<br clear=all />

=Internal Links=
<i>Parent Article:</i> [[Calculus II 10 Conics Parametric Equations and Polar Coordinates]]<br />