Week 6 Activity

Section 6.3 Week 6 Activity

Subsection 6.3.1 Parametric Curves - Tangents and Lengths

Activity 6.3.1.

This parametric curve is two revolutions of the logarithmic spiral.

\begin{align*} \gamma(t) = (e^t \cos t, e^t \sin t) \amp \amp t \in [0, 4\pi] \end{align*}

Calculate \(\frac{dy}{dx}\) and identify some points with vertical tangents. Then calculate the length of the curve.

Solution.

Figure 6.3.1.

First I calcualte the derivatives of the two components. Then I divide by \(y\) derivative by the \(x\) derivative to get an expression for the slope.

\begin{align*} \frac{dx}{dt} \amp = e^t \cos t - e^t \sin t \\ \frac{dy}{dt} \amp = e^t \sin t + e^t \cos t \\ \frac{dy}{dx} \amp = \frac{ e^t \sin t + e^t \cos t }{ e^t \cos t - e^t \sin t} = \frac{\sin t + \cos t}{\cos t - \sin t} \end{align*}

There are vertical tangents whenever the denominator of \(\frac{dy}{dx}\) approaches zero but the numerator does not. This happens when \(\cos t = \sin t\text{,}\) which happens at \(\frac{\pi}{4}\) plus any multiple of \(\pi\text{.}\)

I put the derviatives of the components into the length formula and then calculate the resulting integral. After cancelling of terms which easily cancel, I use the trig identity \(\sin^2 t + \cos^2 t = 1\) to simplify the integral, leaving a reasonable exponential integral.

\begin{align*} L \amp = \int_0^{4\pi} e^t \sqrt{ \cos^2 t - 2 \cos t \sin t + \sin^2 t + \sin^2 t + 2\sin t \cos t + \cos^2 t} dt \\ \amp = \int_0^{4\pi} e^t \sqrt{ 2(\cos^2 t + \sin^2 t)} dt\\ \amp = \int_0^{4\pi} \sqrt{2} e^t dt = \sqrt{2}e^t \Big|_0^{4\pi} \\ \amp = \sqrt{e^{4\pi}} - \sqrt{2} = \sqrt{2} \left( e^{4\pi} - 1 \right) \end{align*}

Activity 6.3.2.

This parametric curve is the Archimedian spiral.

\begin{align*} \gamma(t) = (t \cos t, t \sin t) \amp \amp t \in [0, 4\pi] \end{align*}

Calculate \(\frac{dy}{dx}\) and identify some points with vertical tangents. Then calculate the length of the curve.

In the arclength calculation, I will get to an integral which needs a trig substitution. Instead of doing that trig substitution, I can use the form found on the table of integrals.

\begin{equation*} \int \sqrt{a^2 + x^2} dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \ln \left( x + \sqrt{a^2 + x^2} \right) + c \end{equation*}

Solution.

Figure 6.3.2.

First I calcualte the derivatives of the two components. Then I divide by \(y\) derivative by the \(x\) derivative to get an expression for the slope.

\begin{align*} \frac{dx}{dt} \amp = -t \sin t + \cos t \\ \frac{dy}{dt} \amp = t \cos t + \sin t \\ \frac{dy}{dx} \amp = \frac{t \cos t + \sin t}{-t \sin t + \cos t} \end{align*}

There are vertical tangents whenever the denominator of \(\frac{dy}{dx}\) approaches zero but the numerator does not. This happens twice every revolution, when the denominators vanishes. precise value of the parameter \(t\) at those points is challenging.

I put the derviatives of the components into the length formula and then calculate the resulting integral. Again, some term will cancel and I can use trig identities to simplify some of the remianing term. When I get to the square root integral, I just used the antiderivative given in the question instead of doing the whole process of the trig substitution.

\begin{align*} L \amp = \int_0^{4\pi} \sqrt{ t^2 \sin^2 t - 2t \cos t \sin t + \cos^2 t + t^2 \cos^2 t + 2t \cos t \sin t + \sin^2 t} dt \\ \amp = \int_0^{4\pi} \sqrt{t^2 (\cos^2 t + \sin^2 t) + (\cos^2 t + \sin^2 t)} dt \\ \amp = \int_0^{4\pi} \sqrt{t^2 + 1} \\ \amp = \frac{t}{2} \sqrt{1 + t^2} + \frac{1}{2} \ln \left| t + \sqrt{1 + t^2} \right| \Bigg|_0^{4\pi} \\ \amp = 2\pi \sqrt{1 + 16\pi^2} + \frac{1}{2} \ln \left( 4\pi + \sqrt{1 + 16\pi^2} \right) \end{align*}

Activity 6.3.3.

This parametric curve is a parabola.

\begin{align*} \gamma(t) = (t, at^2) \amp \amp t \in [0,5] \end{align*}

Calculate \(\frac{dy}{dx}\) and identify some points with vertical tangents. Then calculate the length of the curve. You can use the same antiderivative as the previous question.

Solution.

Figure 6.3.3.

First I calcualte the derivatives of the two components. Then I divide by \(y\) derivative by the \(x\) derivative to get an expression for the slope.

\begin{align*} \frac{dx}{dt} \amp = 1 \\ \frac{dy}{dt} \amp = 2at \\ \frac{dy}{dx} \amp = 2at \end{align*}

There are vertical tangents whenever the denominator of \(\frac{dy}{dx}\) approaches zero but the numerator does not. However, this denominator never vanishes, so there are no vertical tangents.

I put the derviatives of the components into the length formula and then calculate the resulting integral. Again, I use the a known antiderivative from the tables for the square root integral instead of doing the whole trig substitution. I need to do a little bit of algebra, factoring \(4a^2\) out of the square root, to make the square root integral fit the expression on the table.

\begin{align*} L \amp = \int_0^5 \sqrt{1 + 4a^2 t^2} dt \\ \amp = \int_0^5 2a \sqrt{\frac{1}{4a^2} + t^2} dt \\ \amp = 2a \left. \frac{1}{2} t \sqrt{\frac{1}{4a^2} + t^2} + \frac{1}{8a^2} \ln \left( \sqrt{\frac{1}{4a^2} + t^2} + t \right) \right|_0^5 \\ \amp = 5a \sqrt{\frac{1}{4a^2} + 25} + \frac{1}{8a^2} \ln \left( \sqrt{ \frac{1}{4a^2} + 25} + 5 \right) - \frac{1}{8a^2} \ln \left( \sqrt{\frac{1}{4a^2}} \right) \\ \amp = \frac{5}{2} \sqrt{1 + 100a^2} + \frac{1}{8a^2} \ln \left( \frac{1}{2a} \sqrt{1 + 100a^2} + 5 \right) + \frac{1}{8a^2} \ln (2a) \end{align*}

Activity 6.3.4.

This parametric curve is called the astroid.

\begin{align*} \gamma(t) = (\cos^3 t, \sin^3 t) \amp \amp t \in [0, 2\pi] \end{align*}

Calculate \(\frac{dy}{dx}\) and identify some points with vertical tangents. Then calculate the length of the curve.

Solution.

Figure 6.3.4.

First I calcualte the derivatives of the two components. Then I divide by \(y\) derivative by the \(x\) derivative to get an expression for the slope.

\begin{align*} \frac{dx}{dt} \amp = 3 \cos^2 t \sin t \\ \frac{dy}{dt} \amp = - 3 \sin^2 t \cos t \\ \frac{dy}{dx} \amp = \frac{-\sin t}{\cos t} = - \tan t \end{align*}

There are vertical tangents whenever the denominator of \(\frac{dy}{dx}\) approaches zero but the numerator does not. This happens when cosine vanishes, which is the two parameter value \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\text{.}\) (These are the known asymptote values for tangent.)

I put the derviatives of the components into the length formula and then calculate the resulting integral. I use the trig identity \(\sin^2 t + \cos^2 t = 1\) to simplify the integrand.

\begin{align*} L \amp = \int_0^{2\pi} 3 \sqrt{\cos^4 t \sin^2 t + \sin^4 t \cos^2 t} dt \\ \amp = 3 \int_0^{2\pi} \sqrt{ \sin^2 t \cos^2 t (\cos^2 t + \sin^2 t)} dt \\ \amp = \int_0^{2\pi} 3 |\sin t \cos t| dt \\ \amp = \int_0^{2\pi} \frac{3}{2} |\sin 2t| dt \end{align*}

This is a bit tricky due to the absolue value. However, I can use a symmetry argument. The expression \(\sin 2t\) makes two periods over this domain. Each period has an area above the axis and an area below, each for half a period. The absolute value puts these areas above the axis. Over two periods, there are four such areas, all with the same shape. So I can integrate over the first quarter of this domain and multiply by 4. That lets me drop the absolute value.

\begin{align*} \amp = 4 \int_0^{\frac{\pi}{2}} \frac{3}{2} \sin 2t dt = 6 \int_0^{\frac{\pi}{2}} \sin 2t dt = 6 \left. \frac{-\cos 2t}{2} \right|_0^{\frac{\pi}{2}} = 6 \left( \frac{1}{2} + \frac{1}{2} \right) = 6 \end{align*}

Activity 6.3.5.

This parametric curve is the cardiod.

\begin{align*} \gamma(t) = (2 \cos t - \cos 2t,2 \sin t - \sin 2t) \amp \amp t \in [0,2\pi] \end{align*}

Calculate \(\frac{dy}{dx}\) and identify some points with vertical tangents. Then calculate the length of the curve.

Solution.

Figure 6.3.5.

First I calcualte the derivatives of the two components. Then I divide by \(y\) derivative by the \(x\) derivative to get an expression for the slope.

\begin{align*} \frac{dx}{dt} \amp = -2 \sin t + 2 \sin 2t \\ \frac{dy}{dt} \amp = 2 \cos t - 2 \cos 2t \\ \frac{dy}{dx} \amp = \frac{\cos t - \cos2t}{-\sin t + \sin 2t} \end{align*}

There are vertical tangents whenever the denominator of \(\frac{dy}{dx}\) approaches zero but the numerator does not. I can use a double angle identity to change the denominator to \((-\sin t + 2 \sin t \cos t)\) and then factor out of sine to get \(\sin t(-1 + 2\cos t)\text{.}\) The zeros of the demonimator are the zeros of the sine function (all multiples of \(\pi\)) and \(\cos t = \frac{1}{2}\) ( in this range \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\)). I have to reject \(t = 0\text{,}\) since that is also a zero of the numerator and the limit of the slope is not infinity, but there is a vertical tangent at \(t = \pi\text{.}\) Similarly, since they are not zero of the numerator, there are vertical tangents at \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\) as well.

I put the derviatives of the components into the length formula and then calculate the resulting integral. The integral simplifies using trig identities. After taking the square root, I can use the same argument as the previous question to get rid of the absolute value by integration over a quarter of the domain and multiplying by 4.

\begin{align*} \amp L = \\ \amp \int_0^{2\pi} \sqrt{ 4 \sin ^2 t - 8 \sin t \sin 2t + 4 \sin^2 2t + 4 \cos^2 2t - 8 \cos t \cos2t + 4\cos^2 t} dt \\ \amp = \int_0^{2\pi} \sqrt{ 4 + 4 - 8 \sin t (2 \sin t \cos t) - 8 \cos t (\cos^2 t - \sin^2 t)} dt\\ \amp = \int_0^{2\pi} \sqrt{8 - 16 \sin^2 t \cos t - 8 \cos^3 t + 8 \sin^2 t \cos t} dt \\ \amp = \int_0^{2\pi} \sqrt{8 - 8 \cos t (\sin^2 t + \cos^2 t)} dt \\ \amp = \sqrt{8} \int_0^{2\pi} \sqrt{1-\cos t} dt = \sqrt{8} \int_0^{2\pi} \sqrt{ 2 \sin^2 2t} dt \\ \amp 4 \int_0^{2\pi} |\sin 2t| dt = 4 \cdot 4 \int_0^{\frac{\pi}{2}} \sin 2t dt \\ \amp = 16 \left. \frac{-\cos 2t}{2} \right|_0^{\frac{\pi}{2}} = 8 \left( \cos 0 - \cos \frac{\pi}{2} \right) = 8 \end{align*}

Subsection 6.3.2 Parametric Curves - Parametrization by Arclenth

Activity 6.3.6.

This parametric curve is the quadratic spiral.

\begin{align*} \gamma(t) = (t^2 \cos t, t^2 \sin t) \amp \amp t \in [0, \infty) \end{align*}

Parametrize the curve by arclength.

Solution.

Figure 6.3.6.

I follow the three part algorithm presented in the notes and the videos. First I setup the integral to calculate the arclength suntion. I need the two derivatives of the components, which I then put into the square root term of the integral. The integral gets pretty messy.

\begin{align*} \frac{dx}{dt} \amp = 2 t \cos t - t^2 \sin t \\ \frac{dy}{dt} \amp = 2 t \sin t + t^2 \cos t \end{align*}

I’ll calculate the sum of the square of the derivative by itself and then insert it into the arclength integral. After expanding the squares, several terms cancel. Then, after factoring and grouping, trig identities help me make more simplifications.

\begin{align*} \left( \frac{dx}{dt} \right)^2 \amp = 4t^2 \cos^2 t - 4t^3 \cos t \sin t + t^4 \sin^2 t \\ \left( \frac{dy}{dt} \right)^2 \amp = 4t^2 \sin^2 t + 4t^3 \cos t \sin t + t^4 \cos t du \\ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dx} \right)^2 \amp = 4t^2 (\cos^2 t + \sin^2 t) + t^4(\cos^2 t + \sin^2 t)\\ \amp = 4t^2 + t^4 = t^2(4+t^2) \\ s(t) \amp = \int_0^t u \sqrt{4+u^2} du \end{align*}

Since there is a \(u\) outside the square root, this looks likes a reasonable substitution using \(v = 4 + u^2\) with \(d = 2u du\text{.}\) I also change the bounds so that I don’t have to do the reverse substitution.

\begin{align*} s \amp = \int_4^{4+t^2} \frac{\sqrt{v}}{2} dv = \frac{v^{\frac{3}{2}}}{3} \Bigg|_4^{4+t^2} = \frac{(4+t^2)^\frac{3}{2}}{3} - \frac{8}{3} \end{align*}

Then I invert the arclength function.

\begin{align*} s + \frac{8}{3} \amp = \frac{(4+t^2)^\frac{3}{2}}{3}\\ 3s + 8 \amp = (1+t^2)^\frac{3}{2} \\ \left( 3s + 8 \right)^\frac{2}{3} \amp = 4+t^2 \\ \left( 3s + 8 \right)^\frac{2}{3} - 4 \amp = t^2 \\ \sqrt{\left( 3s + 8 \right)^\frac{2}{3} -4} \amp = t \\ t(s) \amp = \sqrt{\left( 3s + 8 \right)^\frac{2}{3} -4} \end{align*}

Finally, I replace the original parameter \(t\) with \(t(s)\) from the inversion calculation.

\begin{align*} \gamma(s) \amp = \left( \left( \left( 3s + 8 \right)^\frac{2}{3} -4 \right) \cos \left( \sqrt{\left( 3s + 8 \right)^\frac{2}{3} -4} \right), \right.\\ \amp \left. \left( \left( 3s + 8 \right)^\frac{2}{3} -4 \right) \sin \left( \sqrt{\left( 3s + 8 \right)^\frac{2}{3} -4} \right) \right) \end{align*}

Activity 6.3.7.

This parametric curve is an alternate parametrization of the circle.

\begin{align*} \gamma(t) = \left( \frac{\cos t - \sin t}{\sqrt{2}}, \frac{\cos t + \sin t}{\sqrt{2}} \right) \amp \amp t \in \left[ \frac{-\pi}{4}, \frac{7\pi}{4} \right] \end{align*}

Parametrize the curve by arclength.

Solution.

I follow the three part algorithm presented in the notes and the videos. First I setup the integral to calculate the arclength suntion. I start with the derivatives and then put them into the square root term in the arclength integral.

\begin{align*} \frac{dx}{dt} \amp = \frac{1}{\sqrt{2}} \left( - \sin t - \cos t \right) \\ \frac{dy}{dx} \amp = \frac{1}{\sqrt{2}} \left( - \sin t + \cos t \right) \\ \left( \frac{dx}{dt} \right)^2 \amp = \frac{1}{2} \left( \sin^2 t + 2\cos t \sin t + \cos^2 t \right) \\ \left( \frac{dy}{dx} \right)^2 \amp = \frac{1}{2} \left( \sin^2 t - 2 \cos t \sin t + \cos^2 t \right) \\ \left( \frac{dx}{dy} \right)^2 + \left( \frac{dy}{dx} \right)^2 \amp = \frac{1}{2} \left( 2\sin^2 t + 2\cos^2 t \right) = \frac{2}{2} = 1 \end{align*}

When calculating the sums of the square roots, I get some nice cancellations. What remains can be reduced to a constant with trig identities. That leaves a reasonably easy arclength integral.

\begin{align*} s(t) \amp = \int_{\frac{-\pi}{4}}^t du = u \Bigg|_{\frac{-\pi}{4}}^t = t + \frac{\pi}{4} \\ s(t) \amp = t + \frac{\pi }{4} \end{align*}

Then I invert the arclength function.

\begin{align*} s - \frac{\pi }{4} \amp = t \implies t = s + \frac{\pi}{4} \end{align*}

Finally, I replace the original parameter \(t\) with \(t(s)\) from the inversion calculation. I could directly replace \(t\text{,}\) but if I make use of the some identities for sums and differences, I get a more succinet form of the reparametrized curve.

\begin{align*} \gamma(t) \amp = \left( \frac{\cos t - \sin t}{\sqrt{2}}, \frac{\cos t + \sin t}{\sqrt{2}} \right) = \left( \frac{1}{\sqrt{2}} \sqrt{2} \sin \left( \frac{\pi}{4} - t \right), \frac{1}{\sqrt{2}} \sqrt{2} \sin \left( \frac{\pi}{4} + t \right) \right) \end{align*}

Now I replace \(t\) with \(s + \frac{\pi}{4}\text{.}\)

\begin{align*} \gamma (s) \amp = \left( \sin \left( \frac{\pi}{4} - t \right), \sin \left( \frac{\pi}{4} + t \right) \right) \\ \amp = \left( \sin \left( \frac{\pi}{4} - s + \frac{\pi}{4} \right), \sin \left( \frac{\pi}{4} + s - \frac{\pi}{4} \right) \right)\\ \amp = \left( \sin \left( -s + \frac{\pi}{2} \right), \sin (s) \right) = (\cos (-s), \sin (s)) = (\cos s, \sin s) \end{align*}

At the end of all this work, I have just recovered the most basic parametrization of the circle. However, this makes sense, since each curve has only one parametrization by arclength. Starting with any parametrization of the circle, the algorithm should produce this same answer as the parametrization by arclength.

Activity 6.3.8.

This parametric curve is an alternate parametrization of the half-circle.

\begin{align*} \gamma(t) = \left( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right) \amp \amp t \in [0, \infty) \end{align*}

Parametrize the curve by arclength.

Solution.

I follow the three part algorithm presented in the notes and the videos. First I setup the integral to calculate the arclength suntion. I calculate the two derivatives and them put them into the square root term in the arclength integral.

\begin{align*} \frac{dx}{dt} \amp = \frac{(1+t^2)(-2t) - 2t(1-t^2)}{(1+t^2)^2} = \frac{-4t}{(1+t^2)^2} .\\ \frac{dy}{dt} \amp = \frac{(1+t^2)(2) - 2t(2t)}{(1+t^2)^2} = \frac{2-2t^2}{(1+t^2)^2} \\ s(t) \amp = \int_0^t \frac{1}{(1+u^2)^2} \sqrt{16u^2 + 4 - 8u^2 + 4u^4} du \\ \amp = \int_0^t \frac{1}{(1+u^2)^2} \sqrt{4 + 8 u^2 + 4u^4} du \\ \amp = \int_0^t \frac{1}{(1+u^2)^2} 2 \sqrt{(1+u^2)^2} du = \int_0^t \frac{2}{1+u^2} du \\ \amp = 2 \arctan u \Big|_0^t = 2 \arctan t \end{align*}

Then I invert the arclength function.

\begin{align*} t \amp = \tan \frac{s}{2} \end{align*}

Finally, I replace the original parameter \(t\) with \(t(s)\) from the inversion calculation.

\begin{align*} \gamma(s) \amp = \left( \frac{1-\tan^2 \left( \frac{s}{2} \right) }{1 + \tan^2 \left( \frac{s}{2} \right)}, \frac{2 \tan \left( \frac{s}{2} \right)}{1 + \tan^2 \left( \frac{s}{2} \right)} \right) = \left( \frac{1-\tan^2 \left( \frac{s}{2} \right) }{\sec^2 \left( \frac{s}{2} \right)}, \frac{2 \tan \left( \frac{s}{2} \right)}{\sec^2 \left( \frac{s}{2} \right)} \right) \\ \amp = \left( \cos^2 \left( \frac{s}{2} \right) - \sin^2 \left( \frac{s}{2} \right), 2 \sin \left( \frac{s}{2} \right) \cos \left( \frac{s}{2} \right) \right) = ( \cos s, \sin s) \end{align*}

After this use of the double-angle identities, I do recover the simple parametriztaion of the circle.

Activity 6.3.9.

This parametric curve is the quadrifolium.

\begin{align*} \gamma(t) = (\cos t \sin 2t, \sin t \sin 2t) \amp \amp t \in [0, 2\pi] \end{align*}

Parametrize the curve by arclength. (Go as far as you can with the process for this question; the integral that results is not actually solvable by elementary functions.)

Solution.

Figure 6.3.7.

I follow the three part algorithm presented in the notes and the videos. First I setup the integral to calculate the arclength suntion. I calculate the derivatives of the coordinates and put those into the square root in the arclength integral. Some trig identities help to simplify the expressions.

\begin{align*} \frac{dx}{dt} \amp = - \sin t \sin 2t + 2 \cos t \cos 2t \\ \frac{dy}{dt} \amp = \cos t \sin 2t + 2 \sin t \cos 2t \\ \left( \frac{dx}{dt} \right)^2 \amp = \sin^2 t \sin^2 2t - 4 \sin t \cos t \sin 2t \cos 2t + 4 \cos^2 t \cos^2 2t \\ \left( \frac{dy}{dt} \right)^2 \amp = \cos^2 t \sin^2 2t + + 4 \cos t \sin t \cos 2t \sin 2t + 4 \sin^2 t \cos^2 2t \end{align*}

When I add the squares of the two derivatives, the middle terms cancel. Let me work with the remaining four terms to simplify with trig identities. I notice that I can factor out the terms with \((2t)\) inside the trig functions and use the square identity for sine a cosine.

\begin{align*} \amp \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 }\\ \amp = \sqrt{ \sin^2 t \sin^2 2t + 4 \cos^2 t \cos^2 2t + \cos t \sin 2t + 4 \sin^2t \cos^2 2t}\\ \amp = \sqrt{\sin^2 2t(\sin^2 t + \cos^2 t) + 4 \cos^2 2t (\sin^2 t + \cos^2 t)} \\ \amp = \sqrt{\sin^2 2t + \cos^2 2t + 3 \cos^2 2t}\\ \amp = \sqrt{1 + 3 \cos^2 2t}\\ s(t) \amp = \int_0^t ds = \int_0^t \sqrt{1 + 3 \cos^2 2u} du \end{align*}

This is an elliptic integral, of the same style as the elliptic integral that results from the calculation of the perimeter of an ellipse. Such integrals do not have elementary anti-derivatives, so short of inventing a new function, I am not able to proceed. Theoretically, the inverse of the function does exist and I could invent a name for it to finish the process. But that name doesn’t do much without an investigation of the properties of the unknown function.

Subsection 6.3.3 Conceptual Review Questions

How does a parametric curve differ from a locus?
How can the same shape have multiple parametrizations?
What does it mean to reparametrize a curve?
What is the parametrization by arclength and why is it important?

Prev Top Next