Question #133046
Use cos(t) and sin(t), with positive coefficients, to parametrize the intersection of the surfaces x^2+y^2=64 and z=3x^4.

r(t)=<_,_,_>
1
Expert's answer
2020-09-19T16:10:31-0400

Use the following parametrizationfor the curve r generatedby the intersection:r(t)=(x(t),y(t),z(t)),t[0,2π)Also, note thatr(t):RR3is a vectorvalued function of a real variable.To reach this result,we must consider the curvesthat these equationsdefine on certain planes.The equationx2+y2=64defines a circle of radius8centered on the z-axison the planez=p1,wherep1Ris any constant.The equationz=3x4definesa quartic curve onany planey=p2,wherep2Ris another constant.The surfaces are, therefore,those obtained by translating the circle along thezaxisand the quartic curve alongtheyaxis.To obtain a parametrizationfor the intersection curver(t),we must find equationsforx,yandzas functions oftthat both obeys equationsgiven in the problem.Consider the standardparametrization for a circleXof radiusr.x2+y2=r2X(t)=(rcos(t),rsin(t)),t[0,2π)Comparing this with the firstgiven equation, we getr=8,x=8cost&y=8sintNow, we already havean expression forx(t).For the second condition,we make:z=3x4=3(8cost)4=12288cos4t(y(t),z(t))=(8sint,12288cos4t)And thus, we havethe parametrization given asr(t)=(8cost,8sint,12288cos4t)\textsf{Use the following parametrization}\\ \textsf{for the curve r generated}\\ \textsf{by the intersection}:\\ r(t)=(x(t), y(t), z(t)), \\t \in [0,2π)\hspace{0.1cm} \textsf{Also, note that}\hspace{0.1cm} r(t): \mathbb{R} \rightarrow \mathbb{R}^3 \hspace{0.1cm} \textsf{is a vector} \\ \textsf{valued function of a real variable.}\\ \textsf{To reach this result,}\\\textsf{we must consider the curves}\\\textsf{that these equations}\\\textsf{define on certain planes.}\\ \textsf{The equation}\hspace{0.1cm} x^2 +y^2 = 64 \\\textsf{defines a circle of radius}\\ 8 \hspace{0.1cm} \textsf{centered on the \textit{z-axis}}\\ \textsf{on the plane} \hspace{0.1cm} z = p_1, where \\p_1 \in \mathbb{R} \hspace{0.1cm}\textsf{is any constant.}\\ \textsf{The equation} \hspace{0.1cm} z = 3x^4 \hspace{0.1cm} \textsf{defines}\\ \textsf{a quartic curve on}\\\textsf{any plane} \hspace{0.1cm} y = p_2, \hspace{0.1cm}\textsf{where}\\ p_2 \in \mathbb{R} \hspace{0.1cm}\textsf{is another constant.}\\ \textsf{The surfaces are, therefore}, \\\textsf{those obtained by translating}\\\textsf{ the circle along the}\hspace{0.1cm} z-axis \\\textsf{and the quartic curve along}\\ \textsf{the}\hspace{0.1cm} y-axis.\\ \textsf{To obtain a parametrization}\\ \textsf{for the intersection curve}\hspace{0.1cm} r(t),\\ \textsf{we must find equations}\\ \textsf{for} \hspace{0.1cm} x, y\hspace{0.1cm} \textsf{and}\hspace{0.1cm} z\hspace{0.1cm}\textsf{as functions of} \hspace{0.1cm} t\\ \textsf{that both obeys equations}\\\textsf{given in the problem.}\\\textsf{Consider the standard}\\ \textsf{parametrization for a circle}\\ X\hspace{0.1cm}\textsf{of radius}\hspace{0.1cm} r.\\ x^2 + y^2= r^2\\\Rightarrow X(t) = (rcos(t),rsin(t)), t \in [0,2π)\\ \textsf{Comparing this with the first}\\\textsf{given equation, we get}\\ r = 8, x = 8\cos{t} \hspace{0.1cm}\&\hspace{0.1cm} y = 8\sin{t}\\\textsf{Now, we already have}\\\textsf{an expression for}\hspace{0.1cm} x(t).\\\textsf{For the second condition},\\ \textsf{we make:}\hspace{0.1cm} z =3x^4 = 3(8\cos{t})^4 = 12288\cos^4{t}\\\therefore (y(t), z(t)) = (8\sin{t}, 12288\cos^4{t}) \\\textsf{And thus, we have}\\ \textsf{the parametrization given as} \\ r(t) = (8\cos{t}, 8\sin{t}, 12288\cos^4{t})


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS