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):R→R3is 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,wherep1∈Ris any constant.The equationz=3x4definesa quartic curve onany planey=p2,wherep2∈Ris another constant.The surfaces are, therefore,those obtained by translating the circle along thez−axisand the quartic curve alongthey−axis.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=r2⇒X(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)
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