Answer:
[tex]s(t)=(9 cost,9 sint , 32805 cos^4 t) , t\in (0,2\pi)[/tex]
Step-by-step explanation:
We are given that two surfaces
[tex]x^2+y^2=81[/tex]
and [tex]z=5x^4[/tex]
We have to find the parametrize the intersection of the given surfaces using cost and sin t with positive coefficient.
We know that intersection curve S(t) is given by
S(t)=(x(t),y(t),z(t)),[tex]t\in(0,2\pi)[/tex]
[tex]x^2+y^2=(9)^2[/tex]
Compare with the equation of circle [tex](x-h)^2+(y-k)^2=r^2[/tex]
Where (h,k) is center of circle and r is the radius of the circle.
The center and radius of given circle is (0,0) and 9.
[tex]x=9cost,y=9 sint[/tex]
because cost takes along x - axis and sin t takes along y- axis.
When we substitute these values then we get
[tex]81cos^2t+81sin^2t=81(cos^2t+sin^2t)=81 [/tex] ([tex]sin^2t+cos^2t)=1[/tex])
Hence, [tex]x=9 cost, y= 9sint [/tex]
Substitute the value of x in second equation of surface
[tex]z=5(9cost)^4=32805 cos^4t[/tex]
Hence, the intersection curve s(t) is given by
[tex]s(t)=(9 cost,9 sint , 32805 cos^4 t) , t\in (0,2\pi)[/tex]
