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Use cos(t) and sin(t), with positive coefficients, to parametrize the intersection of the surfaces x^2+y^2=81 and z=5x^4.

User Cowbaymoo
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2 Answers

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x = 9cos(t)
y = 9sin)t)
z =5[9cos(t)]^4
User Satria Suria
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5 votes

Answer:


s(t)=(9 cost,9 sint , 32805 cos^4 t) , t\in (0,2\pi)

Explanation:

We are given that two surfaces


x^2+y^2=81

and
z=5x^4

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)),
t\in(0,2\pi)


x^2+y^2=(9)^2

Compare with the equation of circle
(x-h)^2+(y-k)^2=r^2

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.


x=9cost,y=9 sint

because cost takes along x - axis and sin t takes along y- axis.

When we substitute these values then we get


81cos^2t+81sin^2t=81(cos^2t+sin^2t)=81 (
sin^2t+cos^2t)=1)

Hence,
x=9 cost, y= 9sint

Substitute the value of x in second equation of surface


z=5(9cost)^4=32805 cos^4t

Hence, the intersection curve s(t) is given by


s(t)=(9 cost,9 sint , 32805 cos^4 t) , t\in (0,2\pi)

User Ghan
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8.1k points

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