Answer :
- The vector function r(t) is written as: [tex]r(t)=x_t i+y_t j+z_t k[/tex]
- The intersection of the two surfaces is represented by the vector function r(t) is [tex]r(t)=(4 \cos (t)) i+(4 \sin (t)) i+(16 \cos (t) \cdot \sin (t)) i[/tex].
Given:
[tex]\begin{aligned}&x^2+y^2=16 \\&z=x y\end{aligned}[/tex]
Express 16 as [tex]4^{2}[/tex]: [tex]x^2+y^2=16[/tex]
[tex]x^2+y^2=4^2\\x^2+y^2=4^2 \times 1[/tex]
Trignometry,
[tex]\cos ^2(t)+\sin ^2(t)=1[/tex]
Now, substitute [tex]\cos ^2(t)+\sin ^2(t)[/tex] for 1:
[tex]\begin{aligned}&x^2+y^2=4^2 \times 1 \\&x^2+y^2=4^2 \times\left[\cos ^2(t)+\sin ^2(t)\right]\end{aligned}\\x^2+y^2=4^2 \times \cos ^2(t)+4^2 \times \sin ^2(t)[/tex]
Law of indicates:
[tex]\begin{aligned}&x^2+y^2=[4 \times \cos (t)]^2+[4 \times \sin (t)]^2 \\&x^2+y^2=[4 \cos (t)]^2+[4 \sin (t)]^2\end{aligned}\\x^2=[4 \cos (t)]^2 \text { and } y^2=[4 \sin (t)]^2[/tex]
Taking positive square roots as follows:
[tex]x=4 \cos (t), y=4 \sin (t)[/tex]
Recall that, z = xy.
Now, we have:
[tex]\begin{aligned}&z=4 \cos (t) \times 4 \sin (t) \\&z=16 \cos (t) \cdot \sin (t)\end{aligned}[/tex]
Now, substitute the values:
[tex]r(t)=x_t i+y_t j+z_t k[/tex]
So, the vector r(t) is: [tex]r(t)=(4 \cos (t)) i+(4 \sin (t)) i+(16 \cos (t) \cdot \sin (t)) i[/tex]
Therefore, the vector function r(t) is written as: [tex]r(t)=x_t i+y_t j+z_t k[/tex]
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