Question

Find a vector function, r(t), that represents the curve of intersection of the two surfaces. the cylinder x2 + y2 = 16 and the surface z = xy

Answer 1

Given the cylinder, choosing cylindrical coordinates wouldn't be a bad idea. We take

$\begin{cases}x=4\cos t\\y=4\sin t\end{case}$

so that the hyperbolic sheet $z=xy$ becomes

$z=(4\cos t)(4\sin t)=16\sin t\cos t=8\sin2t$

The cylinder is obtained by taking $0\le t\le2\pi$. So the vector function parameterizing the given intersection is

$\mathbf r(t)=\langle4\cos t,4\sin t,8\sin2t\rangle$

I've attached a plot of the surfaces (cylinder is orange, hyperbolic sheet is blue) and their intersection highlighted in red.

Answer 2

A vector function r(t) is represented as: $r(t) = x_ti + y_tj + z_tk$

The vector function r(t) that represents the intersection of the two surfaces is  $r(t) = (4\cos(t))i +(4\sin(t))i + (16\cos(t) \cdot \sin(t))i$

Given

$x^2 + y^2 = 16$

$z = xy$

Express 16 as $4^2$ in $x^2 + y^2 = 16$

$x^2 + y^2 = 4^2$

Multiply by 1

$x^2 + y^2 = 4^2 \times 1$

In trigonometry

$\cos^2(t) + \sin^2(t) = 1$

So, we substitute $\cos^2(t) + \sin^2(t)$ for 1

$x^2 + y^2 = 4^2 \times 1$

$x^2 + y^2 = 4^2 \times [\cos^2(t) + \sin^2(t)]$

Remove bracket

$x^2 + y^2 = 4^2 \times \cos^2(t) + 4^2 \times \sin^2(t)$

Apply law of indices

$x^2 + y^2 = [4 \times \cos(t)]^2 + [4 \times \sin(t)]^2$

$x^2 + y^2 = [4\cos(t)]^2 + [4\sin(t)]^2$

By comparison:

$x^2 =[4\cos(t)]^2$ and $y^2 = [4\sin(t)]^2$

Take positive square roots

$x =4\cos(t)$ and $y = 4\sin(t)$

Recall that:

$z = xy$

So, we have:

$z = 4\cos(t) \times 4\sin(t)$

$z = 16\cos(t) \cdot \sin(t)$

Substitute the values of x, y and z.

$r(t) = x_ti + y_tj + z_tk$

Hence, the vector function r(t) is:

$r(t) = (4\cos(t))i +(4\sin(t))i + (16\cos(t) \cdot \sin(t))i$

Read more about vector functions at:

https://brainly.com/question/4515610