Question

Find a function r(t) that describes the curve where the surfaces intersect. x^2 y^2=16 z=2x 3y

Answer 1

I'm assuming the surfaces are the cylinder
`x^2+y^2=16` and the plane
`z=2x+3y`.

Let
`x=4\cos(t)` and
`y=4\sin(t)`. Then

`z=2x+3y=8\cos(t)+12\sin(t)`

We can simplify this using the identity

`R \cos(\alpha + \beta) = R \cos(\alpha)\cos(\beta) - R \sin(\alpha) \sin(\beta)`

Let
`\alpha=t`, then

`R \cos(t + \beta) = R \cos(t)\cos(\beta) - R \sin(t) \sin(\beta) = 8\cos(t)+12\sin(t)`

`\implies \begin{cases}R\cos(\beta) = 8 \\ R\sin(\beta) = -12\end{cases}`

We have

`\left(R\cos(\beta)\right)^2 + \left(R\sin(\beta)\right)^2 = 8^2 + (-12)^2 \implies R^2 = 208 \implies R = 4√(13)`

and

`(R\sin(\beta))/(R\cos(\beta)) = -\frac{12}8 \implies \tan(\beta) = -\frac32 \implies \beta = -\arctan\left(\frac32\right)`

Then the vector function for the curve is

`\vec r(t) = 4\cos(t)\,\vec\imath + 4\sin(t)\,\vec\jmath + 4√(13)\,\cos\left(t - \arctan\left(\frac32\right)\right)\,\vec k`

with
`0\le t\le2\pi`.