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

$\left(R\cos(\beta)\right)^2 + \left(R\sin(\beta)\right)^2 = 8^2 + (-12)^2 \implies R^2 = 208 \implies R = 4\sqrt{13}$

and

$\dfrac{R\sin(\beta)}{R\cos(\beta)} = -\dfrac{12}8 \implies \tan(\beta) = -\dfrac32 \implies \beta = -\arctan\left(\dfrac32\right)$

Then the vector function for the curve is

$\vec r(t) = 4\cos(t)\,\vec\imath + 4\sin(t)\,\vec\jmath + 4\sqrt{13}\,\cos\left(t - \arctan\left(\dfrac32\right)\right)\,\vec k$

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