So the surface $S$ is given to be $z=x\sin(x+y)$ and the point $(2,-2,0)$ is given. When solving for a tangent plane, I got the equation:
$z = 2(x-2) + 2(y+2)$
So the vector-function of a space curve is $r(t) = <a(t),b(t),c(t)>$ on the surface $S$. And it is supposed that $c(t) = a(t)\sin(a(t)+b(t))$ for all real numbers of $t$. $r(0)$ is said to equal $<2,-2,0>$.
How can I prove that $r'(0)$ is PARALLEL to the tangent plane equation found above?
Also, is the tangent line to $C$ at $r(0)$ on the tangent plane? Can this be inferred from some general fact about space curves, tangent lines, and tangent planes?
Start with the equation:
$$ c = a \sin(a+b) $$
Everything ($c$, $a$, and $b$) are functions of $t$, but I will just write $a$ instead of $a(t)$, etc...
Now take the $t$-derivative of both sides of the equation (using product and chain rules):
$$ c' = a' \sin(a+b) + a\cos(a+b) \cdot (a'+b') $$
Now we plug in $t=0$, remembering that $a(0) = 2$ and $b(0) = -2$:
$$ c'(0) = 2a'(0) + 2b'(0) $$
The derivative $r'(0)$ is the vector $(a'(0), \, b'(0), \, c'(0))$. By the above equation, $r'(0)$ satisfies the tangent plane equation: $$ z = 2(x-2) + 2(y+2) = 2x + 2y $$