Parametric curve and differentials

Question

The parametric curve C is within the intersection of the xy plane and the surface $x^2+y^2+13z^2+10xy+3xz+2yz+4z=12$. Suppose your parametric curve is $\mathbf{C}=(x=t,y=f(t))$ where f is a continuous, differentiable function.

Evaluate $f(1)$ and explain why the following equality always holds: $10tf'(t)+10f(t)+2t+2f(t)f'(t)=0$

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I'm not sure how to start this problem. Is C just the intersection between the surface and the plane? How do I link this between $f$?

(Any extra examples with the related topic would be appreciated)

Answer 1

Yes, the curve $\mathbf{C}$ is just the intersection between the surface and the $xy$-plane, that is $z=0$.

Hence, if $x=t\in\mathbb{R}$ and $y=f(t)$ we obtain $$t^2+f^2(t)+13\cdot 0+10tf(t)+3t\cdot 0+2f(t)\cdot 0+4\cdot 0=12$$ that is $$t^2+f^2(t)+10tf(t)=12.$$ What do we obtain by taking the derivative with respect to $t$ of the two sides of the above equation?

Note also $f(1)$ can be obtained by solving the equation $1+f^2(1)+10f(1)=12$.

Answer 2

If $(x,y,z)$ is a point in the intersection of the xy plane and the surface , then $z=0$. Hence

$x^2+y^2+10xy=12$.

If $x(t)=t$ and $y(t)=f(t)$ is a parametrization of $C$, then

$t^2+f(t)^2+10tf(t)=12$.

Can you take it from here ?