접평면과 선형근사

$\mathbf{r}(t)=\left< x(t)\,,\,y(t)\,,\,z(t)\right>$ intersects $z^3 + xyz -2 = 0$ when

$$z(t)^3+x(t)\,y(t)\,z(t)-2=0\iff \big(t^2\big)^3 + \big(t^3)(t)\big(t^2\big)-2=0 \iff 2t^6=2\iff t=1$$

since $t$ is required to be positive. The direction vector for the curve at $t=1$ is

$$\mathbf{r}'(1)=3\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+2\,\hat{\mathbf{k}}$$

A normal vector for the surface at $\mathbf{r}(1)=\left< 1,1,1\right>$ is

$${ \boldsymbol{\nabla} }(z^3+xyz)\big|_{(1,1,1)}=[yz\hat{\boldsymbol{\imath}}+xz\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}]_{(1,1,1)} =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+4\hat{\mathbf{k}}$$

The angle $\theta$ between the curve and the normal vector to the surface is determined by

$$\begin{aligned} \big|\left< 3,1,2\right>\big|\,\big|\left< 1,1,4\right>\big|\cos\theta =\left< 3,1,2\right> \cdot\left< 1,1,4\right> &\iff \sqrt{14}\sqrt{18}\cos\theta=12 \\ &\iff \sqrt{7\times 36}\cos\theta=12 \\ &\iff \cos\theta=\frac{2}{\sqrt{7}} \\ &\iff \theta=40.89^\circ \end{aligned}$$

The angle between the curve and the surface is $90-40.89=49.11^\circ$ (to two decimal places).