Problem: Given the surface $\overrightarrow{r}_{uv}=(u \cos v, u \sin v, u^2)$. Find the angle of two curves $v=u+1$, $v=3-u$.
Approach Here is what I've done so far: $$ \begin{gathered} \overrightarrow{r_{u}^{\prime}}=(\cos v, \sin v, 2 u) \\ \overrightarrow{r_{v}^{\prime}}=(-u \sin v, u \cos v, 0) \end{gathered} $$ Parametrized two curves $\rho_{1}: v=u+1$ và $\rho_{2}: v=3-u$ $$ \overrightarrow{\rho_{1}}:\left\{\begin{array}{c} u_{1}(t)=t \\ v_{1}(t)=t+1 \end{array} \text { and } \overrightarrow{\rho_{2}}:\left\{\begin{array}{c} u_{2}(s)=s \\ v_{2}(s)=3-s \end{array} .\right.\right. $$ The derivatives are $$ \begin{aligned} &\left\{\begin{array} { l } { u _ { 1 } ^ { \prime } ( t ) = 1 } \\ { v _ { 1 } ^ { \prime } ( t ) = 1 } \end{array} \text { and } \left\{\begin{array}{c} u_{2}^{\prime}(s)=1 \\ v_{2}^{\prime}(s)=-1 \end{array} .\right.\right. \end{aligned} $$ The intersection of two curves on $(S)$ is $\rho_{1}(1)=\rho_{2}(1)$. Hence $$ \begin{aligned} &\overrightarrow{\rho_{1}^{\prime}}(1)=u_{1}^{\prime}(1) \cdot \overrightarrow{r_{u}^{\prime}}+v_{1}^{\prime}(1) \cdot \overrightarrow{r_{v}^{\prime}}=(\cos v-u \sin v, \sin v+u \cos v, 2 u) \\ &\overrightarrow{\rho_{2}^{\prime}}(1)=u_{2}^{\prime}(1) \cdot \overrightarrow{r_{u}^{\prime}}+v_{2}^{\prime}(1) \cdot \overrightarrow{r_{v}^{\prime}}=(\cos v+u \sin v, \sin v-u \cos v, 2 u). \end{aligned} $$ This implies $$ \left\langle \overrightarrow{\rho_{1}^{\prime}}(1), \overrightarrow{\rho_{2}^{\prime}}(1)\right\rangle=\cos ^{2} v-u^{2} \sin ^{2} v+\sin ^{2} v-u^{2} \cos ^{2} v+4 u^{2}=3 u^{2}+1 $$ and $$ \begin{align} \left\|\overrightarrow{\rho_{1}^{\prime}}(1)\right\|&=\sqrt{\cos ^{2} v+u^{2} \sin ^{2} v-2 u \cos v \sin v+\sin ^{2} v+u^{2} \cos ^{2} v+2 u \sin v \cos v+4 u^{2}} \\ &=\sqrt{5 u^{2}+1} \end{align} $$ $$ \begin{align} \left\|\overrightarrow{\rho_{2}^{\prime}}(1)\right\|&=\sqrt{\cos ^{2} v+u^{2} \sin ^{2} v+2 u \cos v \sin v+\sin ^{2} v+u^{2} \cos ^{2} v-2 u \sin v \cos v+4 u^{2}} \\ &=\sqrt{5 u^{2}+1} \end{align} $$ We have $$ \cos (\alpha)=\frac{\left\langle\overrightarrow{\rho_{1}^{\prime}}\left(t_{0}\right), \overrightarrow{\rho_{2}^{\prime}}\left(t_{0}\right)\right\rangle}{\left\|\overrightarrow{\rho_{1}^{\prime}}\left(t_{0}\right)\right\| \cdot\left\|\overrightarrow{\rho_{2}^{\prime}}\left(t_{0}\right)\right\|}=\frac{3u^2 +1 }{5u^2 +1} $$
Question: So the angle between two curves is a function as $u$. Is this possible? Or should I replace $u$ with another value?
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