 # At what point do the curves r_1(t)=t,4-t,35+t^2 and r_2(s)=7-s,s-3,s^2Z allhvasstH 2021-10-29 Answered
At what point do the curves ${r}_{1}\left(t\right)=t,4-t,35+{t}^{2}$ and ${r}_{2}\left(s\right)=7-s,s-3,{s}^{2}$ intersect? (x,y,z)= Find angle of intersection, $\theta$, correct to the nearest degree.
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Consider the following curves:
${r}_{1}\left(t\right)=$
${r}_{2}\left(s\right)=<7-s,s-3,{s}^{2}>$
Find the point of intersection of the curves and angle of intersection $\theta$.
The parametric equations corresponding to the curve ${r}_{1}\left(t\right)$ are as follows:
x=t
$y=4-t$
$z=35+{t}^{2}$
The parametric equations corresponding to the curve ${r}_{2}\left(s\right)$ are as follows:
$x=7-s$
$y=s-3$
$z={s}^{2}$
At the point of intersection of two curves, the corresponding coordinates of parametric lines are same.
$t=7-s⇒s+t=7$
$4-t=s-3⇒s+t=7$
$35+{t}^{2}={s}^{2}⇒{s}^{2}-{t}^{2}=35$
That is,
${s}^{2}-{t}^{2}=35$
$\left(s-t\right)\left(s+t\right)=35$
$\left(s-t\right)7=35$
$s-t=5$
Solve the equations $s-t=5$ and $s+t=7$, to find the values of s and t.
$s-t+s+t=5+7⇒2s=12⇒s=6$
The corresponding value of t is as follows:
$t=7-s⇒t=7-6⇒t=1$
Thus, the values are t=1 and s=6
Substitute t=1 in ${r}_{1}\left(t\right)$, to find the point of intersection.
${r}_{1}\left(t\right)=<1,4-1,35+{1}^{2}\ge <1,3,36>$
${r}_{1}\left(t\right)=<1,3,36>$
The angle between the normal vectors is given as $\mathrm{cos}\theta =\frac{{r}_{1}^{\prime }\left(t\right)\cdot {r}_{2}^{\prime }\left(s\right)}{|{r}_{1}^{\prime }\left(t\right)||{r}_{2}^{\prime }\left(s\right)|}$
The direction vector of the line ${r}_{1}\left(t\right)$ is $$
The direction vector of the line ${r}_{2}\left(s\right)$ is $<-s,s,{s}^{2}>$
At t=1,
${r}_{1}^{\prime }\left(t\right)=<1,-1,2t>$
${r}_{1}^{\prime }\left(1\right)=<1,-1,2>$
At s=6,