At what point do the curves r_1(t)=t,4-t,35+t^2 and r_2(s)=7-s,s-3,s^2 intersect? (x,y,z)= Find angle of intersection, \theta, correct to the nearest degree.

Dolly Robinson 2021-05-30 Answered
At what point do the curves \(r_1(t)=t,4-t,35+t^2\) and \(r_2(s)=7-s,s-3,s^2\) intersect? (x,y,z)= Find angle of intersection, \(\theta\), correct to the nearest degree.

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Expert Answer

curwyrm
Answered 2021-05-31 Author has 24067 answers
Consider the following curves:
\(r_1(t)=\)
\(r_2(s)=<7-s,s-3,s^2>\)
Find the point of intersection of the curves and angle of intersection \(\theta\).
The parametric equations corresponding to the curve \(r_1(t)\) are as follows:
x=t
\(y=4-t\)
\(z=35+t^2\)
The parametric equations corresponding to the curve \(r_2(s)\) 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.
\(t=7-s\Rightarrow s+t=7\)
\(4-t=s-3\Rightarrow s+t=7\)
\(35+t^2=s^2\Rightarrow s^2-t^2=35\)
That is,
\(s^2-t^2=35\)
\((s-t)(s+t)=35\)
\((s-t)7=35\)
\(s-t=5\)
Solve the equations \(s-t=5\) and \(s+t=7\), to find the values of s and t.
\(s-t+s+t=5+7\Rightarrow2s=12\Rightarrow s=6\)
The corresponding value of t is as follows:
\(t=7-s\Rightarrow t=7-6\Rightarrow t=1\)
Thus, the values are t=1 and s=6
Substitute t=1 in \(r_1(t)\), to find the point of intersection.
\(r_1(t)=<1,4-1,35+1^2>=<1,3,36>\)
\(r_1(t)=<1,3,36>\)
The angle between the normal vectors is given as \(\cos\theta=\frac{r_1'(t)\cdot r_2'(s)}{|r_1'(t)||r_2'(s)|}\)
The direction vector of the line \(r_1(t)\) is \(\)
The direction vector of the line \(r_2(s)\) is \(<-s,s,s^2>\)
At t=1,
\(r_1'(t)=<1,-1,2t>\)
\(r_1'(1)=<1,-1,2>\)
At s=6,
\(r_2'(s)=<-1,1,2s>\)
\(r_2'(6)=<-1,1,12>\)
\(\cos\theta=\frac{r_1'(t)\cdot r_2'(s)}{|r_1'(t)||r_2'(s)|}\)
\(=\frac{r_1'(1)\cdot r_2'(6)}{|r_1'(1)||r_2'(6)|}\)
\(=\frac{<1,-1,2>\cdot<-1,1,12>}{\sqrt{1+1+4}\cdot\sqrt{1+1+144}}\)
\(=\frac{-1-1+24}{\sqrt{6}\cdot\sqrt{146}}\)
\(=\frac{22}{\sqrt{6}\cdot\sqrt{146}}\)
\(=\frac{11}{\sqrt{219}}\)
Thus, \(\theta=\cos^{-1}\left(\frac{11}{\sqrt{219}}\right)\)
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