 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.

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it curwyrm
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, $$\theta=\cos^{-1}\left(\frac{11}{\sqrt{219}}\right)$$