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 r1(t)=t,4t,35+t2 and r2(s)=7s,s3,s2 intersect? (x,y,z)= Find angle of intersection, θ, correct to the nearest degree.
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Expert Answer

Layton
Answered 2021-10-30 Author has 89 answers
Consider the following curves:
r1(t)=<t,4t,35+t2>
r2(s)=<7s,s3,s2>
Find the point of intersection of the curves and angle of intersection θ.
The parametric equations corresponding to the curve r1(t) are as follows:
x=t
y=4t
z=35+t2
The parametric equations corresponding to the curve r2(s) are as follows:
x=7s
y=s3
z=s2
At the point of intersection of two curves, the corresponding coordinates of parametric lines are same.
t=7ss+t=7
4t=s3s+t=7
35+t2=s2s2t2=35
That is,
s2t2=35
(st)(s+t)=35
(st)7=35
st=5
Solve the equations st=5 and s+t=7, to find the values of s and t.
st+s+t=5+72s=12s=6
The corresponding value of t is as follows:
t=7st=76t=1
Thus, the values are t=1 and s=6
Substitute t=1 in r1(t), to find the point of intersection.
r1(t)=<1,41,35+12<1,3,36>
r1(t)=<1,3,36>
The angle between the normal vectors is given as cosθ=r1(t)r2(s)|r1(t)||r2(s)|
The direction vector of the line r1(t) is <t,t,t2>
The direction vector of the line r2(s) is <s,s,s2>
At t=1,
r1(t)=<1,1,2t>
r1(1)=<1,1,2>
At s=6,

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