   Chapter 13.2, Problem 34E

Chapter
Section
Textbook Problem

# At what point do the curves r1(t) = ⟨t, 1 − t, 3 + t2⟩ and r2(s) = ⟨3 − s, s − 2, s2⟩ intersect? Find their angle of intersection correct to the nearest degree.

To determine

To find: The point of intersection of the curves r1(t)=t,1t,3+t2 and r2(s)=3s,s2,s2.

Explanation

Find the values of scalar parameters and substitute in any of the vector functions of curves to obtain the point of intersection of two curves.

As the two curves are intersected, equate x-, y-, and z-components of both the vectors and solve the expressions to obtain the values of scalar parameters t and s.

Equate x-component of vectors as follows.

t=3s (1)

Equate y-component of vectors as follows.

1t=s2 (2)

Equate z-component of vectors as follows.

3+t2=s2 (3)

Solve equation (2) and (3) as follows.

Rewrite the expression in equation (2) as follows.

s=1t+2=3t

Substitute 3t for s in equation (3),

3+t2=(3t)2

Solve the expression to obtain the value of scalar parameter t.

3+t2=9+t26t3+t29t2+6t=06t6=0t=1

Substitute 1 for t in the expression 3t to obtain the value of scalar parameter s.

s=31=2

The values of scalar parameters t, and s are 1, and 2 respectively.

Calculation of point of intersection:

As the x-, -, y-, and z-components of vector r1(t) are t, (1t), and (3+t2), consider the point of intersection of curves is at (t,1t,3+t2).

Substitute 1 for t in the point (t,1t,3+t2),

Point of intersection=(1,11,3+12)=(1,0,4)

Thus, the point of intersection of the curves r1(t)=t,1t,3+t2 and r2(s)=3s,s2,s2 is (1,0,4)_.

To find: The angle of intersection of curves r1(t)=t,1t,3+t2 and r2(s)=3s,s2,s2 at the point of intersection.

Solution:

The angle of intersection of curves r1(t)=t,1t,3+t2 and r2(s)=3s,s2,s2 at the point of intersection is 55°_.

The angle of intersection of two curves is the angle between the two tangent vectors to the curves.

As the point of intersection is obtained at t=1 and s=2, the values of scalar parameters t and s are 1 and 2 respectively.

Formula used:

Write the expression to find the angle between two tangent vectors r1(t) and r2(s) at the point of intersection.

θ=cos1(r1(1)r2(2)|r1(1)||r2(2)|) (4)

Here,

r1(1) is the tangent vector of curve r1(t), which is derivative of vector r1(t) at t=1 and

r2(2) is the tangent vector of curve r2(2), which is derivative of vector r2(t) at s=2.

Calculation of tangent vector r1(t):

To find the derivative of the vector function, differentiate each component of the vector function.

Differentiate each component of the vector function r1(t)=t,1t,3+t2 as follows.

ddt[r1(t)]=ddt(t),ddt(1t),ddt(3+t2)

Use the following formulae to compute the expression.

ddttn=ntn1ddt[u(t)+v(t)]=ddt[u(t)]+ddt[v(t)]ddt(constant)=0

Compute the expression ddt[r1(t)]=ddt(t),ddt(1t),ddt(3+t2) by using the formulae as follows.

r1(t)=ddt(t),[ddt(1)+ddt(t)],[ddt(3)+ddt(t2)]=1,[0+(1)],(0+2t)

r1(t)=1,1,2t (5)

Calculation of tangent vector r1(t) at the point of intersection:

Substitute 1 for t in equation (5),

r1(1)=1,1,2(1)=1,1,2

Calculation of tangent vector r2(s):

Differentiate each component of the vector function r2(s)=3s,s2,s2 as follows

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