Two plane curves are defined by the vector functions r1 (t) = (t, 2t – 4) andr2 (u) = (2u, u²). Find the angle a between the curves at the point of their intersection.

r1t=t,2t-4 and r2u=2u,u2 For the intersection point, t=2u ...1u=t22t-4=u2 ...2 Convert the…

Answered: Two plane curves are defined by the…

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter1: Vectors

Section1.3: Lines And Planes

Problem 33EQ

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Two plane curves are defined by the vector functions r1 (t) = (t, 2t – 4) and
r2 (u) = (2u, u²). Find the angle a between the curves at the point of their intersection.

Expert Solution

Step 1: Consider the provided vector functions,

$r_{1} (t) = <t, 2 t - 4> a n d r_{2} (u) = <2 u, u^{2}>$

For the intersection point,

$\begin{array}{rcl} t & = & 2 u . . . (1) \\ u & = & \frac{t}{2} \\ 2 t - 4 & = & u^{2} . . . (2) \end{array}$

Convert the second vector in form of t.

$r_{2} (t) = <t, {\frac{t}{4}}^{2}>$

Substitute the value of u into equation (2).

$\begin{array}{rcl} 2 t - 4 & = & {(\frac{t}{2})}^{2} \\ 8 t - 16 & = & t^{2} \\ t^{2} - 8 t + 16 & = & 0 \\ {(t - 4)}^{2} & = & 0 \\ t & = & 4 \end{array}$

Now, differentiate the both curves with respect to t.

$r'_{1} (t) = <1, 2> a n d r'_{2} (t) = <1, \frac{t}{2}>$

The value of derivative at $t = 4$ is,

$\begin{array}{rcl} r'_{1} (4) & = & <1, 2> \\ r'_{2} (4) & = & <1, \frac{4}{2}> \\ = & <1, 2> \end{array}$

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