Chapter 11: Q. 40 (page 880)

Using the definitions of normal plane and rectifying plane in Exercises 20 and 21 , respectively , find the equations of these planes at specified points for the vector functions in Exercises 40-42. Note: There are the same functions as in exercises 35,37and 39
$r (t) = <t, t^{2}, \frac{2}{3} t^{3}> at t = 1$

Short Answer

Expert verified

The equation of normal plane is

3 x + 6 y + 6 z = 13

The equation of rectifying plane is $6 x + 3 y - 6 z = 5$

The binomial vector is $B (1) = <\frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}>$

Step by step solution

Step 1. Given information

We have been given a vector function $r (t) = <t, t^{2}, \frac{2}{3} t^{3}> at t = 1$ and we have to find out the binomial vector , normal plane and rectifying plane .

Step 2. Finding the binomial vector

The objective is to find the equation of normal plane and rectifying plane to $r (t)$ at $t = 1$

For this T(t) ,N(t) and B(t) should be calculated first

Consider

\begin{aligned} r (t) = <t, t^{2}, \frac{2}{3} t^{3}> \\ r^{'} (t) = <1, 2 t, 2 t^{2}> \\ ∥r^{'} (t)∥ = \sqrt{1 + (2 t)^{2} + {(2 t^{2})}^{2}} \\ = \sqrt{1 + 4 t^{2} + 4 t^{4}} T (t) = \frac{r^{'} (t)}{∥r^{'} (t)∥} \\ = 2 t^{2} + 1 \\ = \frac{<1, 2 t, 2 t^{2}>}{2 t^{2} + 1} \\ = <\frac{1}{2 t^{2} + 1}, \frac{2 t}{2 t^{2} + 1}, \frac{2 t^{2}}{2 t^{2} + 1}> \\ T (1) = <\frac{1}{3}, \frac{2}{3}, \frac{2}{3}> \end{aligned}

The unit tangent vector to r(t) at t=1 is $<\frac{1}{3}, \frac{2}{3}, \frac{2}{3}>$

By the quotient rule we have

$\begin{aligned} T^{'} (t) = <\frac{- 4 t}{{(2 t^{2} + 1)}^{2}}, \frac{- 4 t^{2} + 2}{{(2 t^{2} + 1)}^{2}}, \frac{4 t}{{(2 t^{2} + 1)}^{2}}> \\ ∥T^{'} (t)∥ = \sqrt{\frac{16 t^{2} + 16 t^{4} + 4 - 16 t^{2} + 16 t^{2}}{{(2 t^{2} + 1)}^{4}}} \\ = \frac{\sqrt{16 t^{4} + 16 t^{2} + 4}}{{(2 t^{2} + 1)}^{2}} \\ = \frac{2 (2 t^{2} + 1)}{{(2 t^{2} + 1)}^{2}} \\ = \frac{2}{2 t^{2} + 1} \end{aligned}$

$\begin{aligned} N (t) = \frac{T (t)}{∥T^{'} (t)∥} \\ = <\frac{- 4 t}{{(2 t^{2} + 1)}^{2}}, \frac{- 4 t^{2} + 2}{{(2 t^{2} + 1)}^{2}}, \frac{4 t}{{(2 t^{2} + 1)}^{2}}> \frac{2 t^{2} + 1}{2} \\ = <\frac{- 2 t}{2 t^{2} + 1}, \frac{- 2 t^{2} + 1}{2 t^{2} + 1}, \frac{2 t}{2 t^{2} + 1}> \\ N (t) = <\frac{- 2}{3}, \frac{- 1}{3}, \frac{2}{3}> \end{aligned}$

Thus the principle normal unit vector to $r (t) a t t = 1$

$N (1) = <\frac{- 2}{3}, \frac{- 1}{3}, \frac{2}{3}>$

now ,

$\begin{aligned} B (1) = T (1) \times N (1) \\ = <\frac{1}{3}, \frac{2}{3}, \frac{2}{3}> \times <\frac{- 2}{3}, \frac{- 1}{3}, \frac{2}{3}> \\ i b |\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{- 2}{3} & \frac{- 1}{3} & \frac{2}{3} \end{array}| \\ = i (\frac{4}{9} + \frac{2}{9}) - j (\frac{2}{9} + \frac{4}{9}) + k (\frac{- 1}{9} + \frac{4}{9}) \\ = \frac{2}{3} i - \frac{2}{3} j + \frac{1}{3} k \\ = <\frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}> \end{aligned}$

The binomial vector to r(t) at t=1 is $B (1) = <\frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}>$

Step 3. Finding the normal plane

Normal plane

The plane determined by the vectors ) and containing the point r(t) is called the normal point C at r(t). Thus the equation of normal plane at t=1 is

First computing

Thus , the equation of normal plane to is

Step 4.Finding rectifying plane

The plane determined by the vectors $T (t_{0}) and B (t_{0})$ and containing the point $r (t_{0})$

is called the rectifying plane for c at r. Thus, the equation of the rectifying plane

(T (1) \times B (1)) \cdot ⟨ x - x (1), y - y (1), z - z (1) ⟩ = 0

\begin{aligned} First computing N (1) \times B (1) : \\ = |\begin{array}{ccc} i & j & k \\ \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{- 2}{3} & \frac{1}{3} \end{array}| \\ = i (\frac{2}{9} + \frac{4}{9}) - j (\frac{1}{9} - \frac{4}{9}) + k (\frac{- 2}{9} - \frac{4}{9}) \\ = \frac{2}{3} i + \frac{1}{3} j - \frac{2}{3} k \\ = <\frac{2}{3}, \frac{1}{3}, \frac{- 2}{3}> \\ Evaluating (T (1) \times B (1)) \cdot ⟨ x - x (1), y - y (1), z - z (1) ⟩ = 0 \\ \Rightarrow <\frac{2}{3}, \frac{1}{3}, \frac{- 2}{3}> \cdot <x - 1, y - 1, z - \frac{2}{3}> = 0 \\ \Rightarrow \frac{2}{3} (x - 1) + \frac{1}{3} (y - 1) - \frac{2}{3} (z - \frac{2}{3}) = 0 \\ \Rightarrow 2 x - 2 + y - 1 - 2 z + \frac{4}{3} = 0 \\ \Rightarrow 2 x + y - 2 z - \frac{5}{3} = 0 \\ \Rightarrow 6 x + 3 y - 6 z - 5 \end{aligned}

Thus the equation of the rectifying plane to r(t) at t=1 is $6 x + 3 y - 6 z = 5$

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Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the binomial vector

Step 3. Finding the normal plane

Step 4.Finding rectifying plane

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Using the definitions of normal plane and rectifying plane in Exercises 20 and 21 , respectively , find the equations of these planes at specified points for the vector functions in Exercises 40-42. Note: There are the same functions as in exercises 35,37and 39r(t)=t,t2,23t3att=1

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the binomial vector

Step 3. Finding the normal plane

Step 4.Finding rectifying plane

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Mechanics Maths

Geometry

Statistics

Logic and Functions

Applied Mathematics

Study anywhere. Anytime. Across all devices.