Chapter 11: Q. 41 (page 880)

Using the definitions of the normal plane and rectifying plane in Exercises 20 and 21, respectively, find the equations of these planes at the specified points for the vector functions in Exercises 40–42. Note: These are the same functions as in Exercises 35, 37, and 39.
$r (t) = <e^{t}, e^{- t}, \sqrt{2} t) at t = 0$

Short Answer

Expert verified

The equation of normal plane is $x - y + \sqrt{2} z = 0$

The equation of rectifying plane is $x + y = 2$

Step by step solution

Step 1. Given information

$r (t) = <e^{t}, e^{- t}, \sqrt{2} t) at t = 0$

Step 2. Calculating unit tangent vector

$\begin{aligned} r (t) = <e^{t}, e^{- t}, \sqrt{2} t> \\ r^{'} (t) = <e^{t}, - e^{- t}, \sqrt{2}> \\ ∥r^{'} (t)∥ = \sqrt{{(e^{t})}^{2} + {(- e^{- t})}^{2} + (\sqrt{2})^{2}} \\ = \sqrt{2^{2 t} + e^{- 2 t} + 2} \\ = e^{t} + \frac{1}{e^{t}} \\ = \frac{2^{2 t} + 1}{e^{t}} T (t) = \frac{r^{'} (t)}{∥r^{'} (t)∥} \\ = \frac{<e^{t}, - e^{- t}, \sqrt{2}>}{\frac{2^{2 t} + 1}{e^{t}}} \\ = <\frac{2^{2 t}}{e^{2 t} + 1}, \frac{- 1}{e^{2 t} + 1}, \frac{\sqrt{2} e^{t}}{e^{2 t} + 1}> \\ At t = 0, T (t) = T (0) = <\frac{e^{0}}{e^{0} + 1}, \frac{- 1}{e^{0} + 1}, \frac{\sqrt{2} e^{0}}{e^{0} + 1}> \\ = <\frac{1}{2}, \frac{- 1}{2}, \frac{\sqrt{2}}{2}> \end{aligned}$

Step 3. Calculating the principle unit normal vector

$T (t) = <\frac{e^{2 t}}{e^{2 t} + 1}, \frac{- 1}{e^{2 t} + 1}, \frac{\sqrt{2} e^{2 t}}{e^{2 t} + 1}>$

$\begin{aligned} T^{'} (t) = <\frac{2 e^{2 t}}{{(e^{2 t} + 1)}^{2}}, \frac{2 e^{2 t}}{{(e^{2 t} + 1)}^{2}}, \frac{- \sqrt{2} e^{3 t} + \sqrt{2} e^{t}}{{(e^{2 t} + 1)}^{2}}> \\ ∥T^{'} (t)∥ = \sqrt{\frac{4 e^{4 t} + 4 e^{4 t} + {(- \sqrt{2} e^{3 t} + \sqrt{2} e^{t})}^{2}}{{(e^{2 t} + 1)}^{4}}} \\ = \sqrt{{\frac{4 e^{4 t} + 4 e^{4 t} + 2 e^{6 t} + 2 e^{2 t} - 4 e^{4 t}}{(e^{2 t} + 1})}^{4}} \\ = \frac{\sqrt{2 e^{6 t} + 4 e^{4 t} + 2 e^{2 t}}}{{(e^{2 t} + 1)}^{2}} \\ = \frac{\sqrt{2} (e^{3 t} + e^{t})}{{(e^{2 t} + 1)}^{2}} \end{aligned}$

$\begin{aligned} N (t) = \frac{T^{'} (t)}{∥T^{'} (t)∥} \\ = <\frac{2 e^{2 t}}{{(2 e^{2 t} + 1)}^{2}}, \frac{2 e^{2 t}}{{(e^{2 t} + 1)}^{2}}, \frac{- \sqrt{2} e^{3 t} + \sqrt{2} e^{t}}{{(2 e^{2 t} + 1)}^{2}}> \frac{{(2 e^{2 t} + 1)}^{2}}{\sqrt{2} (e^{3 t} + e^{t})} \\ = <\frac{2 e^{2 t}}{\sqrt{2} (e^{3 t} + e^{t})}, \frac{2 e^{2 t}}{\sqrt{2} (e^{3 t} + e^{t})}, \frac{- \sqrt{2} e^{3 t} + \sqrt{2} e^{t}}{\sqrt{2} (e^{3 t} + e^{t})}> \\ N (0) = <\frac{2 e^{0}}{\sqrt{2} (e^{0} + e^{0})}, \frac{2 e^{0}}{\sqrt{2} (e^{0} + e^{0})}, \frac{- \sqrt{2} e^{0} + \sqrt{2} e^{0}}{\sqrt{2} (e^{0} + e^{0})}> \\ N (0) = <\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0> \\ N (0) = <\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0> \end{aligned}$

Step 4. Finding the binomial vector

$\begin{aligned} B (0) = T (0) \times N (0) \\ = <\frac{1}{2}, \frac{- 1}{2}, \frac{\sqrt{2}}{2}> \times <\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0> \end{aligned}$

role="math" localid="1649572675802" $\begin{aligned} = |\begin{array}{ccc} i & j & k \\ \frac{1}{2} & \frac{- 1}{2} & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \end{array}| \\ = i [\frac{- 2}{4}] - j [\frac{- 2}{4}] + k [\frac{2 \sqrt{2}}{4}] \\ = <\frac{- 1}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}> \end{aligned}$

Step 5. Finding equation for normal plane

$[N (0) \times B (0)] \cdot ⟨ x - x (0), y - y (0), z - z (0) ⟩ = 0$

We will first calculate, $N (0) \times B (0) :$

$\begin{aligned} |\begin{array}{ccc} i & j & k \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ \frac{- 1}{2} & \frac{1}{2} & \frac{\sqrt{2}}{2} \end{array}| \\ = i (\frac{2}{4}) - j (\frac{2}{4}) + k (\frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4}) \\ = <\frac{1}{2}, \frac{- 1}{2} \cdot \frac{\sqrt{2}}{2}> \end{aligned}$

Now,

$\begin{aligned} <\frac{1}{2}, \frac{- 1}{2}, \frac{\sqrt{2}}{2}> \cdot ⟨ x - 1, y - 1, z ⟩ = 0 \\ \Rightarrow \frac{1}{2} (x - 1) - \frac{1}{2} (y - 1) + \frac{\sqrt{2}}{2} z = 0 \\ \Rightarrow x - 1 - y + 1 + \sqrt{2} z = 0 \\ x - y + \sqrt{2} z = 0 \end{aligned}$

Step 6. Finding equation for rectifying plane

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

We will first calculate, $T (0) \times B (0)$

$\begin{aligned} |\begin{array}{ccc} i & j & k \\ \frac{1}{2} & \frac{- 1}{2} & \frac{\sqrt{2}}{2} \\ \frac{- 1}{2} & \frac{1}{2} & \frac{\sqrt{2}}{2} \end{array}| \\ = I [\frac{- \sqrt{2}}{2} - \frac{\sqrt{2}}{4}] - [\frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4}] + k [\frac{1}{4} - \frac{1}{4}] \\ = \frac{- 2 \sqrt{2}}{4} i - \frac{2 \sqrt{2}}{4} j \\ = \frac{- \sqrt{2}}{2} i - \frac{\sqrt{2}}{2} j \\ = <\frac{- \sqrt{2}}{2}, \frac{- \sqrt{2}}{2}, 0> \end{aligned}$

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

$\begin{aligned} <\frac{- \sqrt{2}}{2}, \frac{- \sqrt{2}}{2}, 0> \cdot ⟨ x - 1, y - 1, z ⟩ = 0 \\ \Rightarrow \frac{- \sqrt{2}}{2} (x - 1) - \frac{\sqrt{2}}{2} (y - 1) = 0 \\ \Rightarrow x - 1 + y - 1 = 0 \\ \Rightarrow x + y = 2 \end{aligned}$

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Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculating unit tangent vector

Step 3. Calculating the principle unit normal vector

Step 4. Finding the binomial vector

Step 5. Finding equation for normal plane

Step 6. Finding equation for rectifying plane

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Using the definitions of the normal plane and rectifying plane in Exercises 20 and 21, respectively, find the equations of these planes at the specified points for the vector functions in Exercises 40–42. Note: These are the same functions as in Exercises 35, 37, and 39.r(t)=et,e−t,2tatt=0

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculating unit tangent vector

Step 3. Calculating the principle unit normal vector

Step 4. Finding the binomial vector

Step 5. Finding equation for normal plane

Step 6. Finding equation for rectifying plane

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Logic and Functions

Discrete Mathematics

Calculus

Mechanics Maths

Statistics

Study anywhere. Anytime. Across all devices.