Chapter 11: Q. 37 (page 880)

For each of the vector-valued functions in Exercises 35–39,
find the unit tangent vector, the principal unit normal vector,
the binormal vector, and the equation of the osculating plane
at the specified value of t.
$r (t) = <e^{t}, e^{- t}, \sqrt{2} t> at t = 0$

Short Answer

Expert verified

The unit tangent vector to r(t) $t = 0 is <\frac{1}{2}, \frac{- 1}{2}, \frac{\sqrt{2}}{2}>$

The principle normal unit vector to r(t) $N (0) = <\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0>$

The binomial vector at $t = 0 is <\frac{- 1}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2})$

The equation of osculating plane to r(t) at t=0 $- x + y + \sqrt{2} z = 0$

Step by step solution

Step 1. Given information

WE have been given the vector valued function $r (t) = <e^{t}, e^{- t}, \sqrt{2} t> at t = 0$

find the unit tangent vector, the principal unit normal vector,

the binormal vector, and the equation of the osculating plane at specified value of t

Step 2. Finding the unit tangent vector

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

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

the unit tangent vector to r(t) at t=0 $<\frac{1}{2}, \frac{- 1}{2}, \frac{\sqrt{2}}{2}>$

Step 3. Finding principle normal unit vector

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

By using quotient rule

\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{- 2 \sqrt{2} e^{3 t} + \sqrt{2} e^{t}}{{(2 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}}{{(2 e^{2 t} + 1)}^{4}}} \\ = \sqrt{\frac{4 e^{4 t} + 4 e^{4 t} + 2 e^{2 t} - 4 e^{4 t}}{{(e^{2 t} + 1)}^{4}}} \\ = \sqrt{\frac{2 e^{6 t} + 4 e^{4 t} + 2 e^{2 t}}{{(e^{2 t} + 1)}^{2}}} \\ = \frac{\sqrt{2} \sqrt{{(e^{3 t} + e^{t})}^{2}}}{{(e^{2 t} + 1)}^{2}} \\ = \frac{\sqrt{2} (e^{3 t} + e^{t})}{{(e^{2 t} + 1)}^{2}} \end{aligned}

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

The principle normal unit vector to r(t) $N (0) = <\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0>$

Step 4. Finding the binomial vector

$B (0) = T (0) \times N (0)$

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

Step 5. Finding osculating plane

The osculating plane at r(t=0) is defined by

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

The equation of osculating plane to r(t) at t=0 $- x + y + \sqrt{2 z} = 0$

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For each of the vector-valued functions in Exercises 35–39,
find the unit tangent vector, the principal unit normal vector,
the binormal vector, and the equation of the osculating plane
at the specified value of t.
$r (t) = <e^{t}, e^{- t}, \sqrt{2} t> at t = 0$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the unit tangent vector

Step 3. Finding principle normal unit vector

Step 4. Finding the binomial vector

Step 5. Finding osculating plane

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For each of the vector-valued functions in Exercises 35–39,find the unit tangent vector, the principal unit normal vector,the binormal vector, and the equation of the osculating planeat the specified value of t.r(t)=et,e−t,2tatt=0

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the unit tangent vector

Step 3. Finding principle normal unit vector

Step 4. Finding the binomial vector

Step 5. Finding osculating plane

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Theoretical and Mathematical Physics

Statistics

Decision Maths

Probability and Statistics

Logic and Functions

Study anywhere. Anytime. Across all devices.

For each of the vector-valued functions in Exercises 35–39,
find the unit tangent vector, the principal unit normal vector,
the binormal vector, and the equation of the osculating plane
at the specified value of t.
$r (t) = <e^{t}, e^{- t}, \sqrt{2} t> at t = 0$