Q. 29 Binormal vectors and osculating ... [FREE SOLUTION]

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Chapter 11: Q. 29 (page 901)

Binormal vectors and osculating planes: Find the binormal vector and equation of the osculating plane for the given function at the specified value of t.
$r (t) = 鉄� t, 2 t \sin t, 2 t \cos t 鉄�, t = 蟺$

Short Answer

Expert verified

The binomial vector is,

$\frac{1}{(4 蟺^{2} + 5) \sqrt{4 蟺^{4} + 17 蟺^{2} + 20}} <- 8 蟺^{4} - 26 蟺^{2} - 20, - 4 蟺^{3} - 5 蟺, - 8 蟺^{2} - 10>$ .

The equation of the osculating plane is,

$(8 蟺^{4} + 26 蟺^{2} + 20) (x - 蟺) + (4 蟺^{3} + 5 蟺) y + (8 蟺^{2} + 10) (z + 2 蟺) = 0$

Step by step solution

Step 1. Given Information

We are given,

r (t) = 鉄� t, 2 t \sin t, 2 t \cos t 鉄�, t = 蟺

Step 2. Finding the binormal vector.

Finding the binormal vector,

$r (t) = 鉄� t, 2 t \sin t, 2 t \cos t 鉄�, t = 蟺 r^{'} (t) = 鉄� 1, 2 (t \cos t + \sin t), 2 - t \sin t + \cos t 鉄� ||r^{'} (t)|| = \sqrt{1 + 4 (t \cos t + \sin t)^{2} + 4 (- t \sin t + \cos t)^{2}} = \sqrt{1 + 4 (t^{2} + 1)} = \sqrt{4 t^{2} + 5}$

The unit tangent vector is given by,

$T (t) = \frac{r^{'} (t)}{||r^{'} (t)||} = \frac{1}{\sqrt{4 t^{2} + 5}} 鉄� 1, 2 (t \cos t + \sin t), 2 (- t \sin t + \cos t) 鉄�$

At $t = 蟺$ ,

$T (t) = T (蟺) = \frac{1}{\sqrt{4 蟺^{2} + 5}} 鉄� 1, 2 (- t), 2 (- 1) 鉄� = <\frac{1}{\sqrt{4 蟺^{2} + 5}}, \frac{- 2 蟺}{\sqrt{4 蟺^{2} + 5}}, \frac{- 2}{\sqrt{4 蟺^{2} + 5}}>$

Step 3. Finding the binormal vector.

By using the Quotient Rule for derivatives,

$T^{'} (t) = \frac{1}{{(4 t^{2} + 5)}^{\frac{3}{2}}} <\begin{array}{l} - 4 t, 2 (4 t^{2} + 5) (- t \sin t + 2 \cos t) - 8 t (t \cos t + \sin t), \\ 2 (4 t^{2} + 5) (- t \cos t - 2 \sin t) - 8 t (- t \sin t + \cos t) \end{array}> ||T^{'} (t)|| = \sqrt{\frac{16 t^{2} + 4 {(4 t^{2} + 5)}^{2} (t^{2} + 4) + 64 t^{2} (t^{2} + 1) - 32 t^{2} (4 t^{2} + 5)}{{(4 t^{2} + 5)}^{3}}} ||T^{'} (t)|| = \sqrt{\frac{64 t^{6} + 352 t^{4} + 660 t^{2} + 400}{{(4 t^{2} + 5)}^{3}}} ||T^{'} (t)|| = \sqrt{\frac{4 (16 t^{6} + 88 t^{4} + 165 t^{2} + 100)}{{(4 t^{2} + 5)}^{3}}} ||T^{'} (t)|| = \frac{2 \sqrt{16 t^{6} + 88 t^{4} + 165 t^{2} + 100}}{{(4 t^{2} + 5)}^{\frac{3}{2}}} ||T^{'} (t)|| = \frac{2 \sqrt{(4 t^{2} + 5) (4 t^{4} + 17 t^{2} + 20)}}{{(4 t^{2} + 5)}^{\frac{3}{2}}} N (t) = \frac{T^{'} (t)}{||T^{'} (t)||}$

We need to evaluate N(t) at $t = 蟺$ ,

$<\frac{- 4 蟺}{{(4 蟺^{2} + 5)}^{\frac{3}{2}}}, \frac{- 8 蟺^{2} - 20}{{(4 蟺^{2} + 5)}^{\frac{3}{2}}} \frac{8 蟺^{3} + 18 蟺}{{(4 蟺^{2} + 5)}^{\frac{3}{2}}}>$

At $t = 蟺$ ,

role="math" localid="1649817284636" $||T^{'} (t)|| = ||T^{'} (蟺)|| = \frac{2 \sqrt{(4 蟺^{2} + 5) (4 蟺^{2} + 17 蟺^{2} + 20)}}{{(4 蟺^{2} + 5)}^{\frac{3}{2}}} At t = 蟺, N (t) = \frac{T^{'} (t)}{|T^{'} (t)|} = \frac{1}{\sqrt{(4 蟺^{2} + 5) (4 蟺^{4} + 17 蟺^{2} + 20)}} <- 2 蟺, - 2 (5 + 2 蟺^{2}), 蟺 (9 + 4 蟺^{2})>$

Step 4. Finding the binormal vector.

The binormal vector is given by,

$T (蟺) 脳 N (蟺) = \frac{1}{\sqrt{4 蟺^{2} + 5}} 鉄� 1, - 2 t, - 2 鉄� 脳 \frac{1}{\sqrt{(4 蟺^{2} + 5) (4 蟺^{4} + 17 蟺^{2} + 20)}} <- 2 蟺, - 2 (5 + 2 蟺^{2}), 蟺 (9 + 4 蟺^{2})> = \frac{1}{(4 蟺^{2} + 5) \sqrt{4 蟺^{4} + 17 蟺^{2} + 20}} |\begin{matrix} i & j & k \\ 1 & - 2 蟺 & - 2 \\ - 2 蟺 & - 10 - 4 蟺^{2} & 9 蟺 + 4 蟺^{3} \end{matrix}| = \frac{1}{(4 蟺^{2} + 5) \sqrt{4 蟺^{4} + 17 蟺^{2} + 20}} [\begin{array}{l} i (- 18 蟺^{2} - 8 蟺^{4} - 20 - 8 蟺^{2}) - j (9 蟺 + 4 蟺^{3} - 4 蟺) \\ + k (- 10 - 4 蟺^{2} - 4 蟺^{2}) \end{array}] = \frac{1}{(4 蟺^{2} + 5) \sqrt{4 蟺^{4} + 17 蟺^{2} + 20}} <- 8 蟺^{4} - 26 蟺^{2} - 20, - 4 蟺^{3} - 5 蟺, - 8 蟺^{2} - 10>$

Step 5. Finding the equation of the osculating plane

The equation of the osculating at $r (t = 蟺)$ is defined by,

$B (蟺) - 鉄� x - x (蟺), y - y (蟺), z - z (蟺) 鉄� = 0 \frac{1}{(4 蟺^{2} + 5) \sqrt{4 蟺^{4} + 17 蟺^{2} + 20}} <- 8 蟺^{4} - 26 蟺^{2} - 20, - 4 蟺^{3} - 5 蟺, - 8 蟺^{2} - 10> . 鉄� x - 蟺, y - 0, z + 2 蟺鉄� = 0 (- 8 蟺^{4} - 26 蟺^{2} - 20) (x - 蟺) + (- 4 蟺^{3} - 5 蟺) y + (- 8 蟺^{2} - 10) (z + 2 蟺) = 0 (8 蟺^{4} + 26 蟺^{2} + 20) (x - 蟺) + (4 蟺^{3} + 5 蟺) y + (8 蟺^{2} + 10) (z + 2 蟺) = 0$

Hence, the equation is $(8 蟺^{4} + 26 蟺^{2} + 20) (x - 蟺) + (4 蟺^{3} + 5 蟺) y + (8 蟺^{2} + 10) (z + 2 蟺) = 0$ .

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91影视

Binormal vectors and osculating planes: Find the binormal vector and equation of the osculating plane for the given function at the specified value of t.
$r (t) = 鉄� t, 2 t \sin t, 2 t \cos t 鉄�, t = 蟺$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the binormal vector.

Step 3. Finding the binormal vector.

Step 4. Finding the binormal vector.

Step 5. Finding the equation of the osculating plane

One App. One Place for Learning.

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91影视

Binormal vectors and osculating planes: Find the binormal vector and equation of the osculating plane for the given function at the specified value of t.r(t)=鉄�t,2tsint,2tcost鉄�,t=蟺

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Finding the binormal vector.

Step 3. Finding the binormal vector.

Step 4. Finding the binormal vector.

Step 5. Finding the equation of the osculating plane

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Calculus

Pure Maths

Logic and Functions

Theoretical and Mathematical Physics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Binormal vectors and osculating planes: Find the binormal vector and equation of the osculating plane for the given function at the specified value of t.
$r (t) = 鉄� t, 2 t \sin t, 2 t \cos t 鉄�, t = 蟺$