Chapter 11: Q. 35 (page 901)

Osculating circles: Find the center and radius of the osculating circle to the given vector function at the specified value of t.
$r (t) = ⟨ t, 2 t \sin t, 2 t \cos t ⟩, t = π$

Short Answer

Expert verified

The center of the osculating circle is,

$⟨ - π, 0, - 2 π ⟩ + \frac{5 + 4 π^{2}}{2 (4 π^{4} + 17 π^{2} + 20)} <- 2 π, - 2 (5 + 2 π^{2}), π (9 + π^{2})>$ .

The radius is $\frac{{(4 π^{2} + 5)}^{\frac{3}{2}}}{2 \sqrt{4 π^{4} + 17 π^{2} + 20}}$ .

Step by step solution

Step 1. Given Information

We are given,

$r (t) = ⟨ t, 2 t \sin t, 2 t \cos t ⟩, t = π$

Step 2. Find the center and radius of the osculating circle

Finding the center and radius of the osculating circle,

$r (t) = ⟨ t, 2 t \sin t, 2 t \cos t ⟩ r^{'} (t) = ⟨ 1, 2 (t \cos t + \sin t), 2 (- t \sin t + \cos t) ⟩ r^{''} (t) = ⟨ 0, 2 (- t \sin t + 2 \cos t), 2 (- t \sin t - 2 \sin t) ⟩ r^{'} (t) \times r^{''} (t) = |\begin{matrix} i & j & k \\ 1 & 2 (t \cos t + \sin t) & 2 (- t \sin t + \cos t) \\ 0 & 2 (- t \sin t + 2 \cos t) & 2 (- t \cos t - 2 \sin t) \end{matrix}| = i [2 (t \cos t + \sin t) 2 (- t \cos t - 2 \sin t) - 4 (- t \sin t + \cos t) (- t \sin t + 2 \cos t)] - j [(2 (- t \cos t - 2 \sin t))] + k [2 (- t \sin t + 2 \cos t)] = <- 4 t^{2} - 8, 2 t \cos t + 2 \sin t, - 2 t \sin t + 4 \cos t> ||r^{'} (t) \times r^{''} (t)|| = \sqrt{{(- 4 t^{2} - 8)}^{2} + (2 t \cos t + 2 \sin t)^{2} + (- 2 t \sin t + 4 \cos t)^{2}} At t = π, ||r^{'} (t) \times r^{''} (t)|| = \sqrt{{(- 4 π^{2} - 8)}^{2} + (- 2 π)^{2} + (- 4)^{2}} = \sqrt{16 π^{4} + 64 + 64 π^{2} + 4 π^{2} + 16} = \sqrt{16 π^{4} + 68 π^{2} + 80} ||r^{'} (t)|| = \sqrt{1 + 4 (t \cos t + \sin t)^{2} + 4 (- t \sin t + \cos t)^{2}} = \sqrt{4 t^{2} + 5}$

Step 3. Find the center and radius of the osculating circle

At $t = π,$

$||r^{'} (t)|| = \sqrt{4 π^{2} + 5}$

The curvature at $t = π$ is,

$\frac{||r^{'} (t) \times r^{''} (t)||}{{||r^{'} (t)||}^{3}} = \frac{\sqrt{16 π^{2} + 68 π^{2} + 80}}{{(\sqrt{4 π^{2} + 5})}^{3}} = \frac{2 \sqrt{4 π^{2} + 17 π^{2} + 20}}{{(4 π^{2} + 5)}^{\frac{3}{2}}}$

Thus radius of the osculating circle is

$\frac{1}{k} = \frac{{(4 π^{2} + 5)}^{\frac{3}{2}}}{2 \sqrt{4 π^{4} + 17 π^{2} + 20}}$

Step 4. Find the center and radius of the osculating circle

The center is given by,

$T (t) = \frac{r^{'} (t)}{||r^{'} (t)||} = \frac{1}{\sqrt{4 t^{2} + 5}} ⟨ 1, 2 (t csot + \sin t), 2 (- t \sin t + \cos t) ⟩ At t = π, T (t) = T (π) = \frac{1}{\sqrt{4 π^{2} + 5}} ⟨ 1, - 2 π, - 2 ⟩$

Using the Quotient rule for derivatives,

$T^{'} (t) = \frac{1}{{(4 t^{2} + 5)}^{\frac{3}{2}}} \{\begin{array}{l} - 4 t, 2 (4 t^{2} + 5) (- t \sin t + 2 csot) - 8 t (t \cos t + \sin t), \\ 2 (4 t^{2} + 5) (- t \cos t - 2 \sin t) - 8 t (- t \sin t + \cos t) \end{array}> ||T^{'} (t)|| = \sqrt{\frac{16 t^{2} + 4 {(4 t^{2} + 5)}^{2} (t^{2} + 4) + 64 t^{2} (t^{2} + 1) - 32 t^{2} (4 t^{3} + 5)}{{(4 t^{2} + 5)}^{3}}} ||T^{'} (t)|| = \sqrt{\frac{64 t^{6} + 352 t^{4} + 660 t^{2} + 400}{{(4 t^{2} + 5)}^{3}}} ||T^{'} (t)|| = \frac{2 \sqrt{16 t^{6} + 88 t^{4} + 165 t^{2} + 100}}{{(4 t^{2} + 5)}^{\frac{3}{2}}} ||T^{'} (t)|| = \frac{2 \sqrt{(4 t^{2} + 5) (4 t^{4} + 17 t^{2} + 20)}}{{(4 t^{2} + 5)}^{\frac{3}{2}}} At t = π, T^{'} (π) = <\frac{- 4 π}{{(4 π^{2} + 5)}^{\frac{3}{2}}}, \frac{- 8 π^{2} - 20}{{(4 π^{2} + 5)}^{\frac{3}{2}}} \frac{8 π^{3} + 18 π}{{(4 π^{2} + 5)}^{\frac{3}{2}}}> At t = π, ||T^{'} (π)|| = \frac{2 \sqrt{(4 π^{2} + 5) (4 π^{4} + 17 π^{2} + 20)}}{{(4 π^{2} + 5)}^{\frac{3}{2}}} At t = π, N (t) = \frac{T (π)}{||T^{'} (π)||} = \frac{1}{\sqrt{(4 π^{2} + 5) (4 π^{4} + 17 π^{2} + 20)}} <- 2 π, - 2 (5 + 2 π^{2}), π (9 + 4 π^{2})>$

Step 5. Find the center and radius of the osculating circle

The binomial vector is,

$= \frac{1}{\sqrt{(4 π^{2} + 5) (4 π^{4} + 17 π^{2} + 20)}} |\begin{matrix} i & j & k \\ 1 & - 2 π & - 2 \\ - 2 π & - 10 - 4 π^{2} & 9 π + 4 π^{3} \end{matrix}| = \frac{1}{\sqrt{(4 π^{2} + 5) (4 π^{4} + 17 π^{2} + 20)}} <- 8 π^{4} - 26 π^{2} - 20, - 4 π^{3} - 5 π, - 8 π^{2} - 10>$

The centre of the osculating circle is given by,

$r (t_{0}) + \frac{1}{k} N (t_{0}) = r (π) + \frac{1}{k} N (π) = ⟨ - π, 0, - 2 π ⟩ + \frac{5 + 4 π^{2}}{2 (4 π^{4} + 17 π^{2} + 20)} <- 2 π, - 2 (5 + 2 π^{2}), π (9 + π^{2})>$

Hence. the center is,

role="math" localid="1649843878643" $⟨ - π, 0, - 2 π ⟩ + \frac{5 + 4 π^{2}}{2 (4 π^{4} + 17 π^{2} + 20)} <- 2 π, - 2 (5 + 2 π^{2}), π (9 + π^{2})>$

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Osculating circles: Find the center and radius of the osculating circle to the given vector function at the specified value of t.
$r (t) = ⟨ t, 2 t \sin t, 2 t \cos t ⟩, t = π$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Find the center and radius of the osculating circle

Step 3. Find the center and radius of the osculating circle

Step 4. Find the center and radius of the osculating circle

Step 5. Find the center and radius of the osculating circle

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Osculating circles: Find the center and radius of the osculating circle to the given vector function at the specified value of t.r(t)=⟨t,2tsint,2tcost⟩,t=π

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Find the center and radius of the osculating circle

Step 3. Find the center and radius of the osculating circle

Step 4. Find the center and radius of the osculating circle

Step 5. Find the center and radius of the osculating circle

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Applied Mathematics

Pure Maths

Geometry

Discrete Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

Osculating circles: Find the center and radius of the osculating circle to the given vector function at the specified value of t.
$r (t) = ⟨ t, 2 t \sin t, 2 t \cos t ⟩, t = π$