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Chapter 11: Q. 35 (page 890)

In Exercises 31鈥�35 find the curvature of the given function at the indicated value of x. Then sketch the curve and the osculating circle at the indicated point.
$y = \frac{1}{3} {(x^{2} + 2)}^{3 / 2}, x = 1$

Short Answer

Expert verified

The value of curvature is $\frac{\sqrt{3}}{6}$

Step by step solution

01

Step 1. Given information.

The given function is $y = \frac{1}{3} {(x^{2} + 2)}^{3 / 2}, x = 1$ .

02

Step 2. Curvature.

Curvature is given by,

$k = \frac{|f^{''} (x)|}{{(1 + {(f^{'} (x))}^{2})}^{\frac{3}{2}}} Let y = f (x) = \frac{1}{3} {(x^{2} + 2)}^{\frac{3}{2}} f^{'} (x) = x \sqrt{x^{2} + 2} f^{''} (x) = \frac{x^{2}}{\sqrt{x^{2} + 2}} + \sqrt{x^{2} + 2} = \frac{2 x^{2} + 2}{\sqrt{x^{2} + 2}} k = \frac{|\frac{2 x^{2} + 2}{\sqrt{x^{2} + 2}}|}{{(1 + {(x \sqrt{x^{2} + 2})}^{2})}^{\frac{3}{2}}} At x = 1, k = \frac{\frac{4}{\sqrt{3}}}{(1 + 3)^{\frac{3}{2}}} k = \frac{1}{2 \sqrt{3}} k = \frac{\sqrt{3}}{6}$

03

Step 3. Graph.

The graph is

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Most popular questions from this chapter

For each of the vector-valued functions, find the unit tangent vector.

$r (t) = (\sin 2 t, \cos 2 t, t)$

For constants $伪, 尾$ , and $纬$ , the graph of a vector-valued function of the form

$r (t) = 伪 e^{纬 t} (\cos t) i + 伪 e^{纬 t} (\sin t) j + 尾 e^{纬 t} k, t 鈮� 0$

Let $f (t)$ be a differentiable scalar function and $r (t)$ be a differentiable vector function. Prove that $\frac{d}{d t} (f (t) r (t)) = f' (t) r (t) + f (t) r' (t)$ . (This is Theorem 11.11 (b).)

For each of the vector-valued functions, find the unit tangent vector.

r (t) = (\cos (伪 t), \sin (伪 t))

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