Chapter 11: Q. 35 (page 890)

In Exercises 31–35 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

Step 1. Given information.

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

Step 2. Curvature.

Curvature is given by,

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

Step 3. Graph.

The graph is

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In Exercises 31–35 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

Step by step solution

Step 1. Given information.

Step 2. Curvature.

Step 3. Graph.

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In Exercises 31–35 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=13x2+23/2,x=1

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Curvature.

Step 3. Graph.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Calculus

Theoretical and Mathematical Physics

Pure Maths

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Logic and Functions

Study anywhere. Anytime. Across all devices.

In Exercises 31–35 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.
$y = \frac{1}{3} {(x^{2} + 2)}^{3 / 2}, x = 1$