Chapter 6: Q. 12 (page 539)

Do you think that the function $f (x) = 2 x^{2}$ has twice the arc length of $g (x) = x^{2}$ on the interval [0, 3]? Why or why not?

Short Answer

Expert verified

No, the function $f (x) = 2 x^{2}$ has not twice the arc length of $g (x) = x^{2}$ on the interval $[0, 3]$ because $\sqrt{1 + 16 x^{2}} \neq 2 \sqrt{1 + 4 x^{2}} .$

Step by step solution

Step 1. Given Information.

The given functions are $f (x) = 2 x^{2} and g (x) = x^{2} .$

Step 2. Explanation.

To find that the function $f (x) = 2 x^{2}$ has twice the arc length of $g (x) = x^{2}$ , we will use the formula of arc length of the definite integral which is $\int_{a}^{b} \sqrt{1 + {(f' (x))}^{2}} d x .$

Now, let $L_{1}$ for the functionrole="math" localid="1650707862373" $f (x) = 2 x^{2}, s o f' (x) = 4 x :$

$L_{1} = \int_{0}^{3} \sqrt{1 + (4 x)^{2}} d x L_{1} = \int_{0}^{3} \sqrt{1 + 16 x^{2}} d x$

Let $L_{2}$ for the function $g (x) = x^{2}, s o g' (x) = 2 x :$

$L_{2} = \int_{0}^{3} \sqrt{1 + (2 x)^{2}} d x L_{2} = \int_{0}^{3} \sqrt{1 + 4 x^{2}} d x$

Thus, we can depict that $L_{1} \neq 2 L_{2} .$

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Do you think that the function $f (x) = 2 x^{2}$ has twice the arc length of $g (x) = x^{2}$ on the interval [0, 3]? Why or why not?

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Explanation.

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Do you think that the function f(x)=2x2has twice the arc length of g(x)=x2on the interval [0, 3]? Why or why not?

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Explanation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Applied Mathematics

Calculus

Geometry

Discrete Mathematics

Decision Maths

Study anywhere. Anytime. Across all devices.

Do you think that the function $f (x) = 2 x^{2}$ has twice the arc length of $g (x) = x^{2}$ on the interval [0, 3]? Why or why not?