Chapter 11: Q16P (page 559)

Find the arc length of one arch of $y = s i n x$ .

Short Answer

Expert verified

The length of one arc is $\sqrt{2} E (π, \frac{1}{\sqrt{2}}) \approx 3.82019$ .

Step by step solution

Given Information

The function is $y = \sin x$ .

Definition of the differential element of displacement.

The differential element of displacement is defined as $\int d s = \int \sqrt{{dx}^{2} + {dx}^{2}}$ .

Find the value of Integral

The formula of differential element of displacement is $\int d s = \int \sqrt{d x^{2} + d y^{2}}$ .

$\int d s = \int \sqrt{d x^{2} (1 + \frac{d y^{2}}{d x^{2}})} \int d s = \int \sqrt{(1 + \frac{d y^{2}}{d x^{2}})} d x$

Substitute the value given below in the equation mentioned above.

$y = \sin x \frac{d y}{d x} = \cos x$

The equation becomes as follows:

$\int d s = \int \sqrt{(1 + \cos^{2} x)} d x \int d s = \int \sqrt{(1 + (1 - \sin^{2} x))} d x \int d s = \int \sqrt{(2 - \sin^{2} x)} d x \int d s = \sqrt{2} \int \sqrt{(1 - \frac{\sin^{2} x}{2})} d x$

The formula for the elliptical function is $E (\frac{π}{2}, K) = \int_{0}^{\frac{π}{2}} \sqrt{1 - k^{2} \sin^{2} θ} d θ$ .

The equation becomes as follows:

$\int d s = \sqrt{2} \int \sqrt{(1 - \frac{\sin^{2} x}{2})} d x \int d s = \sqrt{2} E (π, \frac{1}{\sqrt{2}}) \int d s \approx 3.82019$

The length of one arc is $\sqrt{2} E (π, \frac{1}{\sqrt{2}}) \approx 3.82019$ .

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Find the arc length of one arch of $y = s i n x$ .

Short Answer

Step by step solution

Given Information

Definition of the differential element of displacement.

Find the value of Integral

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Find the arc length of one arch of y=sinx.

Short Answer

Step by step solution

Given Information

Definition of the differential element of displacement.

Find the value of Integral

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

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Find the arc length of one arch of $y = s i n x$ .