ABILENECHRISTIANUNIVERSITY

Department of Mathematics

Arclength And Revolutions Section 6.4

Dr. John Ehrke Department of Mathematics

Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS The Idea of Arc The idea of calculating the length along a represents the basic idea of : deal with a hard problem by breaking it up into small problems (actually infinitesimal) and obtain the answer in the . To calculate the arc length of a curve we divide the curve defined on [a, b] into segments s1,..., sn of equal length ∆s. The ∆s can be approximated by the ∆x and ∆y using the as shown in the figure below.

In this case, we have (∆s)2 ≈ (∆x)2 + (∆y)2. As we let ∆x → 0, this approximation becomes better as the curve actually approaches becomes more linear, and we obtain (ds)2 = (dx)2 + (dy)2.

Slide 2/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS A Better Form for Arc Length In our notation we have q (ds)2 = (dx)2 + (dy)2 =⇒ ds = dx2 + dy2.

A variant of this formula can be obtained by factoring out the dx. In this case we obtain, s  dy 2 ds = 1 + dx. dx Integrating both sides of this expression gives us s Z Z b  dy 2 Arc Length = ds = 1 + dx. a dx An alternative form used by many texts, is to let y = f (x) in which case we have

Z b q Arc Length = 1 + f 0(x)2 dx. a We will use these forms interchangeably.

Slide 3/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS of a

Example Show that the circumference of a circle of a is 2πa.

The√ upper half of a circle of radius a centered at (0, 0) is given by the function y = a2 − x2 for −a ≤ x ≤ a.

Slide 4/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Solution continued... The circle has vertical lines at x = ±a and f 0(±a) is undefined, which prevents us from using the arc length formula. So we’ll have to set up the problem√ to avoid this. Instead we consider one-eighth of the circle on the interval [0, a/ 2]. (See the figure on the previous slide.) We determine that f 0(x) = − √ x . The arc a2−x2 length corresponding to this region is given by √ √ s Z a/ 2 q Z a/ 2  x 2 1 + f 0(x)2 dx = 1 + − √ dx 2 2 0 0 a − x √ Z a/ 2 dx = a √ 2 2 0 a − x √  x  a/ 2 = a sin−1 a 0 πa = 4 It follows that the circumference of the full circle is 8(πa/4) = 2πa units.

Slide 5/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Arc Length in Polar Coordinates Observe in the previous calculation we used the term units, when in reality the units are actually . Note that if a = 1, then our answer for the eighth of the circle’s arc length is π/4, but θ = π/4! Recall, the definition of the as presented in most trigonometry courses is the unit of the angle, equal to the central angle whose arc is equal in length to the radius, and is given formula s = rθ. This suggests that maybe this problem is more appropriately described in terms of polar coordinates.

Recall that x = r cos θ and y = r sin θ are the conversion formulas. This means that dr dr dx = cos θ − r sin θ and dy = sin θ + r cos θ. dθ dθ Calculating ds we get !  dx 2  dy 2 ds2 = + dθ dθ dθ !  dr 2  dr 2 = cos θ − r sin θ + sin θ + r cos θ dθ dθ dθ !  dr 2 = r2 + dθ dθ

Slide 6/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Solution via Polar Coordinates Our previous calculation gives the arc length formula in polar coordinates s  dr 2 ds = r2 + dθ. dθ

For a circle of radius a, the formula is given by r(θ) = a, where 0 ≤ θ ≤ 2π. Using the arc length formula we have

Z Z 2π p ds = a2 + 0 dθ 0 Z 2π = a dθ 0 = 2πa

This is consistent with the formula for the circumference of a circle, and was a much easier integration as anticipated.

Slide 7/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS A More Difficult Example

Example √ √ Find the length of the curve y = ln(x + x2 − 1) on the interval [1, 2].

Slide 8/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Solution continued... This graph has a vertical tangent at x = 1, so we will have to adjust our strategy. The easiest trick to use is to express this function in terms of y (since there are no horizontal asymptotes p present). In this case we solve for y = ln(x + x2 − 1) in terms of x, p ey = x + x2 − 1 p ey − x = x2 − 1 e2y − 2eyx = −1 e2y + 1 ey + e−y x = = 2ey 2 √ √ The interval [1, 2] corresponds to the y-interval [0, ln( 2 + 1)]. The arc length is calculated as √ s Z ln( 2+1)  ey − e−y 2 1 + dy = 1. 0 2 Note that 1 ( y + −y), the hyperbolic cosine ( ) and its ( ), appeared in this 2 e e cosh y sinh y problem.

Slide 9/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Made For Parametric Equations Suppose we have a smooth curve described by the parametric equations x = f (t) and y = g(t). The term smooth means f and g have continuous first that are not simultaneously zero. It is a simple exercise to rewrite our arc length formula in this case. We have Z b q Arc Length = dx2 + dy2 a s Z b  dx2 dy2  = + dt2 2 2 a dt dt s Z b  dx 2  dy 2 = + dt. a dt dt Example Recall the formula for the asteroid is given by x = cos3 t, y = sin3 t, 0 ≤ t ≤ 2π. Find the arc length of one trace of the asteroid.

Slide 10/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Solution continued... We use the symmetry and calculate only the arc length of the first quadrant portion. We have

 dx 2 = 9 cos4 t sin4 t dt  dy 2 = 9 sin4 t cos2 t dt s  dx 2  dy 2 + = 3 cos t sin t dt dt

Therefore the length of the first quadrant portion is given by

Z π/2 3 cos t sin t dt = 3/2. 0 This means the total length of the asteroid is 4(3/2) = 6.

Slide 11/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Surface Areas via Revolution In a previous lecture, we learned how to find the length of a curve using the arclength . This was an important step because it allows us to find the surface area created by rotating a curve about an axis.

If we follow the same strategy we used with arc length, we can approximate the original curve with a piecewise . When a segment of this approximation is rotated about an axis, it creates a simpler figure whose surface area approximates the actual surface area. These bands can be considered a portion of a circular cone, and so we will need to know something about the surface areas of cones. Example What is the formula for the lateral surface area of the cone and the truncated cone?

Slide 12/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Surface Area of Cones A circular cone with base radius r and slant height l can be unfolded by cutting along the slant height. The object obtained is a sector of a circle of radius l, pictured below.

We know that in general, the area of a sector a circle such as the one shown above is given by 1 1  2πr  A = l2θ = l2 = πrl. 2 2 l To derive the formula for the surface area of revolution, we need the truncated cone, shown below:

Slide 13/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Truncated Cone Surface Area

The area of the band with slant height l and upper and lower radii r1 and r2 is found by subtracting the areas of the two cones in the previous slide:

A = πr2(l1 + l) − πr1l1 = π [(r2 − r1)l1 + r2l] .

Using similar triangles, we compute

l l + l 1 = 1 r1 r2 which gives r2l1 = r1l1 + r1l ⇐⇒ (r2 − r1)l1 = r1l. Updating our original equation we have

A = π(r1l + r2l) = 2πrl

where r is the average radius of the band.

Slide 14/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS The Surface Area Integral We apply the previous formula to our strategy. Consider the surface shown below which is obtained by rotating the curve y = f (x), a ≤ x ≤ b about the x-axis where f is positive and has a continuous derivative. We divide the interval [a, b] into n subintervals with endpoints x0, x1,... xn and equal width ∆x. If yi = f (xi), then a band with length l, on the interval [xi−1, xi] will have radii yi−1, yi. Using the formula for the surface area of the truncated cone from before, we have A = π(yi−1 + yi) · l.

As ∆x → 0, yi−1 + yi ≈ 2yi, so we have s  2 q dyi 0 2 A = 2πyi 1 + ∆x = 2πf (xi) 1 + f (xi) ∆x. dx The area of the complete is then given by n q Z b q X 0 2 0 2 S = 2πf (xi) 1 + f (xi) ∆x → 2πf (x) 1 + [f (x)] dx. i=0 a This integral (like the arc length integral before) can take on many forms.

Slide 15/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Best Way to Think About This These formulas can be remembered by thinking of 2πy and 2πx as the circumference of a circle traced out by the (x, y) on the curve as it is rotated about the x-axis or y-axis, respectively.

Example

Find the√ surface area of the solid of revolution obtained by rotating the curve f (x) = x on the interval 1 ≤ x ≤ 4 about the x-axis.

Slide 16/17— Dr. John Ehrke— Lecture 8— Fall 2012 ABILENECHRISTIANUNIVERSITYDEPARTMENTOFMATHEMATICS Solution... √ √ For f (x) = x, we have dy/dx = 1/2 x. This gives the surface area r Z 4 √ 1 Z 4 √ S = 2π x 1 + dx = π 4x + 1 dx. 1 4x 1 Substituting u = 4x + 1 and du = 4 dx. Changing limits to u = 17 and u = 5, we obtain π Z 17 √ π  2 17 π √ √ S = u du = u3/2 = (17 17 − 5 5). 4 5 4 3 5 6

Slide 17/17— Dr. John Ehrke— Lecture 8— Fall 2012