M1120 Class 5 Dan Barbasch September 6, 2011 Dan Barbasch () - - PowerPoint PPT Presentation

M1120 Class 5

Dan Barbasch September 6, 2011

Dan Barbasch () M1120 Class 5 September 6, 2011 1 / 1

Course Website

http://www.math.cornell.edu/˜web1120/index.html

Dan Barbasch () M1120 Class 5 September 6, 2011 2 / 1

Problems from Section 6.2

Use both methods, washers/shells whenever possible. Exercise 1: [(6) in the text] Compute the volume of the body generated by rotating the region bounded by y = 9x √ x3 + 9 , x = 0, x = 3 about the y-axis. Exercise 2: [(10) in the text] Compute the volume of the body generated by rotating the region bounded by y = 2 − x2, x = 0, y = x2 about the y-axis. Exercise 3: [(22) in the text] Compute the volume of the body generated by rotating the region bounded by y = √x, y = 0, y = 2 − x about the x-axis.

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Problems from Section 6.2

Use both methods, washers/shells whenever possible. Exercise 4: [(24) in the text] Consider the region bounded by y = x3, y = 8, x = 0. Compute the volume of the body generated by this region when rotated about the

1 y-axis. 2 line x = 3. 3 line x = −2. 4 x-axis. 5 line y = 8. 6 line y = −1. Dan Barbasch () M1120 Class 5 September 6, 2011 3 / 1

Arc Length

Formulas: ds2 = dx2 + dy2. ds =

• 1 +

dy dx 2 dx, ds =

• 1 +

dx dy 2 dy. Length of the curve y = f (x) between x = a and x = b (length of x = g(y) between y = a and y = b.) s = b

a

• 1 + f ′(x)2 dx
• r

s = b

a

• 1 + g′(y)2 dy.

These formulas are based on the Pythagorean theorem and linear approximation: (∆s)2 ≈ (∆x)2 + (∆y)2, ∆y ≈ f ′(x)∆x, ∆x ≈ g′(y)∆y.

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Examples

Example 1: Find the length of the curve x = g(y) = 1 2

• ey + e−y

for 0 ≤ y ≤ T. Answer: The general formula is s = T

• 1 + g′(y)2 dy.

In this case, g′(y) = (ey − e−y)/2, and ds =

• 1 + [(ey − e−y)/2]2 dy.

Simplifying we get 1 + (e2y − 2 + e−2y)/4 = (e2y + 2 + e−2y)/4 = ey + e−y 2 2 . So ds = ey+e−y

2

• dy. Integrating we get

s(T) = T ey + e−y 2 dy = ey − e−y 2

• T

= eT − e−T 2 .

Dan Barbasch () M1120 Class 5 September 6, 2011 5 / 1

Examples

The function s(T) we obtained as an answer above is called the arc length function of the curve. It measures the distance along the curve from some point (where y = 0 here) to a “general point” where y = T. Some times you will see s(b) − s(a) = b

a

ds = b

a

• 1 + f ′(x)2 dx.

(s(a) = 0!)

Dan Barbasch () M1120 Class 5 September 6, 2011 5 / 1

Example 2: A particle has position at time t, (x(t), y(t)) = (1 + 2 cos 2t, 2 − 2 sin 2t). How far does the particle travel between times t = 3π and t = 20π? Answer: dx = x′(t) dt and dy = y′(t) dt. In this case, x′(t) = −4 sin 2t, y′(t) = −4 cos 2t. The arc length element is ds2 = dx2 + dy2 =(−4 sin 2t)2dt2 + (−4 cos 2t)2dt2 =16(sin2 2t + cos2 2t) dt2 = 16dt2. Then s(20π) − s(3π) = 20π

3π

4 dt = 4[20π − 3π]. and s(3π) = 0. This is distance traveled, not displacement.

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Questions: What is the path of the object? What is its position at time t?

Dan Barbasch () M1120 Class 5 September 6, 2011 6 / 1

Questions: What is the path of the object? What is its position at time t? Answers: The object is moving on a circle of radius 2 centered at (1, 2) because (x − 1)2 + (y − 2)2 = 4. Its angular velocity is −2radians.

Dan Barbasch () M1120 Class 5 September 6, 2011 6 / 1

Problems from section 6.3

Find the length of the following curves: Problem 7: y = 3 4x

4 3 − 3

8x

2 3 + 5 for 1 ≤ x ≤ 8.

Problem 8: y = x3 3 + x2 + x + 1 4x + 4 for 0 ≤ x ≤ 2. Problem 18: x = y

• sec2 t − 1 dt for − π

3 ≤ y ≤ π 4 .

Problem 21: y = x

• cos (2t) dt for 0 ≤ x ≤ π

4

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Solution to problem 18:

Write g(y) := x √ sec2 t − 1 dt. The arc length element is ds2 = (1 + g′(y)2) dy2. Since g(y)′ =

• sec2y − 1,

ds2 = (1 + sec2 y − 1) dy2 = sec2 y dy2, s = π/4

−π/3

sec y dy = ln | sec y + tan y||π/4

−π/3

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Surface Area

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Surface Area

Central to surface area of a surface of revolution is the idea of approximating a little rotated piece of the surface by such a frustrum.

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Surface Area

Central to surface area of a surface of revolution is the idea of approximating a little rotated piece of the surface by such a frustrum. Before rotation, the piece of curve has length ds.

Dan Barbasch () M1120 Class 5 September 6, 2011 9 / 1

Surface Area

Central to surface area of a surface of revolution is the idea of approximating a little rotated piece of the surface by such a frustrum. Before rotation, the piece of curve has length ds. After rotation, the area dA of the rotated piece is: dA = 2π radius ds . So the total area is = b

a

2πr ds . The surface area of a region obtained by rotating the curve y = f (x) ≥ 0 between x = a and x = b about the x−axis is A = b

a

2πf (x)

• 1 + f ′(x)2 dx.

Dan Barbasch () M1120 Class 5 September 6, 2011 9 / 1

Problems from Section 6.4

Problem 18: Find the surface area of the region obtained by rotating x = 1 3y

3 2 − y 1 2 about the y-axis for 1 ≤ y ≤ 3

Problem 26:

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Problems from Section 6.4

Problem 30: Find the surface area (without bottom ) of the 90ft NWS Dome

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Hints to the Exercises:

Problem 18: The graph is

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Hints to the Exercises:

Problem 26: Since y = (r/h)x, dy = (r/h)dx, so ds =

• 1 + (r/h)2 dx.

So the area is A = 2π h y

• 1 + (r/h)2 dx =2π

h (r/h)x

• 1 + (r/h)2 dx =

=πr

• r2 + h2.

Verify the arithmetic.

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Hints to the Exercises:

Problem 30: Rotate the curve x =

• R2 − y2 for −r ≤ y ≤ R about the

y−axis.

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