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 9 x √ rotating the region bounded by y = , x = 0, x = 3 about the x 3 + 9 y -axis. Exercise 2: [(10) in the text] Compute the volume of the body generated by rotating the region bounded by y = 2 − x 2 , x = 0, y = x 2 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. Dan Barbasch () M1120 Class 5 September 6, 2011 3 / 1

Problems from Section 6.2 Use both methods, washers/shells whenever possible. Exercise 4: [(24) in the text] Consider the region bounded by y = x 3 , 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: ds 2 = dx 2 + dy 2 . � � � 2 � 2 � dy � dx ds = 1 + dx , ds = 1 + dy . dx 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 . ) � b � b � � 1 + f ′ ( x ) 2 dx 1 + g ′ ( y ) 2 dy . s = s = or a a 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 . Dan Barbasch () M1120 Class 5 September 6, 2011 4 / 1

Examples Example 1: Find the length of the curve x = g ( y ) = 1 e y + e − y � � for 2 0 ≤ y ≤ T . � T � 1 + g ′ ( y ) 2 dy . Answer: The general formula is s = 0 In this case, g ′ ( y ) = ( e y − e − y ) / 2 , and ds = 1 + [( e y − e − y ) / 2] 2 dy . � Simplifying we get � e y + e − y � 2 1 + ( e 2 y − 2 + e − 2 y ) / 4 = ( e 2 y + 2 + e − 2 y ) / 4 = . 2 So ds = e y + e − y dy . Integrating we get 2 � T e y + e − y dy = e y − e − y = e T − e − T T � � s ( T ) = . � 2 2 2 0 � 0 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 � b � b � 1 + f ′ ( x ) 2 dx . s ( b ) − s ( a ) = ds = a a ( 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 2 t , 2 − 2 sin 2 t ) . How far does the particle travel between times t = 3 π and t = 20 π ? Answer: dx = x ′ ( t ) dt and dy = y ′ ( t ) dt . In this case, x ′ ( t ) = − 4 sin 2 t , y ′ ( t ) = − 4 cos 2 t . The arc length element is ds 2 = dx 2 + dy 2 =( − 4 sin 2 t ) 2 dt 2 + ( − 4 cos 2 t ) 2 dt 2 =16(sin 2 2 t + cos 2 2 t ) dt 2 = 16 dt 2 . Then � 20 π s (20 π ) − s (3 π ) = 4 dt = 4[20 π − 3 π ] . 3 π and s (3 π ) = 0 . This is distance traveled, not displacement. 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 ? 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 − 2 radians . 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 3 − 3 4 2 3 + 5 for 1 ≤ x ≤ 8. 4 x 8 x Problem 8: y = x 3 1 3 + x 2 + x + 4 x + 4 for 0 ≤ x ≤ 2. � y � sec 2 t − 1 dt for − π 3 ≤ y ≤ π Problem 18: x = 4 . 0 � x � cos (2 t ) dt for 0 ≤ x ≤ π Problem 21: y = 4 0 Dan Barbasch () M1120 Class 5 September 6, 2011 7 / 1

Solution to problem 18: √ � x sec 2 t − 1 dt . The arc length element is Write g ( y ) := 0 ds 2 = (1 + g ′ ( y ) 2 ) dy 2 . Since g ( y ) ′ = � sec 2 y − 1 , ds 2 = (1 + sec 2 y − 1) dy 2 = sec 2 y dy 2 , � π/ 4 sec y dy = ln | sec y + tan y || π/ 4 s = − π/ 3 − π/ 3 Dan Barbasch () M1120 Class 5 September 6, 2011 8 / 1

Surface Area 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. 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 . 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 = 2 π radius ds . � b So the total area is = 2 π r ds . a 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 � b � 1 + f ′ ( x ) 2 dx . A = 2 π f ( x ) a 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 3 1 2 − y 2 about the y -axis for 1 ≤ y ≤ 3 3 y Problem 26: Dan Barbasch () M1120 Class 5 September 6, 2011 10 / 1

Problems from Section 6.4 Problem 30: Find the surface area (without bottom ) of the 90 ft NWS Dome Dan Barbasch () M1120 Class 5 September 6, 2011 10 / 1

Hints to the Exercises: Problem 18: The graph is Dan Barbasch () M1120 Class 5 September 6, 2011 11 / 1

Hints to the Exercises: 1 + ( r / h ) 2 dx . � Problem 26: Since y = ( r / h ) x , dy = ( r / h ) dx , so ds = So the area is � h � h � � 1 + ( r / h ) 2 dx =2 π 1 + ( r / h ) 2 dx = A = 2 π ( r / h ) x y 0 0 � r 2 + h 2 . = π r Verify the arithmetic. Dan Barbasch () M1120 Class 5 September 6, 2011 11 / 1

Hints to the Exercises: R 2 − y 2 for − r ≤ y ≤ R about the � Problem 30: Rotate the curve x = y − axis. Dan Barbasch () M1120 Class 5 September 6, 2011 11 / 1