DOC PREVIEW
MIT 18 01 - Parametric Equations, Arclength, Surface Area

This preview shows page 1-2 out of 5 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 5 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 5 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 5 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 31 18.01 Fall 2006 Lecture 31: Parametric Equations, Arclength,Surface Area Arclength, continued Example 1. Consider this parametric equation: x = t2 y = t3 for 0 ≤ t ≤ 1 x 3 = (t2)3 = t6; y 2 = (t3)2 = t6 = ⇒ x 3 = y 2 = ⇒ y = x 2/3 0 ≤ x ≤ 1 dsdydxdsdydxFigure 1: Infinitesimal Arclength. (ds)2 = (dx)2 + (dy)2 (ds)2 = (2t dt)2 + (3t2 dt)2 = (4t2 + 9t4)(dt)2 � �� � � �� � (dx)2 (dy)2 � t=1 � 1 �� 1 � Length = ds = 4t2 + 9t4dt = t 4 + 9t2dt t=0 0 0 1 = (4 + 9t2)3/2 ��� = 1 (133/2 − 43/2)27 0 27 Even if you can’t evaluate the integral analytically, you can always use numerical methods. 1Lecture 31 18.01 Fall 2006 Surface Area (surfaces of revolution) ydsa byxFigure 2: Calculating surface area ds (the infinitesimal curve length in Figure 2) is revolved a distance 2πy. The surface area of the thin strip of width ds is 2πy ds. Example 2. Revolve Example 1 (x = t2, y = t3 , 0 ≤ t ≤ 1) around the x-axis. Refer to Figure 3. yxFigure 3: Curved surface of a trumpet. 2� Lecture 31 18.01 Fall 2006 �� 13 �� 1 �2π t t 4 + 9t2 dt 4Area = 2πy ds = 0 ����� �� � = 2π t4 + 9t2 dt y ds 0 Now, we discuss the method used to evaluate t4(4 + 9t2)1/2dt We’re going to ignore the factor of 2π. You can reinsert it once you’re done evaluating the integral. We use the trigonometric substitution 2 2 t = tan u; dt = sec2 u du; tan2 u + 1 = sec2 u3 3 Putting all of this together gives us: � �� �4 � � ��1/2 � �2 4 2 t4(4 + 9t2)1/2 dt = tan u 4 + 9 tan2 u sec2 u du 3 9 3 � �5 �2 = tan4 u(2 sec u)(sec2 u du)3 This is a tan − sec integral. It’s doable, but it will take a long time for you to work the whole thing out. We’re going to stop evaluating it here. Example 3 Let’s use what we’ve learned to find the surface area of the unit sphere (see Figure 4). abrotate the curve by 2π radians xy..Figure 4: Slice of spherical surface (orange peel, only, not the insides). 3Lecture 31 18.01 Fall 2006 For the top half of the sphere, � y = 1 − x2 We want to find the area of the spherical slice between x = a and x = b. A spherical slice has area � x=b A = 2πy ds x=a From last time, dx ds = √1 − x2 Plugging that in yields a remarkably simple formula for A: � b � dx � b A = a 2π 1 − x2 √1 − x2 = a 2π dx = 2π(b − a) Special Cases For a whole sphere, a = −1, and b = 1. 2π(1 − (−1)) = 4π is the surface area of a unit sphere. For a half sphere, a = 0 and b = 1. 2π(1 − 0) = 2π


View Full Document

MIT 18 01 - Parametric Equations, Arclength, Surface Area

Documents in this Course
Graphing

Graphing

46 pages

Exam 2

Exam 2

3 pages

Load more
Download Parametric Equations, Arclength, Surface Area
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Parametric Equations, Arclength, Surface Area and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Parametric Equations, Arclength, Surface Area 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?