Unformatted text preview:

18.01 Calculus Jason Starr Fall 2005 Lecture 19. October 28, 2005 Homework. Problem Set 5 Part I: (d) and (e); Part II: Problems 2 and 3. Practice Problems. Course Reader: 4A-1, 4A-3, 4B-1, 4B-3, 4B-6. 1. Differentials revisited. In a typical applied integration problem, the main difficulty is finding the integrand and the limits of integration. An unknown quantity, for instance area A, depends on some other quantity, for instance the x-coordinate. The method is to allow the independent variable x vary “infinitesimally” from x to x + dx and then use geometry or science to deduce the corresponding variation dA of the unknown quantity. The deduction is usually intuitive rather than rigorous. What is important is whether the deduction leads to the correct solution. If so, the method of Riemann s ums usually gives a rigorous proof of the intuitive answer. But if the solution is incorrect, no argument will prove it correct. 2. Areas between curves. Given an interval a ≤ x ≤ b and two functions f (x) ≥ g(x) defined on the interval, what is the area of the region bounded by the lines x = a, x = b and the curves� � � � �� � � 18.01 Calculus Jason Starr Fall 2005 y = f (x), y = g(x)? This problem can be solved directly: the area is the difference of the area between y = f (x) and the x-axis and the area between y = g(x) and the x-axis. Each of these is a Riemann integral. The differential method asks, what is the infinitesimal change in the area A as x varies from x to x + dx? The infinitesimal region is a rectangle of base dx and height f (x) − g(x). Thus the infinitesimal change in A is, dA = height × base = (f (x) − g(x))dx. Integrating gives, x=b A = dA = f (x) − g(x)dx. x=a Of course this is the same answer as in the last paragraph. But for other applied integral problems, the differential method may be the only method that works. Example. Find the area bounded by the curve y = x(x2 − 3) and a horizontal tangent line. The horizontal tangent lines are the tangent lines to the curve at critical points. Setting the derivative equal to 0 gives, dy dx = 3x 2 − 3 = 3(x − 1)(x + 1). Thus the critical points are x = ±1. The function is odd, so symmetry suggests the area is the same regardless of the choice of critical point. Thus, choose the critical point x = −1. The corresponding value of the function is, y = (−1)((−1)2 − 3) = (−1)(−2) = 2. This is the equation of the horizontal tangent line. Each intersection point (b, f (b)) of the tangent line with y = x(x2 − 3) occurs at a solution x = b of, x(x 2 − 3) = 2. By hypothesis, x = −1 is a solution. Thus the polynomial factors as, 3 x − 3x − 2 = (x + 1)(x 2 − x − 2) = (x + 1)2(x − 2). The remaining intersection point is (2, 2). So the area bounded by the curve y = x(x2 − 3) and the tangent line y = 2 is, x=2 � 2 2 − (x(x 2 − 3))dx = −x 3 + 3x + 2dx. x=−1 −1 Using the Fundamental Theorem of Calculus, this equals, 24 3x2−x+ + 2x 27/4.= 4 2 −1� � � � � � � �� � � � �� � � 18.01 Calculus Jason Starr Fall 2005 3. Volumes of solids of revolutions: the disk method. If the region in the xy-plane bounded by x = a, x = b, y = f (x) and the x-axis is revolved through xyz-space about the x-axis, what is the volume of the resulting solid? The solid is called a solid of revolution. The disk method applies the method of differentials to solve this problem. As x increases from x to x + dx, the corresponding infinitesimal region of the solid is essentially a disk. The width of the disk is dx. The area of the disk is π[f (x)]2 . Thus the infinitesimal volume of the disk is, dV = Area × width = π[f (x)]2dx. Thus the volume is, x=b V = dV = π[f (x)]2dx. x=a Example. Find the volume of a right circular cone whose base has radius R and whose vertex has height H above the base. The cone is the solid of revolution for the plane region bounded by x = 0, the x-axis, and the line containing (0, R) and (H, 0). The equation of this line is, (H − x)R y = . H Thus the area of an infinitesimal disk is, dV = Area × width = π (H − x)2R2 dx,H2 and the volume is, x=H (H − x)2R2 πV = dV = dx. H2 x=0 Making the substitution u = H − x, du = −dx gives, 0u=0 R2 R2 3u2V = π (−du) = π H2 −u . H2 3u=H H Evaluating gives the volume, V = 23.πR H/In particular, the antiderivative of u2 is responsible for the denominator 3 in the formula for the volume. Example. Find the volume of a sphere of radius R. The sphere of radius R is the solid of revolution for the plane region bounded by the x-axis and 2the upper semicircle y = √R2 − x . Thus the volume is, � R Rx=R 3√R2 22]2dx = π(R2 − x )dx = π R2 x − xV = π[ − x. 3−R −R −Rx=� � � � � � � � �� 18.01 Calculus Jason Starr Fall 2005 Evaluating gives the volume, 43/3.πRV = 4. The slice method. A generalization of the disk method is the slice method. The problem is to find the volume of a region bounded by the planes x = a and x = b given the knowledge of the area A(x) of the slice of the solid by the plane containing (x, 0, 0) parallel to the yz-plane. As x increases from x to x + dx, the corresponding infinitesimal region of the solid is essentially a cylinder. The width of the cylinder is dx. And the area is the area A(x) of the slice. Thus the infinitesimal volume of the cylinder is, dV = Area × width = A(x)dx. Thus the volume is, x=b V = dV = A(x)dx. x=a Example. Find the volume of the “corner” region bounded by the xy-plane, the xz-plane, the yz-plane, and the plane containing (L, 0, 0), (0, L, 0) and (0, 0, L). This region is bounded by the planes x = 0 and x = L. The x-slice of the region is a right isosceles triangle. The base and altitude of the triangle both equal f (x), where y = f (x) is the equation of the line passing through (0, L) and (L, 0). This equation is, f (x) = L − x. Thus the slice area is 1 1 A(x) = base × altitude = (L − x)2 . 2 …


View Full Document

MIT 18 01 - Study Notes

Documents in this Course
Graphing

Graphing

46 pages

Exam 2

Exam 2

3 pages

Load more
Download Study Notes
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 Study Notes 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 Study Notes 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?