Lecture 18: Taylor`s approximation revisited Some time ago, we
... (x − y)n+1 f (n+2) (y)dy. (n + 1)! c Here, the last is by integration by parts. We integrate (x − y)n and differentiate f (n+1) (y) − f (n+1) (c). Note that the boundary terms vanish since (x − y)n vanishes at y = x and f (n+1) (y) − f (n+1) (c) vanishes at y = c. Having this remarkable formula for ...
... (x − y)n+1 f (n+2) (y)dy. (n + 1)! c Here, the last is by integration by parts. We integrate (x − y)n and differentiate f (n+1) (y) − f (n+1) (c). Note that the boundary terms vanish since (x − y)n vanishes at y = x and f (n+1) (y) − f (n+1) (c) vanishes at y = c. Having this remarkable formula for ...
Note 5. Surface Integrals • Parametric equations of surfaces A
... j + ∂z k. Let E be a solid region whose boundary surface S has outward orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then ZZ ...
... j + ∂z k. Let E be a solid region whose boundary surface S has outward orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then ZZ ...
Math 156 Applied Honors Calculus II Fall 2016 hw4 , due: Tuesday
... 2. An object of mass m is moving in a straight line subject to a force F (x), where x is the object’s position. Let v(x) be the object’s velocity as a function of position. a) Show that the work done in moving the object from x0 to x1 is equal to the change in R x1 the object’s kinetic energy, W = x ...
... 2. An object of mass m is moving in a straight line subject to a force F (x), where x is the object’s position. Let v(x) be the object’s velocity as a function of position. a) Show that the work done in moving the object from x0 to x1 is equal to the change in R x1 the object’s kinetic energy, W = x ...
Unit 10 PowerPoint Slides
... An area’s centroid is the point located at the “weighted-average” position of all points in the area. For objects of uniform density, the centroid is the same as the object’s center of mass. ...
... An area’s centroid is the point located at the “weighted-average” position of all points in the area. For objects of uniform density, the centroid is the same as the object’s center of mass. ...
29. How to find the total distance traveled by a
... The definite integral of velocity on [a, b] gives the displacement of a particle on [a, b]. To find the position of a particle given its initial position and the velocity function, add the initial position to the displacement (integral of velocity). To find the total distance traveled on [a, b] by a ...
... The definite integral of velocity on [a, b] gives the displacement of a particle on [a, b]. To find the position of a particle given its initial position and the velocity function, add the initial position to the displacement (integral of velocity). To find the total distance traveled on [a, b] by a ...
Test #3 Topics
... In Section 2.1, we used the idea of instantaneous velocity to introduce the concept of a limit. In Section 3.1 (and again in Section 3.5) that idea was developed further to introduce the derivative. To compute the distance traveled by an object moving along a straight line at constant velocity (e.g. ...
... In Section 2.1, we used the idea of instantaneous velocity to introduce the concept of a limit. In Section 3.1 (and again in Section 3.5) that idea was developed further to introduce the derivative. To compute the distance traveled by an object moving along a straight line at constant velocity (e.g. ...
The Evaluation Theorem
... Then hand in the solution to your problem on December 3 (the day of the next exam). If I can’t do the problem that you hand in (on November 29) and you show me how to do it (in what you hand in on December 3), then I will replace your two lowest problem scores on the next exam with 10s. If I can do ...
... Then hand in the solution to your problem on December 3 (the day of the next exam). If I can’t do the problem that you hand in (on November 29) and you show me how to do it (in what you hand in on December 3), then I will replace your two lowest problem scores on the next exam with 10s. If I can do ...
MA 140 Calculus II Credits: 4 Prerequisites MA 130 Description A
... polar and parametric coordinates, calculus using polar and parametric coordinates, conic sections. Theory is augmented by applications from a variety of disciplines. Learning Outcomes At the completion of this course, students will be able to: • Find a wide variety of anti-derivatives analytically • ...
... polar and parametric coordinates, calculus using polar and parametric coordinates, conic sections. Theory is augmented by applications from a variety of disciplines. Learning Outcomes At the completion of this course, students will be able to: • Find a wide variety of anti-derivatives analytically • ...
Chapter 5: Integration
... and it’s given that 400 (x = 4) units are demanded when price is $30/unit. a. Find the demand function p(x). b. At what price will 300 units be demanded? At what price for zero units? c. How many units are demanded at $20/unit? ...
... and it’s given that 400 (x = 4) units are demanded when price is $30/unit. a. Find the demand function p(x). b. At what price will 300 units be demanded? At what price for zero units? c. How many units are demanded at $20/unit? ...