Chapter 15.2: Sample "Work" problems for Math 293

Return to the Math 293 Chapter 15 Section 2 Topics page Revised July 20, 2014
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Chapter 15
Section 15.2: "Work" problems
Before we start, let us recall that there are many ways to present such problems. The Finney, Thomas, demana, and Waits Calculus book has the following table:
Different Ways to Write the Work Integral
Work=`int_(t=a)^(t=b)bbF•bbTds` The definition
Work=`int_(t=a)^(t=b)bbF•dbbr` Compact differential form
Work=`int_a^b bbF•(dbbr)/(dt)dt` Expanded to incude `dt`. Emphasizes the velocity vector `(dbbr)/(dt)`
Work=`int_a^b (M(dg)/(dt)+N(dh)/(dt)+P(dk)/(dt))dt` Emphasizes the component functions
Work=`int_a^b (M(dx)/(dt)+N(dy)/(dt)+P(dy)/(dt))dt` Abbreviates the components of `bbr`.
Work=`int_a^b Mdx+Ndy+Pdz` `dt^'s` canceled. The most common differential form.
The book goes on to lay out the general method of solving such problems

Evaluation
To evaluate the work integral, take these steps:
  1. Evaluate `bbF` on the curve as a function of `t`.
  2. Find `(dbbr)/(dt)`.
  3. Dot `bbF` with `(dbbr)/(dt)`
  4. Integrate from `t=a` to `t=b`

Problem 1: find the work done by the vector field:
`bbF = (y+z-x^2)bbi + (x+z-y^2)bbj +(x+y-z^2)bbk`
over the curve
`bbr=tbbi+t^2bbk + t^3bbk, 0<=t<=1`
from `(0,0,0)` to `(1,1,1)`.
Evaluate `bbF`
in terms of `t`		 `F=(t^2+t^3-(t^2)) bbi + (t+t^3-(t^2)^2) bbj +(t +t^2 -(t^3)^2) bbk`
`F=t^3 bbi + (t+t^3-t^4) bbj + (t+t^2-t^6) bbk`
Find `(dbbr)/(dt)` `(dbbr)/(dt) = d/(dt) (tbbi + t^2bbj t^3bbk)`
`(dbbr)/(dt) = = 1bbi +2t bbj + 3t^2 bbk`
Find `bbF•(dbbr)/(dt)` `bbF•(dbbr)/(dt) = t^3(1) + (t+t^3-t^4)(2t) +(t+t^2-t^6)(3t^2)`
`bbF•(dbbr)/(dt) = t^3 +2t^2 +2t^4 - 2t^5 +3t^3 +3t^4-3t^8`
`bbF•(dbbr)/(dt) = 2t^2 + 4t^3 + 5t^4 - 2t^5 - 3t^8`
Find the
definite integal		 `int_0^1 2t^2 + 4t^3 + 5t^4 - 2t^5 - 3t^8 dt= (2/3)t^3 + t^4 + t^5 - (2/6)t^6 - (3/9)t^9]_(\quad0)^(\quad1)`
`int_0^1 2t^2 + 4t^3 + 5t^4 - 2t^5 - 3t^8 dt= (2/3 + 1 + 1 -1/3 -1/3) -(0) = 2`

Problem 2: find the work done by the vector field:
`bbF = xbbi + zbbj + ybbk`
over the curve
`bbr=3cos(t)i+3sin(t)k + (t/2)k, 0<=t<=pi/6`
from `(0,0,0)` to `((3sqrt(3))/2,3/2,pi/12)`.

Evaluate `bbF`
in terms of `t`		 `bbF = 3cos(t)bbi + (t/2)bbj + 3sin(t)bbk`
Find `(dbbr)/(dt)` `(dbbr)/(dt)= d/(dt)(3cos(t)i+3sin(t)k + (t/2)k)`
`(dbbr)/(dt)= -3sin(t)bbi +3cos(t)bbj + (1/2)k`
Find `bbF•(dbbr)/(dt)` `bbF•(dbbr)/(dt) = (3cost(t))(-3sin(t)) + (t/2)(3cos(t)) + 3sin(t)(1/2)`
`bbF•(dbbr)/(dt) = -9sin(t)cos(t) +(2t)/3cost(t) +3/2sin(t)`
Find the
definite integal		 `int_0^(pi/6) -9sin(t)cos(t) +(2t)/3cost(t) +3/2sin(t) dt = (9/2)cos^2(t) + (3/2)(t sin(t))]_(\quad0)^(\quadpi/6)`
`int_0^(pi/6) -9sin(t)cos(t) +(2t)/3cost(t) +3/2sin(t) dt = (9/2)(3/4)+(3/2)(pi/6)(1/2)-(9/2-0)`
`int_0^(pi/6) -9sin(t)cos(t) +(2t)/3cost(t) +3/2sin(t) dt = pi/8-9/8`

Return to the Math 293 Chapter 15 Section 2 Topics page

©Roger M. Palay
Saline, MI 48176
July, 2014