L3 Integration - Part 2
Completion requirements
Unit 7B
Integrals Part 2
Lesson 3: Integration
Displacement from Velocity
Recall from Lessons 1 and 2, for objects travelling along a line, velocity «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/math» is defined as a change in position (displacement) «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/math» over time «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math», and the velocity function can be determined by differentiating the position function «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/math». That is, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math». As such, it is possible to work backward from the velocity function to determine the position function by taking the antiderivative of the velocity function.Suppose Anna sleds down a hill in the winter. She reaches a velocity of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math»
at the bottom of the hill. From that point on, she coasts in a straight line. If her coasting velocity is given by «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mo»§#8722;«/mo»«mn»2«/mn»«mi»t«/mi»«mo»+«/mo»«mn»10«/mn»«/math»,
what is Anna’s position «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/math», measured from the bottom of the hill, at any time «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math»,
where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»§#8805;«/mo»«mn»0«/mn»«/math»? How far will she coast?
The velocity function «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mo»§#8722;«/mo»«mn»2«/mn»«mi»t«/mi»«mo»+«/mo»«mn»10«/mn»«/math», where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»§#8805;«/mo»«mn»0«/mn»«/math»,
may be written as a differential equation.
Solve the differential equation by finding the indefinite integral.
To determine the value of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«/math», substitute the initial conditions into the position function.
Since Anna’s position is measured from the bottom of the hill, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»0«/mn»«/math» when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»0«/mn»«/math».
Therefore, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mo»§#8722;«/mo»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mi»t«/mi»«/math», where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»§#8805;«/mo»«mn»0«/mn»«/math».
Anna will continue to slide until her velocity is zero.
Find «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»0«/mn»«/math».
Anna will continue to coast for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
Now, find Anna’s position at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«/math».
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», Anna will be «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» from the bottom of the hill.
Since «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«/math» at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»0«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math» and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»25«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»s«/mi»«/math», the change in Anna’s position (or displacement) is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math». Since Anna coasts in the same direction for the entire «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», the distance she coasts before coming to rest will be the same as the displacement.
Another approach to this problem is to start with the graph of the velocity function.
Note the distance travelled is equal to the shaded area between the velocity function and the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math»-axis.
The area can be calculated using the formula for the area of a triangle.
The area can also be calculated from the definite integral.
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», Anna will have coasted «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» from the bottom of the hill.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mo»§#8722;«/mo»«mn»2«/mn»«mi»t«/mi»«mo»+«/mo»«mn»10«/mn»«/math»
Solve the differential equation by finding the indefinite integral.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8747;«/mo»«mfenced»«mrow»«mo»§#8722;«/mo»«mn»2«/mn»«mi»t«/mi»«mo»+«/mo»«mn»10«/mn»«/mrow»«/mfenced»«mi
mathvariant=¨normal¨» «/mi»«mi»d«/mi»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mi»t«/mi»«mo»+«/mo»«mi»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
To determine the value of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«/math», substitute the initial conditions into the position function.
Since Anna’s position is measured from the bottom of the hill, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»0«/mn»«/math» when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»0«/mn»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mi»t«/mi»«mo»+«/mo»«mi»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mn»0«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«msup»«mfenced»«mn»0«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mfenced»«mn»0«/mn»«/mfenced»«mo»+«/mo»«mi»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mn»0«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»C«/mi»«/mtd»«/mtr»«/mtable»«/math»
Therefore, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mo»§#8722;«/mo»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mi»t«/mi»«/math», where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»§#8805;«/mo»«mn»0«/mn»«/math».
Anna will continue to slide until her velocity is zero.
Find «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»0«/mn»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»2«/mn»«mi»t«/mi»«mo»+«/mo»«mn»10«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»0«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»2«/mn»«mi»t«/mi»«mo»+«/mo»«mn»10«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»2«/mn»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»10«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«/mtd»«/mtr»«/mtable»«/math»
Anna will continue to coast for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
Now, find Anna’s position at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mi»t«/mi»«mo»+«/mo»«mi»C«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»s«/mi»«mfenced»«mn»5«/mn»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«msup»«mfenced»«mn»5«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mfenced»«mn»5«/mn»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»25«/mn»«mo»+«/mo»«mn»50«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»25«/mn»«/mtd»«/mtr»«/mtable»«/math»
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», Anna will be «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» from the bottom of the hill.
Since «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«/math» at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»0«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math» and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mn»25«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»s«/mi»«/math», the change in Anna’s position (or displacement) is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math». Since Anna coasts in the same direction for the entire «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», the distance she coasts before coming to rest will be the same as the displacement.
Another approach to this problem is to start with the graph of the velocity function.
Note the distance travelled is equal to the shaded area between the velocity function and the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math»-axis.
The area can be calculated using the formula for the area of a triangle.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»A«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»b«/mi»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mn»5«/mn»«/mfenced»«mfenced»«mn»10«/mn»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»25«/mn»«/mtd»«/mtr»«/mtable»«/math»
The area can also be calculated from the definite integral.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»A«/mi»«mfenced»«mn»5«/mn»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msubsup»«mo»§#8747;«/mo»«mn»0«/mn»«mn»5«/mn»«/msubsup»«mfenced»«mrow»«mo»§#8722;«/mo»«mn»2«/mn»«mi»t«/mi»«mo»+«/mo»«mn»10«/mn»«/mrow»«/mfenced»«mi
mathvariant=¨normal¨» «/mi»«mi»d«/mi»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«msubsup»«mfenced open=¨[¨ close=¨]¨»«mrow»«mo»§#8722;«/mo»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mi»t«/mi»«/mrow»«/mfenced»«mn»0«/mn»«mn»5«/mn»«/msubsup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced»«mrow»«mo»§#8722;«/mo»«msup»«mfenced»«mn»5«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mfenced»«mn»5«/mn»«/mfenced»«/mrow»«/mfenced»«mo»§#8722;«/mo»«mfenced»«mrow»«mo»§#8722;«/mo»«msup»«mfenced»«mn»0«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»10«/mn»«mfenced»«mn»0«/mn»«/mfenced»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»25«/mn»«mo»§#8722;«/mo»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»25«/mn»«/mtd»«/mtr»«/mtable»«/math»
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mn»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math», Anna will have coasted «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» from the bottom of the hill.
- The position «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/math» of an object is the solution to the differential equation generated from the velocity function «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«/math». Therefore, if «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math», then «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mo»=«/mo»«mo»§#8747;«/mo»«mi»v«/mi»«mfenced»«mi»t«/mi»«/mfenced»«mi mathvariant=¨normal¨» «/mi»«mi»d«/mi»«mi»t«/mi»«mo»+«/mo»«mi»C«/mi»«/math». The value of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«/math» is evaluated from initial conditions.
- The distance travelled is the area between the graph of the velocity function and the «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math»-axis over the desired interval.