L3 Area and Volume - Part 3
Completion requirements
Unit 5
Applications of Derivatives
B. Related Rates Problems
Lesson 3: Area and Volume
When solving related rates problems, implicit differentiation is applied. The only difference from what has been done in previous Units is that all variables are differentiated with respect to time. A method for solving related rates problems is
outlined below.
- Determine a relationship between the variable(s) for which the rate(s) of change is (are) given and the variable for which the rate of change must be found.
- Differentiate the relation implicitly with respect to time.
- Substitute the values given into the differentiated relation and solve for the desired rate.
A circular ring is heated so that it expands. If the radius is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»100«/mn»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«mo»/«/mo»«mi
mathvariant=¨normal¨»s«/mi»«/math», determine the rate at which the circumference is increasing.
Step 1:
The circumference of a circle is related to the radius by the following formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mo»=«/mo»«mn»2«/mn»«mo»§#960;«/mo»«mi»r«/mi»«/math»
Step 2:
Differentiate with respect to time.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mi»C«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mrow»«mn»2«/mn»«mo»§#960;«/mo»«mi»r«/mi»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»C«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#960;«/mo»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Step 3:
Since the ring is expanding, the rate of change of the radius is positive. Substitute «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»100«/mn»«/mfrac»«mi
mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
The circumfrence of the ring is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mo»§#960;«/mo»«mn»50«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»C«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#960;«/mo»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»C«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#960;«/mo»«mfenced»«mfrac»«mn»1«/mn»«mn»100«/mn»«/mfrac»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mo»§#960;«/mo»«mn»50«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
The circumfrence of the ring is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mo»§#960;«/mo»«mn»50«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
A circular disk is heated and then cooled. During the cooling process, the radius of the disk is found to be decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»50«/mn»«/mfrac»«mi mathvariant=¨normal¨»
«/mi»«mi»mm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». At what rate is the area of the disk changing when the radius of the disk is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»100«/mn»«mi»mm«/mi»«/math»?
Step 1:
The area and radius of the disk are related by the following formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»A«/mi»«mo»=«/mo»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/math»
Step 2:
Differentiate with respect to time.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mi»A«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mrow»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#960;«/mo»«mi»r«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Step 3:
Since the radius is decreasing with time, the rate of change is negative. Thus, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»50«/mn»«/mfrac»«mi
mathvariant=¨normal¨» «/mi»«mi»mm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». Solve for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math»
when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«mo»=«/mo»«mn»100«/mn»«mi mathvariant=¨normal¨» «/mi»«mi»mm«/mi»«/math».
The area of the disk is decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»4«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»mm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#960;«/mo»«mi»r«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#960;«/mo»«mfenced»«mn»100«/mn»«/mfenced»«mfenced»«mrow»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»50«/mn»«/mfrac»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»4«/mn»«mo»§#960;«/mo»«/mtd»«/mtr»«/mtable»«/math»
The area of the disk is decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»4«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»mm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
Water is being poured into an aquarium that is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»60«/mn»«mi»cm«/mi»«/math» long, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»30«/mn»«mi»cm«/mi»«/math» wide, and «math
xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»40«/mn»«mi»cm«/mi»«/math» deep. Determine a relationship between the rate at which water is being poured and the rate at which the depth is increasing.
Step 1:
Determine the relationship between the rate of change of the volume of water and the rate of change of the depth of water.
As water is poured into the aquarium, it takes the shape of the container. Thus, the rate at which the water is poured into the aquarium is the rate of change of water volume.
As water is poured into the aquarium, it takes the shape of the container. Thus, the rate at which the water is poured into the aquarium is the rate of change of water volume.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«mo»=«/mo»«mi»l«/mi»«mo»§#8729;«/mo»«mi»w«/mi»«mo»§#8729;«/mo»«mi»h«/mi»«/math»
Of the three dimensions, the only one that varies is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»V«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»60«/mn»«mo»§#8729;«/mo»«mn»30«/mn»«mo»§#8729;«/mo»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/math»
| Note: Think about the statement “Of the three dimensions, the only one that varies is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math».” The length and width of the aquarium are fixed. As such, when the relation is differentiated with respect to time, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»l«/mi»«/math» and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»w«/mi»«/math» are constants, as shown by the substitution. |
Step 2:
Differentiate with respect to time.
This is the relationship between the rate of change of the volume and the rate of change of the depth of water.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mi»V«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«mi»h«/mi»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»V«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
This is the relationship between the rate of change of the volume and the rate of change of the depth of water.