L3 Area and Volume - Practice 1
Completion requirements
Unit 5
Applications of Derivatives
B. Related Rates Problems
Lesson 3: Area and Volume
Practice
Once you feel confident with Related Rates: Area and Volume, click on the Practice tab and complete problems 1 to 5. Check your answers by going to the Solutions tab.
Instructions: Click the Download File button to download a printable PDF of the questions. Answer each of the following practice questions on a separate piece of paper. Step by step solutions are provided under the Solutions tab. You will learn the material more thoroughly if you complete the questions before checking the answers.
Click here for a formula sheet for Unit 5.
Instructions: Click the Download File button to download a printable PDF of the questions. Answer each of the following practice questions on a separate piece of paper. Step by step solutions are provided under the Solutions tab. You will learn the material more thoroughly if you complete the questions before checking the answers.
Click here for a formula sheet for Unit 5.
1.
The area of a circle is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«/math», and it is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mo»§#960;«/mo»«mn»2«/mn»«/mfrac»«mi
mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi»min«/mi»«/math». Find the rate at which the radius of the circle is increasing.
2.
The area of a square is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». At what rate
is the side length increasing when the side length is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»200«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math»?
3.
A rectangular container is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» long, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math»
wide, and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» deep. Water is being pumped in at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mo»§#160;«/mo»«msup»«mi
mathvariant=¨normal¨»m«/mi»«mn»3«/mn»«/msup»«mo»/«/mo»«mi»min«/mi»«/math». How fast is the surface of the water rising?
4.
The radius of a spherical balloon is decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». At what rate is the surface area of the balloon decreasing
when the radius is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»14«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math»?
5.
A cylindrical vase has a height of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»53«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math» and a diameter of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»24«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math». It is
being filled with water such that the depth is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». Determine the rate at which the
water is being poured into the vase.
1.
The area of a circle is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«/math», and it is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mo»§#960;«/mo»«mn»2«/mn»«/mfrac»«mi
mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi»min«/mi»«/math». Find the rate at which the radius of the circle is increasing.
Step 1:
The area and radius of the circle 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 area is increasing with time, the rate of change is positive. Thus, «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»«mo»=«/mo»«mfrac»«mo»§#960;«/mo»«mn»2«/mn»«/mfrac»«mi
mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi»min«/mi»«/math». Solve for «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»«/math».
Determine the radius of the circle when the area of the circle is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«/math».
Substitute «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«mo»=«/mo»«mn»3«/mn»«/math» into «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»«mrow»«mn»4«/mn»«mi»r«/mi»«/mrow»«/mfrac»«/math» to determine the rate at which the radius is changing.
The radius of the circle is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»12«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi»cm«/mi»«mo»/«/mo»«mi»min«/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»«mfrac»«mo»§#960;«/mo»«mn»2«/mn»«/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»«mfrac»«mo»§#960;«/mo»«mrow»«mn»2«/mn»«mfenced»«mrow»«mn»2«/mn»«mo»§#960;«/mo»«mi»r«/mi»«/mrow»«/mfenced»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»1«/mn»«mrow»«mn»4«/mn»«mi»r«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Determine the radius of the circle when the area of the circle is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«/math».
«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»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mn»9«/mn»«mo»§#960;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»9«/mn»«mo»§#960;«/mo»«/mrow»«mo»§#960;«/mo»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mn»9«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mn»3«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»r«/mi»«/mtd»«/mtr»«/mtable»«/math»
Substitute «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«mo»=«/mo»«mn»3«/mn»«/math» into «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»«mrow»«mn»4«/mn»«mi»r«/mi»«/mrow»«/mfrac»«/math» to determine the rate at which the radius is changing.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»r«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mrow»«mn»4«/mn»«mi»r«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mrow»«mn»4«/mn»«mfenced»«mn»3«/mn»«/mfenced»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»12«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
The radius of the circle is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»12«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi»cm«/mi»«mo»/«/mo»«mi»min«/mi»«/math».
2.
The area of a square is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». At what rate
is the side length increasing when the side length is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»200«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math»?
Step 1:
The area and side length of a square are related by the following formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»A«/mi»«mo»=«/mo»«msup»«mi»s«/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»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«/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»«mi»s«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Step 3:
Since the area is increasing with time, the rate of change is positive. Thus, «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»«mo»=«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi
mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«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»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math»
when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«mo»=«/mo»«mn»200«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«/math».
The side length is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»40«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«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»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi»s«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mn»10«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mfenced»«mn»200«/mn»«/mfenced»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»10«/mn»«mn»400«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»1«/mn»«mn»40«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
The side length is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»40«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
3.
A rectangular container is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» long, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math»
wide, and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» deep. Water is being pumped in at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mo»§#160;«/mo»«msup»«mi
mathvariant=¨normal¨»m«/mi»«mn»3«/mn»«/msup»«mo»/«/mo»«mi»min«/mi»«/math». How fast is the surface of the water rising?
Step 1:
The volume and height of the water are related by the following formula. Note the height of the water changes with time, but the length and width of the container are constants.
«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»«mi»l«/mi»«mo»§#8729;«/mo»«mi»w«/mi»«mo»§#8729;«/mo»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo»§#8729;«/mo»«mn»2«/mn»«mo»§#8729;«/mo»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/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»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»6«/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»6«/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»
Step 3:
Since the volume is increasing with time, the rate of change is positive. Thus, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»V«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«mo»§#160;«/mo»«msup»«mi»cm«/mi»«mn»3«/mn»«/msup»«mo»/«/mo»«mi»min«/mi»«/math».
The water level is rising at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»4«/mn»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi»min«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«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»6«/mn»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»3«/mn»«mn»2«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»3«/mn»«mrow»«mn»2«/mn»«mfenced»«mn»6«/mn»«/mfenced»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»1«/mn»«mn»4«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
The water level is rising at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»4«/mn»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi»min«/mi»«/math».
4.
The radius of a spherical balloon is decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». At what rate is the surface area of the balloon decreasing
when the radius is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»14«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math»?
Step 1:
The surface area and radius of a balloon are related by the following formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»S«/mi»«mi»A«/mi»«mo»=«/mo»«mn»4«/mn»«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»«mrow»«mi»S«/mi»«mi»A«/mi»«/mrow»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mrow»«mn»4«/mn»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»S«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»8«/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»«mn»2«/mn»«mo»§#160;«/mo»«mi
mathvariant=¨normal¨» «/mi»«mi»cm«/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»S«/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»14«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi»cm«/mi»«/math».
The surface area of the balloon is decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»224«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/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»S«/mi»«mi»A«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»8«/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»8«/mn»«mo»§#960;«/mo»«mfenced»«mn»14«/mn»«/mfenced»«mfenced»«mrow»«mo»§#8722;«/mo»«mn»2«/mn»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»224«/mn»«mo»§#960;«/mo»«/mtd»«/mtr»«/mtable»«/math»
The surface area of the balloon is decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»224«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
5.
A cylindrical vase has a height of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»53«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math» and a diameter of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»24«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math». It is
being filled with water such that the depth is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». Determine the rate at which
the water is being poured into the vase.
Step 1:
The volume and height of the water in the vase are related by the following formula.
«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»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#960;«/mo»«mo»§#8729;«/mo»«msup»«mfenced»«mn»12«/mn»«/mfenced»«mn»2«/mn»«/msup»«mo»§#8729;«/mo»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»144«/mn»«mo»§#960;«/mo»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/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»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»144«/mn»«mo»§#960;«/mo»«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»144«/mn»«mo»§#960;«/mo»«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»
Step 3:
Since the height is increasing with time, the rate of change is positive. Thus, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi
mathvariant=¨normal¨» «/mi»«mi»cm«/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»V«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math».
The water is being poured into the vase at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»288«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»3«/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»V«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»144«/mn»«mo»§#960;«/mo»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»144«/mn»«mo»§#960;«/mo»«mo»§#8729;«/mo»«mfenced»«mn»2«/mn»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»288«/mn»«mo»§#960;«/mo»«/mtd»«/mtr»«/mtable»«/math»
The water is being poured into the vase at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»288«/mn»«mo»§#960;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»3«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».