Unit 5

Applications of Derivatives

B. Related Rates Problems

Lesson 3: Area and Volume

The previous example shows the importance of identifying variables and constants in the problem. If a quantity remains constant throughout the problem, its value can be substituted before differentiation. If a quantity varies, substitution must take place after differentiation.

Example 4

Helium is pumped into a spherical balloon at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»100«/mn»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»3«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». How fast is the radius increasing when the radius of the balloon is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math»?

Solution

Step 1:

The volume and radius of the balloon are related by the following formula.

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«mo»=«/mo»«mfrac»«mn»4«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»3«/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»V«/mi»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«mrow»«mfrac»«mn»4«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»3«/mn»«/msup»«/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»«mfrac»«mn»4«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«mfenced»«mrow»«mn»3«/mn»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfenced»«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»4«/mn»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«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 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»«mn»100«/mn»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»3«/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»r«/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»5«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi»cm«/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»4«/mn»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«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»«mn»100«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mo»§#960;«/mo»«msup»«mfenced»«mn»5«/mn»«/mfenced»«mn»2«/mn»«/msup»«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»«mn»100«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»100«/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»«mn»1«/mn»«mo»§#960;«/mo»«/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»

The radius of the balloon is increasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mo»§#960;«/mo»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».

Example 5

A cylindrical tank has a radius of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math». Water is pumped in at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi mathvariant=¨normal¨»m«/mi»«mn»3«/mn»«/msup»«mo»/«/mo»«mi»min«/mi»«/math». How fast is the water level rising when the water is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»4«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» deep?

Solution

Step 1:

The volume and depth of the water are related by the following formula. Note the depth of the water changes with time, but the radius of the tank is a constant.

«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»«mn»2«/mn»«mn»2«/mn»«/msup»«mo»§#8729;«/mo»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mo»§#960;«/mo»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/math»

Step 2:

Differentiate with respect to time.

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»«mn»3«/mn»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«msup»«mi mathvariant=¨normal¨»m«/mi»«mn»3«/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»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/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»4«/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»«mn»3«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/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»«mfrac»«mn»3«/mn»«mrow»«mn»4«/mn»«mo»§#960;«/mo»«/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»«/mtable»«/math»

The water’s depth is increasing at a constant rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»3«/mn»«mrow»«mn»4«/mn»«mo»§#960;«/mo»«/mrow»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi»min«/mi»«/math».