L5 Related Motion - Part 4
Completion requirements
Unit 5
Applications of Derivatives
B. Related Rates Problems
Lesson 5: Related Motion
Engine oil is added to the engine of a car. The funnel used has a radius of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math» and the height of the cone-shaped portion is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»15«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math».
The oil is drained from the funnel at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»12«/mn»«mo»§#160;«/mo»«msup»«mi»cm«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math». When the oil level is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mo»§#160;«/mo»«mi»cm«/mi»«/math»
from the bottom, find the rate at which the oil level is falling.
The volume of a cone is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«mi»h«/mi»«/math».
Step 1:
Draw and label a diagram.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«/math» be the volume of oil, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msup»«mi»cm«/mi»«mn»3«/mn»«/msup»«/math», at time «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math».
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math» be the oil level, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»cm«/mi»«/math», and let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«/math» be the radius of the oil surface, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»cm«/mi»«/math».
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«/math» be the volume of oil, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msup»«mi»cm«/mi»«mn»3«/mn»«/msup»«/math», at time «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math».
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math» be the oil level, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»cm«/mi»«/math», and let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«/math» be the radius of the oil surface, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»cm«/mi»«/math».
Step 2:
State the given and required related rates.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨»«mtr»«mtd»«msub»«mfenced open=¨¨ close=¨|¨»«mfrac»«mrow»«mi»d«/mi»«mi»V«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mfenced»«mrow»«mi»h«/mi»«mo»=«/mo»«mn»5«/mn»«/mrow»«/msub»«mo»=«/mo»«mo»§#8722;«/mo»«mn»12«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«msup»«mi»cm«/mi»«mn»3«/mn»«/msup»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«mtr»«mtd»«mspace/»«/mtd»«/mtr»«mtr»«mtd»«msub»«mfenced open=¨¨ close=¨|¨»«mfrac»«mrow»«mi»d«/mi»«mi»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mfenced»«mrow»«mi»h«/mi»«mo»=«/mo»«mn»5«/mn»«/mrow»«/msub»«mo»=«/mo»«mi mathvariant=¨normal¨»?«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 3:
Write an equation.
Using similar triangles, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«/math» can be written in terms of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math».
Use the diagram from Step 1 to draw two similar triangles to represent the problem.
Substitute «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»2«/mn»«mi»h«/mi»«/mrow»«mn»3«/mn»«/mfrac»«/math» into «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«mi»h«/mi»«/math».
Differentiate with respect to «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«mi»h«/mi»«/math»
Using similar triangles, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«/math» can be written in terms of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math».
Use the diagram from Step 1 to draw two similar triangles to represent the problem.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mn»10«/mn»«mn»15«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»r«/mi»«mi»h«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mn»15«/mn»«mi»r«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»10«/mn»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»r«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»10«/mn»«mi»h«/mi»«/mrow»«mn»15«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»r«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»2«/mn»«mi»h«/mi»«/mrow»«mn»3«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Substitute «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»r«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»2«/mn»«mi»h«/mi»«/mrow»«mn»3«/mn»«/mfrac»«/math» into «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»V«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«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»«mfrac»«mn»1«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»r«/mi»«mn»2«/mn»«/msup»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mfenced»«mfrac»«mrow»«mn»2«/mn»«mi»h«/mi»«/mrow»«mn»3«/mn»«/mfrac»«/mfenced»«mn»2«/mn»«/msup»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»3«/mn»«/mfrac»«mo»§#960;«/mo»«mfenced»«mfrac»«mrow»«mn»4«/mn»«msup»«mi»h«/mi»«mn»2«/mn»«/msup»«/mrow»«mn»9«/mn»«/mfrac»«/mfenced»«mi»h«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»4«/mn»«mn»27«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»h«/mi»«mn»3«/mn»«/msup»«/mtd»«/mtr»«/mtable»«/math»
Differentiate with respect to «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math».
«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»27«/mn»«/mfrac»«mo»§#960;«/mo»«msup»«mi»h«/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»27«/mn»«/mfrac»«mo»§#960;«/mo»«mo»§#8729;«/mo»«mn»3«/mn»«msup»«mi»h«/mi»«mn»2«/mn»«/msup»«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»«mfrac»«mrow»«mn»12«/mn»«mo»§#960;«/mo»«msup»«mi»h«/mi»«mn»2«/mn»«/msup»«/mrow»«mn»27«/mn»«/mfrac»«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»«mfrac»«mrow»«mn»4«/mn»«mo»§#960;«/mo»«msup»«mi»h«/mi»«mn»2«/mn»«/msup»«/mrow»«mn»9«/mn»«/mfrac»«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 4:
Substitute the values of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»h«/mi»«/math» and «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»
into the derivative and 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».
The oil level is falling at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»27«/mn»«mrow»«mn»25«/mn»«mo»§#960;«/mo»«/mrow»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math» when the oil level is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mi»cm«/mi»«/math» from the bottom of the conical portion of the filter.
«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»«mfrac»«mrow»«mn»4«/mn»«mo»§#960;«/mo»«msup»«mi»h«/mi»«mn»2«/mn»«/msup»«/mrow»«mn»9«/mn»«/mfrac»«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»«mo»§#8722;«/mo»«mn»12«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»4«/mn»«mo»§#960;«/mo»«msup»«mfenced»«mn»5«/mn»«/mfenced»«mn»2«/mn»«/msup»«/mrow»«mn»9«/mn»«/mfrac»«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»«mo»§#8722;«/mo»«mn»12«/mn»«mo»§#8729;«/mo»«mn»9«/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»h«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mo»§#8722;«/mo»«mn»12«/mn»«mo»§#8729;«/mo»«mn»9«/mn»«/mrow»«mrow»«mn»4«/mn»«mo»§#960;«/mo»«msup»«mfenced»«mn»5«/mn»«/mfenced»«mn»2«/mn»«/msup»«/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»«mo»§#8722;«/mo»«mfrac»«mn»27«/mn»«mrow»«mn»25«/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 oil level is falling at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»27«/mn»«mrow»«mn»25«/mn»«mo»§#960;«/mo»«/mrow»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»cm«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math» when the oil level is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5«/mn»«mi»cm«/mi»«/math» from the bottom of the conical portion of the filter.
In the dark, a «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math» tall jogger passes a «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»6«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math»
tall streetlight at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»5«/mn»«mn»4«/mn»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
a.
At what rate is the shadow of the jogger increasing in length?
b.
At what rate is the tip of the jogger’s shadow moving?
a.
Find the rate at which the jogger’s shadow is increasing in length.
Step 1:
Draw and label a diagram.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«/math» be the length of the jogger’s shadow.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math» be the distance from the streetlight to the jogger.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«/math» be the distance from the streetlight to the tip of the jogger’s shadow.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»s«/mi»«/math» be the length of the jogger’s shadow.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math» be the distance from the streetlight to the jogger.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«/math» be the distance from the streetlight to the tip of the jogger’s shadow.
Step 2:
State the given and required related rates.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mn»5«/mn»«mn»4«/mn»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«mtr»«mtd»«mspace/»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mi mathvariant=¨normal¨»?«/mi»«mo»§#160;«/mo»«mi»and«/mi»«mi mathvariant=¨normal¨» «/mi»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mi mathvariant=¨normal¨»?«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 3:
Write an equation.
Use similar triangles to find «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».
Use the diagram from Step 1 to draw two similar triangles to represent the problem.
Differentiate with respect to «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«mo»=«/mo»«mi»x«/mi»«mo»+«/mo»«mi»s«/mi»«/math»
Use similar triangles to find «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».
Use the diagram from Step 1 to draw two similar triangles to represent the problem.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mn»6«/mn»«mrow»«mi»x«/mi»«mo»+«/mo»«mi»s«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»2«/mn»«mi»s«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mn»6«/mn»«mi»s«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mfenced»«mrow»«mi»x«/mi»«mo»+«/mo»«mi»s«/mi»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mn»6«/mn»«mi»s«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi»x«/mi»«mo»+«/mo»«mn»2«/mn»«mi»s«/mi»«/mtd»«/mtr»«mtr»«mtd»«mn»4«/mn»«mi»s«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi»x«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»s«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»x«/mi»«/mtd»«/mtr»«/mtable»«/math»
Differentiate with respect to «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math».
«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»s«/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»1«/mn»«mn»2«/mn»«/mfrac»«mi»x«/mi»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Step 4:
Substitute the value of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math» into the derivative and 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».
The jogger’s shadow is increasing in length at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»5«/mn»«mn»8«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»=«/mo»«mi mathvariant=¨normal¨» «/mi»«mn»0«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»625«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/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»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mfrac»«mn»5«/mn»«mn»4«/mn»«/mfrac»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»5«/mn»«mn»8«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
The jogger’s shadow is increasing in length at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»5«/mn»«mn»8«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»=«/mo»«mi mathvariant=¨normal¨» «/mi»«mn»0«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»625«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
b.
Find the rate at which the tip of the jogger’s shadow is moving.
The speed at which the tip of the shadow is moving is found by taking the derivative of the relation «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«mo»=«/mo»«mi»x«/mi»«mo»+«/mo»«mi»s«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»+«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math», where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math» is the rate at which the tip of the jogger’s shadow is moving, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math» is the rate at which the jogger is running, and «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» is the rate at which the length of the jogger’s shadow is increasing.
The tip of the jogger’s shadow is moving at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»15«/mn»«mn»8«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»=«/mo»«mi mathvariant=¨normal¨» «/mi»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»875«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
Note: The shadow is moving faster than the man because it is growing as it and the man moves.
The speed at which the tip of the shadow is moving is found by taking the derivative of the relation «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«mo»=«/mo»«mi»x«/mi»«mo»+«/mo»«mi»s«/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»+«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math», where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math» is the rate at which the tip of the jogger’s shadow is moving, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math» is the rate at which the jogger is running, and «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» is the rate at which the length of the jogger’s shadow is increasing.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»+«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»5«/mn»«mn»4«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»5«/mn»«mn»8«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»15«/mn»«mn»8«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
The tip of the jogger’s shadow is moving at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»15«/mn»«mn»8«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»=«/mo»«mi mathvariant=¨normal¨» «/mi»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»875«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math».
Note: The shadow is moving faster than the man because it is growing as it and the man moves.