Unit 5

Applications of Derivatives

B. Related Rates Problems

Lesson 5: Related Motion

Related Motion

You have probably been unconsciously dealing with related rates for a long time. If you are in a car approaching a railroad crossing at the same time a train is approaching, what considerations do you intuitively make?

The most important safety consideration for drivers in this situation is the rate at which the distance between the car and the train is changing. This rate of change is dependent on the rates at which the car and the train are travelling. The rate of change of the distance between the car and train is also affected by the distances of the car and the train from the crossing.

Watch the video A Pythagoras Question to see how a related motion problem is solved.

In Lesson 4, steps to solve related rates problems were formulated. Those same steps will be applied throughout this Lesson.

Draw and label a diagram. Include the values that never change (constants) and include variables for the values that do change.
List all given and required rates as derivatives with respect to time.
Write an equation that relates the variables from Step 2, and take the derivative. This equation may stem from a well-known geometric formula for area or volume, the Pythagorean Theorem, the properties of similar triangles, or trigonometry. The chain rule is then used to differentiate the equation with respect to time.
Substitute the given information into the differentiated equation, and solve for the unknown rate.
Note: A common error is made when the given numerical information is substituted too soon. If a value is true only for a particular instant, it can only be used after differentiation.

Example 1

A straight, level road crosses a railroad track at a right angle. A car is on the road «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» from the crossing, travelling at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»80«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math» toward the crossing. At the same time, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» from the crossing, a train is travelling at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»100«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math» toward the crossing. At what rate is the distance between the car and the train changing at that instant?

Solution

Step 1:

Draw and label a diagram.

Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math» be the distance, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»km«/mi»«/math», between the car and the crossing.

Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«/math» be the distance, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»km«/mi»«/math», between the train and the crossing.

Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»z«/mi»«/math» be the distance, in «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»km«/mi»«/math», between the car and the train.

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»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mfenced»«mtable»«mtr»«mtd»«mi»x«/mi»«mo»=«/mo»«mn»1«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»y«/mi»«mo»=«/mo»«mn»2«/mn»«/mtd»«/mtr»«/mtable»«/msub»«mo»=«/mo»«mo»§#8722;«/mo»«mn»80«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mfenced open=¨¨ close=¨|¨»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mfenced»«mtable»«mtr»«mtd»«mi»x«/mi»«mo»=«/mo»«mn»1«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»y«/mi»«mo»=«/mo»«mn»2«/mn»«/mtd»«/mtr»«/mtable»«/msub»«mo»=«/mo»«mo»§#8722;«/mo»«mn»100«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mfenced open=¨¨ close=¨|¨»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mfenced»«mtable»«mtr»«mtd»«mi»x«/mi»«mo»=«/mo»«mn»1«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»y«/mi»«mo»=«/mo»«mn»2«/mn»«/mtd»«/mtr»«/mtable»«/msub»«mo»=«/mo»«mi mathvariant=¨normal¨»?«/mi»«/mtd»«/mtr»«/mtable»«/math»

Note: Since both the car and the train are moving toward the crossing, their distances from the crossing are decreasing. Thus, their rates of change are negative.

Step 3:

Write an equation.

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msup»«mi»z«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«msup»«mi»y«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«msup»«mi»z«/mi»«mn»2«/mn»«/msup»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mfenced»«mo»+«/mo»«mfrac»«mi»d«/mi»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mfenced»«msup»«mi»y«/mi»«mn»2«/mn»«/msup»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mn»2«/mn»«mi»z«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mi»x«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»+«/mo»«mn»2«/mn»«mi»y«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»z«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»x«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»+«/mo»«mi»y«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»

Find «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»z«/mi»«/math» before proceeding.

Solve for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»z«/mi»«/math» when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mn»1«/mn»«/math» and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«mo»=«/mo»«mn»2«/mn»«/math».

Step 4:

Substitute the values of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»,«/mo»«mo»§#160;«/mo»«mi»y«/mi»«mo»,«/mo»«mo»§#160;«/mo»«mi»z«/mi»«mo»,«/mo»«mo»§#160;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/math», and «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» into the derivative and solve for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»d«/mi»«mi»z«/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»«mi»z«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»x«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«mo»+«/mo»«mi»y«/mi»«mo»§#8729;«/mo»«mfrac»«mrow»«mi»d«/mi»«mi»y«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfenced»«msqrt»«mn»5«/mn»«/msqrt»«/mfenced»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced»«mn»1«/mn»«/mfenced»«mfenced»«mrow»«mo»§#8722;«/mo»«mn»80«/mn»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mn»2«/mn»«/mfenced»«mfenced»«mrow»«mo»§#8722;«/mo»«mn»100«/mn»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mo»§#8722;«/mo»«mn»280«/mn»«/mrow»«msqrt»«mn»5«/mn»«/msqrt»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mo»§#8722;«/mo»«mn»280«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#8722;«/mo»«mn»56«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mtd»«/mtr»«/mtable»«/math»

Since the distance between the car and the train is decreasing, the rate should be a negative value.

The distance between the train and the car is decreasing at a rate of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»56«/mn»«msqrt»«mn»5«/mn»«/msqrt»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math», or approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»125«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math», when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mn»1«/mn»«/math» and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»y«/mi»«mo»=«/mo»«mn»2«/mn»«/math».