The minimum distance between them and determine whether it occur before or after the first of them has reached the watering hole.

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Answer 1

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Chapter 2.9, Problem 46E

To determine

The minimum distance between them and determine whether it occur before or after the first of them has reached the watering hole.

Expert Solution & Answer

Answer to Problem 46E

The occurrence of minimum distance between them happens after cheetah has reacted the watering hole at t=3.33sec.

It happens before gazelle has reacted the watering hole at t=4sec.

Explanation of Solution

Given:

The speed of the Cheetah is 30m/s towards the popular watering hole and the misguided gazelle is running due east towards the same spot at rate of 20m/s.

Cheetah starts from 100m and gazelle starts from 80m.

The rate of change of the distance between them at t =0, t =2, t =3 and t =4.

Considering both cheetah and gazelle started at the same time and they travel for t seconds.

As, the cheetah is running in the south direction for the speed of 30m/s for t seconds , the distance travelled is 30t.

As, it begins from the distance of 100m, the position of the cheetah is given by the following equation:

y(t)=30t−100

Also, the gazelle is running in the east direction at the speed of 20m/s for t seconds, the distance travelled is 20t.

As it begins from the distance of 80m, the position of gazelle:

x(t)=20t−80

As, the both of them are moving in the direction which are perpendicular to each-other, the distance between them is calculated by Pythagoras theorem.

r(t)= ( x( t ) )2+ ( y( t ) )2= ( 20t−80 )2+ ( 30t−100 )2.............(1)

Evaluating the rate of change of the distance between them by differentiating:

ddtr(t)=ddt ( 20t−80 )2+ ( 30t−100 )2=12 ( 20t−80 ) 2 + ( 30t−100 ) 2 ddt[(20t− 80 2)+(30t−100)]=[2( 20t−80) d dt( 20t−80)+2( 30t−100) d dt( 30t−100)]2 ( 20t−80 ) 2 + ( 30t−100 ) 2 =20( 20t−80)+30( 30t−100)2 ( 20t−80 ) 2 + ( 30t−100 ) 2

Therefore, the rate of change of the distance between them at t=0 is calculated as follows:

dr(0)dt=20( 0−80)+30( 0−100) ( 0−80 ) 2 + ( 0−100 ) 2 =20( −80)+30( −100) ( −80 ) 2 + ( −100 ) 2 =−35.91m/s

The rate of change of the distance between them at t=2 is calculated as follows

dr(2)dt=20( 40−80)+30( 60−100) ( 40−80 ) 2 + ( 60−100 ) 2 =20( 40−80)+30( 60−100) ( −40 ) 2 + ( −40 ) 2 =−35.35m/s

The rate of change of the distance between them at t=3 is calculated as follows

dr(3)dt=20( 60−80)+30( 90−100) ( 60−80 ) 2 + ( 90−100 ) 2 =20( −20)+30( −10) ( −20 ) 2 + ( −10 ) 2 =−31.30m/s

The rate of change of the distance between them at t=4 is as follows:

dr(4)dt=20( 80−80)+30( 120−100) ( 80−80 ) 2 + ( 120−100 ) 2 =20(0)+30( 20) ( 0 ) 2 + ( 20 ) 2 =30m/s

The minimum distance between the animals is calculated by equating the derivative to zero.

20( 20t−80)+30( 30t−100)2 ( 20t−80 ) 2 + ( 30t−100 ) 2 =020(20t−80)+30(30t−100)=0200(2t−8)+300(3t−10)=0

It gives:

( 2t−8)( 3t−10)=−300200( 2t−8)( 3t−10)=−324t−16=−9t+30t=4613

The occurrence of minimum distance between them happens after cheetah has reacted the watering hole at t=3.33sec

But this happens before gazelle has reacted the watering hole at t=4sec.

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