We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Help

0
214
1 I'm guessing we make formulas and use the distance formula, but my memory is a bit foggy on this.

Mar 21, 2018

### Best Answer

#1
+3

Let the point that  Car B  started at be  S .

After  t  hours, the distance between  B  and  S  =  70km/hour * t hours  =  70t km

Affter  t  hours, the distance between  A  and  S  =  40 km - 40km/hour * t hours  =  (40 - 40t) km So after  t  hours, the distance between  A  and  B   =   √[  (70t)2  +  (40 - 40t)2  ]

distance between  A  and  B   =   √[  4900t2  +  1600 - 3200t + 1600t2  ]

distance between  A  and  B   =   √[  6500t2 - 3200t + 1600  ]

distance between  A  and  B   =   √[  100(65t2 - 32t + 16)  ]

distance between  A  and  B   =   10√[  65t2 - 32t + 16  ]

Let the distance between  A  and  B  be  y .

y   =   10√[  65t2 - 32t + 16  ]

We want to find what value of  t  produces the minimum value of  y  in that equation.

dy/dt   =   d/dt( 10√[  65t2 - 32t + 16  ]  )

dy/dt   =   10 d/dt √[  65t2 - 32t + 16  ]

dy/dt   =   10 (1/2) ( 65t2 - 32t + 16 )^(-1/2) d/dt( 65t2 - 32t + 16 )

dy/dt   =   10 (1/2) ( 65t2 - 32t + 16 )^(-1/2) (130t - 32)

Set  dy/dt  equal to zero.

0   =   10 (1/2) ( 65t2 - 32t + 16 )^(-1/2) (130t - 32)

Divide both sides by 10, multiply both sides by 2, multiply both sides by  ( 65t2 - 32t + 16 )^(1/2)

0   =  (130t - 32)

32  =  130t

t  =  16/65       hours

This is the only value of  t  that makes  dy/dt  be  0, so this is the value of  t  that minimizes  y .

When  t  =  16/65 ...

y   =   10√[  65(16/65)2 - 32(16/65) + 16  ]

y   =   56√ / 13     kilometers

Mar 21, 2018

### 1+0 Answers

#1
+3
Best Answer

Let the point that  Car B  started at be  S .

After  t  hours, the distance between  B  and  S  =  70km/hour * t hours  =  70t km

Affter  t  hours, the distance between  A  and  S  =  40 km - 40km/hour * t hours  =  (40 - 40t) km So after  t  hours, the distance between  A  and  B   =   √[  (70t)2  +  (40 - 40t)2  ]

distance between  A  and  B   =   √[  4900t2  +  1600 - 3200t + 1600t2  ]

distance between  A  and  B   =   √[  6500t2 - 3200t + 1600  ]

distance between  A  and  B   =   √[  100(65t2 - 32t + 16)  ]

distance between  A  and  B   =   10√[  65t2 - 32t + 16  ]

Let the distance between  A  and  B  be  y .

y   =   10√[  65t2 - 32t + 16  ]

We want to find what value of  t  produces the minimum value of  y  in that equation.

dy/dt   =   d/dt( 10√[  65t2 - 32t + 16  ]  )

dy/dt   =   10 d/dt √[  65t2 - 32t + 16  ]

dy/dt   =   10 (1/2) ( 65t2 - 32t + 16 )^(-1/2) d/dt( 65t2 - 32t + 16 )

dy/dt   =   10 (1/2) ( 65t2 - 32t + 16 )^(-1/2) (130t - 32)

Set  dy/dt  equal to zero.

0   =   10 (1/2) ( 65t2 - 32t + 16 )^(-1/2) (130t - 32)

Divide both sides by 10, multiply both sides by 2, multiply both sides by  ( 65t2 - 32t + 16 )^(1/2)

0   =  (130t - 32)

32  =  130t

t  =  16/65       hours

This is the only value of  t  that makes  dy/dt  be  0, so this is the value of  t  that minimizes  y .

When  t  =  16/65 ...

y   =   10√[  65(16/65)2 - 32(16/65) + 16  ]

y   =   56√ / 13     kilometers

hectictar Mar 21, 2018