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A forest ranger is in a forest 2 miles from a straight road. A car is located c miles down the road. The forest ranger can walk 2 miles per hour in the forest and travel 4 miles per hour along the road. In the picture below, the forest ranger walks in a straight line to a point x unit from the end of the road on the left, and then along the road.
The total travel time for the ranger to get to the car is the sum of the travel time in the forest and the travel time on the road. Write the total time T as a function of x, using the letter c for the distance from the end of the road on the left to the car.
T(x) = ?
Toward what point on the road should the ranger walk in order to minimize the travel time to the car if...
(a) c = 9 miles? (Numerical Answer ONLY)
x = ?
(b) c = 1/2 miles? Remember that the point must lie between the left end of the highway and the car. (Numerical Answer ONLY)
x = ?

Respuesta :

sqdancefan

Answer:

  • T(x) = (√(4+x^2))/2 +(c -x)/4
  • (a) x = 1.155 miles
  • (b) x = 1/2 mile

Step-by-step explanation:

The straight line distance from the starting point to the point x on the road is given by the Pythagorean theorem:

  d1 = √(2²+x²)

The time required to walk that distance in the woods (at 2 miles per hour) is ...

  time = distance / speed

  t1 = √(4+x²)/2

The remaining distance along the road to the car is ...

  d2 = c - x . . . . . for c > x

and the corresponding time at 4 mph is ...

  t2 = (c -x)/4

Then the total time is ...

  T(x) = t1 +t2

  T(x) = √(4+x²)/2 +(c -x)/4

__

(a) The value of x that minimizes total time is the one that makes the derivative zero.

  T'(x) = x/(2√(4+x²)) -1/4

  0 = (2x -√(4+x²))/(4√(4+x²))

  √(4+x²) = 2x . . . . . fraction is zero when numerator is zero; add radical

  4 +x² = 4x² . . . . . . square both sides

  4/3 = x² . . . . . . . . . subtract x², divide by 3

  x = (2/3)√3 ≈ 1.1547

If c = 9 miles, x = 1.1547 miles.

__

(b) For c < (2/3)√3, the shortest travel time will be along the straight-line path to the car.

If c = 1/2 mile, x = 1/2 mile.

To answer this question we know that  t = d/v and express time as a function of the parameters we know

The solution is:

T = √ ( x² + 4 ) / 2  + ( c - 4 )/4

a) T = 4.85 h

b) T= 3.6225 h

Total time required by the ranger to get to the car, is the sum of the time walking through the forest at 2 m/h and the time walking along the road at 4 m/h.

According to that:

T = T₁ + T₂               and   t = d / v

T₁ = L/2         and     T₂ = ( c - x ) /4

L = √ ( x² + 4 )          T₁ =  √ ( x² + 4 )   /2

T = √ ( x² + 4 ) / 2  + ( c - x )/4          ( 1)

Now to answer letter b we get the derivative of T with respect to x

dT/dx = ( 1/2) × (1/2)×2×x / √( x² + 4) - 1/4

dT/dx =  0            ⇒     (1/2) ×( x / √x² + 4   - 1/4  = 0

x / 2× √x² + 4 - 1/4 = 0

2×x -  √(x² + 4)  = 0

2×x = √(x² + 4)

Squaring both sides of the equation

4×x² = x² + 4

Solving for x

3×x² = 4

x² = 4/3

x = ±√4/3       we dismiss the negative value

x = 2/√3 miles      x = 1.15 miles

This value of x will minimize the time

Now if c = 9 miles then

Pluggin that value in (1)

T = (√(4/3) + 4 )/2 + ( 9 - 2/√3)/ 4           √3 = 1.73

T = √(16/3) / 2 + 1.385

T = 3.46 + 1.385

T = 4.85 h

In the case of c = 0.5 miles and x = 1.15 miles, c < x     ( x for minimum time)

That means the ranger overpassed the car and need to come back                   ( 1.15 - 0.5 ) = 0.65  at 4 m/h   T₂ = 0.1625 h   T₁ = 4.85

T= 3.6225 h

Related link :https://brainly.com/question/24618879

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