without the help of a graph, how i prove that the route through the points (0.689,−0.311) and (−0.689,−0.311) is the fastest route.?or maybe the question is: without the help of a graph how do i find those point??? QuestionDonald plays Monopoly and draws the "Go straight to jail" card. After an outburst of rage,Donald decides that he wants to go to prison as soon as possible, but discovers thatthere is a swamp between him and the prison. Donald is in point (2, 0) andthe prison in (-2, 0) and the swamp is within the curve, |x| + ly| = 1.all targets in Duck City Units. Donald moves with a speed 1 outside the swamp andconstant speed v> 0 in the swamp.What can the speed v in the swamp be if:(i) The fastest route involves starting by going straight to a point (a, b) at the edgeof the swamp with o <| b| <1 (illustrated in the figure)? In that case, findb as a function of v anddraw the graph(2sqr(b^2+(2-a)^2) + 2a/vwe know this is the fastest rout, but why??Expert AnswerSolution:The time taken to reach the prison is 2/b2 + (2 – a)? + 24.The function is r (b) = 2V1 + 2b + 2b?2(1-b)2(1-(0))At b =0, t (0) = 2y1 +2 (O) + 2(0)? += 2+2This is the time taken when longest distance in the swamp is covered. In swamp, he needs tomaintain the constant speed.At b = 1, 1 (1) = 2V1+ 2 (1) + 2(1) +2(1-(1))2V5.2(1-(-1))At b = -1, 1 (–1) = 2/1 + 2(-1) + 2(–1)² += 2 + .These are the extreme values of time.Conclusion:2(1-b)For 0 < |b| < 1, 1 (b) = 2y1+ 26 + 262 +This is the fastest route because it include the value of b for which the the time taken isminimum.As the graph suggests, the minimum value of t occurs when b=-0.311.(-0.311, 2.823)That is, the route through the points (0. 689, -0. 311) and (-0. 689, -0. 311) is the fastest

The time t in terms of b is tb=21+2b+2b2+21-bv. Differentiate t with respect to b:…

Answered: without the help of a graph, how i…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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without the help of a graph, how i prove that the route through the points (0.689,−0.311) and (−0.689,−0.311) is the fastest route.?

or maybe the question is: without the help of a graph how do i find those point???

Question
Donald plays Monopoly and draws the "Go straight to jail" card. After an outburst of rage,
Donald decides that he wants to go to prison as soon as possible, but discovers that
there is a swamp between him and the prison. Donald is in point (2, 0) and
the prison in (-2, 0) and the swamp is within the curve, |x| + ly| = 1.
all targets in Duck City Units. Donald moves with a speed 1 outside the swamp and
constant speed v> 0 in the swamp.
What can the speed v in the swamp be if:
(i) The fastest route involves starting by going straight to a point (a, b) at the edge
of the swamp with o <| b| <1 (illustrated in the figure)? In that case, findb as a function of v and
draw the graph
(2sqr(b^2+(2-a)^2) + 2a/v
we know this is the fastest rout, but why??
Expert Answer
Solution:
The time taken to reach the prison is 2/b2 + (2 – a)? + 24.
The function is r (b) = 2V1 + 2b + 2b?
2(1-b)
2(1-(0))
At b =0, t (0) = 2y1 +2 (O) + 2(0)? +
= 2+2
This is the time taken when longest distance in the swamp is covered. In swamp, he needs to
maintain the constant speed.
At b = 1, 1 (1) = 2V1+ 2 (1) + 2(1) +
2(1-(1))
2V5.
2(1-(-1))
At b = -1, 1 (–1) = 2/1 + 2(-1) + 2(–1)² +
= 2 + .
These are the extreme values of time.
Conclusion:
2(1-b)
For 0 < |b| < 1, 1 (b) = 2y1+ 26 + 262 +
This is the fastest route because it include the value of b for which the the time taken is
minimum.
As the graph suggests, the minimum value of t occurs when b=-0.311.
(-0.311, 2.823)
That is, the route through the points (0. 689, -0. 311) and (-0. 689, -0. 311) is the fastest

Expert Solution

Solution:

The time t in terms of b is $t (b) = 2 \sqrt{1 + 2 b + 2 b^{2}} + \frac{2 (1 - b)}{v}$ .

Differentiate t with respect to b:

$t' (b) = \frac{2 (2 + 4 b)}{2 \sqrt{1 + 2 b + 2 b^{2}}} - \frac{2}{v} = \frac{(2 + 4 b)}{\sqrt{1 + 2 b + 2 b^{2}}} - \frac{2}{v}$

Set the derivative to zero and v =2:

$\frac{(2 + 4 b)}{\sqrt{1 + 2 b + 2 b^{2}}} - \frac{2}{2} = 0 \frac{(2 + 4 b)}{\sqrt{1 + 2 b + 2 b^{2}}} = 1 \sqrt{1 + 2 b + 2 b^{2}} = 2 + 4 b [S q u a r e b o t h s i d e s] 1 + 2 b + 2 b^{2} = {(2 + 4 b)}^{2}$

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