At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

a woman at a point a on the shore of a circular lake with radius 1.2 mi wants to arrive at the point B diametrically opposite A on the other side of the lake in the shortest possible time (see the figure). she can walk at the rate of 5 mi/h from C to B and row a boat at 3.5 mi/h from A to C . setting T (x) function to the time by interlaced AC, x is interlacted AC. how to move for shortest.

A Woman At A Point A On The Shore Of A Circular Lake With Radius 12 Mi Wants To Arrive At The Point B Diametrically Opposite A On The Other Side Of The Lake In class=

Sagot :

The angle θ to the diameter that she should row is; θ = 56.44°

How to find the angle of an arc?

It should be noted that;

a = 90° - θ. Thus;

b = 2θ

For any circle, the angle subtended (in radians) is equal to the arc length (d_w ) divided by the radius (1.2). Thus; d_w = 4θ .

The time required to walk this distance is given by;

t_w = 4θ/5 = 0.8θ

The distance rowed is given by: cos θ = d_r/2.4

d_r = 2.4cos θ while the time needed for it is;

t_r = 2.4cos θ/3.5

t_r = 0.6857θ

Thus, total time is;

T(θ) = t_r + t_w

T(θ) = 0.6857θ + 0.8θ

T(θ) = 1.4857θ

Now we need the critical numbers:

T'(θ) = 1 - 1.2 sin θ

At T'(θ) = 0, we have;

1 - 1.2 sin θ = 0

1/1.2 = sin θ

θ = sin⁻¹(1/1.2)

θ = 56.44°

Read more about angle of an Arc at; https://brainly.com/question/2005046

#SPJ1

View image AFOKE88
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.