Calc. Practice Problems #15 of 16?

Question

Hello. I have just uploded onto Photobucket 16 calculus practice problems, all of which will be of similar content to my eventual final exam. Now, an answer to these questions would be nice, but is not necessary. Instead, I am concerned with and interested in THE METHOD as to how one could arrive to the correct answer, as well as an EXPLANATION as to why each step needs to be performed in relation to the ones that follow. Anybody who can break down and explain the contents of these problems to me in a SIMPLISTIC way would just be amazing in my book. So thank you very much should you decide to help enlighten me, as I appreciate this more then the words in this post could even begin to convey!

Question #15: http://www.i138.photobucket.com/albums/q271/Link3324/math15.jpg

By the way, This is NOT for any kind of required assignment. It is a guide to work from and to understand before the actual test. Any help would be greatly appreciated. THANK YOU!

Answer 1

hello again
the time that would take the worman to walk is
distance traveled/speed = (pi/2)*d/4mi/hour = Pi*d/8
similarly the time that takes to row is
distance traveled /speed= d/2mi/hr so the question is which one
is shorter in another words Pi*d/8 since pi/8<1/2
so the woman better walk since she gets there faster
please ignor the dumass above
good luck

Answer 2

Circular lake of radius r.

Rates:

Rowing r1 = 2 mph
Walking r2 = 4 mph

Distances:

Rowing d1 = 2r
Walking d2 = πr

Times:

Rowing t1 = d1 / r1 = 2r / 2 = 1 hour
Walking t2 = d2 / r2 = πr / 4 hour < t1

So walking is faster. But is it faster if you do part of each? Please note that if you row for just a little ways (along a chord of the circle) , the course rowed will approximate the course and distance walked much more closely than rowing across the diameter of the lake. Consequently, the walking time will be faster than rowing to any other point along the circumference of the lake as well.

More formally, if you row part way and walk the remainder:

Let θ, be the degrees of arc subtended by the chord c that was rowed.

d1 = 2rsin(θ/2)
d2 = (π - θ)r

t = t1 + t2 = d1/r1 + d2/r2
= 2rsin(θ/2)/2 + (π - θ)r/4
= rsin(θ/2) + (π - θ)r/4

dt/dθ = (1/2)rcos(θ/2) - r/4 = 0
(1/2)rcos(θ/2) = r/4
cos(θ/2) = 1/2
θ/2 = π/3
θ = 2π/3

Check to see if this is a minimum or a maximum with the second derivative.

d^2t/dθ^2 = -(1/4)rsin(θ/2) = -(1/4)rsin(π/3) < 0
Therefore it is a maximum, not what we want. So the minimum is at one of the extremes. As we have seen above, the minimum time is achieved by walking the entire distance.

Answer 3

Can you allow for part rowing and part walking? If so, then it's better to row and then walk.

This is a hard problem to solve in that case though because you would have to express the distance in two variables (the distance over water and the distance over land). After that you'll have to solve for an equation in terms of one variable and optimize that function.

Answer 4

Walking time is πD/4.
Rowing time is D/2
πD/4 > D/2, so
rowing would take the least time.