Minimization Problem - rowing and walking

Minimizing Time it Takes to Row and Walk to a Given Point

 

We're in a rowboat two miles from shore on a cold, rainy river. There's a tent and a nice campfire three miles down the river, and one mile inland. We can row at two miles an hour, and walk at four miles an hour. We need to find the spot on the shore that will get us to camp in the shortest time.

First, as always, draw a picture.

 

We ask:

1.       What do we want to do?

We want to get to camp as fast as possible. In math terms, that means we want to minimize the time it takes to get from where we are to camp.

2.       How do we accomplish this?

We find the derivative of the equation for time, and set it to zero.

3.       What formulas can help us?

Time = distance / rate

Distance = sqrt(a^2 + b^2)

4.       What do we know already?

The time it will take to row to the spot x miles down the river is sqrt(x^2 + 4)/2.

The time it will take to walk from point x to camp is sqrt((3 –x)^2 + 1)/4

 

5.       What is the problem we need to solve?

We need the derivative of time. We set up the equation for total time we need:

 

T = sqrt(x^2 + 4)/2 + sqrt(x^2 + 4)/2

We find the derivative of the time equation:

dT = x/ (2sqrt(x^2 + 4))  -  (3 –x)/ (4sqrt( (3 –x)^2 + 1))

 

 

We set dT to zero:

 

x  (2sqrt(x^2 + 4))  -  (3 –x) / (4sqrt( (3 –x)^2 + 1) ) = 0

 

x / ( 2sqrt(x^2 + 4) )  =  (3 –x) / (4sqrt( (3 –x)^2 + 1) )

 

 

At this point, the algebra gets a tiny bit intricate. Go slow. I got into trouble by rushing.

We cross-multiply, and eventually we end up with:

 

12x^4 – 72x^3 + 108x^2 + 96x – 144 = 0

 

We can divide the whole thing by 12, ending up with:

 

x^4 – 6x^3 + 9x^2 + 8x – 12 = 0

 

This is where I love Synthetic Division. We know that the only possible integer answers are 0, 1, 2, and 3. My policy is to always try 1 first in Synthetic Division.

 

1 is the answer. Rowing to a point 1 mile down the river will get us to camp quickest.

In math terms, x = 1 will minimize the value of time (T).

 

Since we know that x has to be between 0 and 3, we can always plug numbers in that range into the derivative and see which number makes the derivative equal to zero. It works easily in this case because the answer is an integer. But that's cheating, kind of, so you didn't hear it from me.

 

 

 

 

True Story-No Such Thing as a Non-Math Person

Back in 1940, in a one-room school in Oklahoma, a little old lady teacher said to my Dad, (in front of the whole class)  "You will  nev-ah und-uh-stand math-a-matics." Luckily, my Dad didn't believe her. He moved on and got a degree in Civil Engineering and a Professional Engineer's license. He's run water desalinization and reclamation facilities all over the world.

My point is, nobody has the ability, let alone the right, to judge your potential. There is no such thing as a "math person" or "not a math person." Learning math is a holistic process. Whatever kind of brain you have is the right kind. If you take care of your mind and body, you can learn math. In return, math will help sustain and support that mind-body balance. It will also amaze you at what you can do.

If you amaze someone else in the process, that's a plus.

Related Rates: Cone-shaped pile

Related Rate

Find the rate of change of the height of a cone, knowing the rate of change of the volume.

 

A cone-shaped pile of dirt is growing at the rate of 10 cubic feet/minute. The diameter of its base is 3 times its height.

How fast is the height of the pile growing when it's 15feet high?

We'll ask some questions.

What do we want? - We want the rate of change of the height of the cone. We'll call the height h, and its rate of change dh/dt.

What do we know already? -We know the rate of change of the volume of the cone. We'll call the volume V, and its rate of change dV/dt. We also know that dV/dt = 10 cubic feet/minute.

What geometric formulas can help us? We know the formula for the volume of the cone. V = (1/3)(pi)(r^2)h

What substitutions can we make? (or,  How can we get rid of that r? ) – We have the relationship 2r = 3h, or r = 3h/2.

(The first time I did this problem,  I momentarily forgot that the diameter is twice the radius.)

Now we can set up our geometric equation.

V = (pi/3)(r^2)h

Substitute 3h/2 for r.

V = (3pi/4)(h^3)

We take the derivative of each side, over time.

dV/dt =(9pi/4)(h^2)(dh/dt)

We know the values of dV/dt and h, so we substitute those values into our equation.

10 cubic feet/minute = (9*15^2 feet squared *pi) /4* dh/dt

Solve for dh/dt.

8/(405pi) feet/minute = dh/dh

 

Related rate problems are just algebra equations. If I'd known that when I first encountered them back in the day, I'd have liked them a lot better a lot sooner. I also should have brushed up on my solid geometry, or at least learned some of it first.

 

Source:
 Larsen, Hostetler and Edwards,  Calculus, Fifth Edition)

Shell Method: Between x^2 and 4x - x^2 around y-axis

Shell Method

Volume of the space between y = x^2 and y = 4x – x^2 when it is revolved around the  y axis.

 

First of all, draw a picture.

 

The shaded area is the area we're interested in.

We ask some questions to organize our thought process.                                                                                                                                                              

1.       What is the area of the rectangle? – The width of the rectangle is dx, the height will be 4x – x² –x², or 4x  − 2x² (The higher curve minus the lower curve.)

2.       What is the radius?- With the shell method, the radius is the distance between the rectangle and the line we're revolving around. In this case, it will be x.

3.       Will our integral be in terms of x or y? – the integral will be in terms of x, because the radius is along the x. (With the shell method, if we're revolving around a vertical line, everything will be x, if around a horizontal line, everything will be y.)

4.       What are our boundaries of x? – We will integrate between 0 and 2 (See diagram, or set x² equal to 4x – x².

What we want is 2pi times the integral of the area of the rectangle times the radius. In this case it will be:

2pi times the integral of( x times 4x minus 2x² times dx )= 2pi times (8/3) = 16pi/3.

 

 

Disk/Washer Method: y = x ^ 1/2 around

Disk/Washer Method

 y = x ^1/2 around the y-axis

Our mission is to find the volume that results when the curve y = x ^(1/2) between y = 0, x = 0 and x = 4 around the y-axis.

First of all draw a picture. I can't emphasize this enough.

 

We ask a few questions to organize our thoughts.

Are we integrating over x or y?

We're revolving around y, so our integral has to be in terms of y. We'll have to integrate over the y-values, 0 to 2.

One radius or two radii?

Two radii. The outer one is 4, the inner is the x value of the curve y = x ^ 1/

 

How do we set up our integral?

The outer radius is 4, so I wouldn't even bother with an integral. The "outer shape" is a cylinder of volume 32pi.

The inner radius is the x-value of the curve y = x ^ 1/2 . Since we have to integrate in terms of y, we solve for x in terms of y. Our inner radius is y^2. We integrate over the y-values, from 0 to 2.

 

The volume of the shape will be:

 

32pi -  pi times ( the integral of y^4 between y = 0 and y = 2), which is 32pi - (32pi/5) Our answer is 128pi/5.

 

Basically, what we want is the volume of a cylinder, minus the volume of a shape kind of like this:

 

 

 

 

 

Greeting

I am an artist who loves math. It is my mission in life to help other people to love math too. I am dedicated to the proposition that math intrinsically fun, creative and not harmful to humans in any way.

I have a degree in math (also one in art) and am constantly studying, refreshing and re-evaluating my math perspective. Anytime I run across a math problem that I think is cool, I'm going to post it. (By cool, I mean "seemed difficult but wasn't once I saw something I hadn't before." I'll share the thought processes that worked for me, such as the questions I asked myself, and how I answered them. I hope it will clear up some hassles for other people. Those hassles are usually the result of forgetting what we already know.

There's one bit of advice I must share. Don't skip prerequisites for math classes. You won't save any time. I learned that the hard way.

I haven't yet figured out how to put math symbols into this post so that they look decent, so I'm just using plain words. That might not be such a bad thing. If anything is unclear, please comment.

I welcome requests for math problems through the comment section or at hypatia31@gmail.com.