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Rates of Paddling


Date: 04/25/2000 at 07:23:11
From: Stan
Subject: A Confusing Word Problem

Please help me! I have this word problem:

At my usual rowing rate, I can travel 12 miles downstream on a certain 
river in 6 hours less than it takes me to travel the same distance 
upstream. But if I could double my usual rowing rate for this 24 mile 
round trip, the downstream 12 miles would then take only 1 hour less 
than the upstream 12 miles. How fast is this river's current?

I can't figure it out! I tried all ways of making variables - it just 
doesn't make sense to me! Please show me how to figure this out and 
what the correct ANSWER is.

Thank you so much.


Date: 04/25/2000 at 12:47:17
From: Doctor Rick
Subject: Re: A Confusing Word Problem

Hi, Stan.

Yes, this problem takes some thinking. You need to understand it 
before you can solve it.

The first step, as usual, is to define the appropriate variables:

     v = my usual rowing rate, in miles per hour

     c = the river's current, in miles per hour

Now, in order to write equations, you need to apply the rate equation.

     distance = rate * time

     time = distance / rate

How long does it take to row 12 miles downstream? Your speed relative 
to the shore is the sum of your rowing rate and the river's current, 
or (v+c). Using the rate equation, the time (in hours) is (12 miles)/
(v+c miles/hour).

How long does it take to row 12 miles upstream? Your speed relative to 
the shore is your rowing rate MINUS the river's current, because the 
current takes you backward, slowing you down.

I've given you enough guidance to get you going. See what you can do 
with it now. If you get the equations but you have trouble solving 
them, write back and tell me what you've worked out.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/25/2000 at 13:15:08
From: Stan
Subject: Re: A Confusing Word Problem

Hi! Okay, I've gotten these equations: [Thanks for your help!]

     downstream: t = 12 mi/(v+c miles/hr)
and
     upstream:   t = 12 mi/(v-c miles/hr) 

Tell me if they're wrong.

It makes more sense now; but how do I solve now? I have an up and 
downstream how does that help?

Please help; I'm very determined to figure this out!  :)

-Stan
P.S. Thanks for your help so far!


Date: 04/25/2000 at 13:44:30
From: Doctor Rick
Subject: Re: A Confusing Word Problem

Hi again, Stan!

Good work, now you have a way to express mathematically how long it 
takes to travel upstream or downstream. You can drop the units now, 
since the units are identified in the definitions of the variables:

     Time to travel 12 miles downstream (in hours) = 12/(v+c)
     Time to travel 12 miles upstream (in hours)   = 12/(v-c)

Next, use these expressions to translate the given information into 
equations. For instance,

   At my usual rowing rate, I can travel 12 miles downstream on a 
   certain river in six hours less than it takes me to travel the same 
   distance upstream.

Another way of saying this is

   The time it takes to travel 12 miles downstream is 6 hours less 
   than the time it takes to travel 12 miles upstream.

Can you write this as an equation?

The other information you are given concerns travel times if the 
rowing rate is doubled. In other words, replace v in your expressions 
with (2v). Then you have what you need to write a second equation.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/25/2000 at 14:15:26
From: Stan
Subject: Re: A Confusing Word Problem

Hi again!

All right: I wrote two equations - are they right? Please correct them 
if they aren't.

   1. [12/(v+c)]-6 hrs = 12/(v-c)

   2. [12/(2v+c)]-1 hr = 12/(2v-c)

Now, how do I solve all of this mumbo-jumbo to get the right answer?
Thanks again for your help so far.  :)

Stan


Date: 04/25/2000 at 17:08:05
From: Doctor Rick
Subject: Re: A Confusing Word Problem

Hi again!

You've got the equations almost correct. Be careful, though: your 
first equation says that the time to go UPstream (that's the right 
side of the equation) is 6 hours less than the time to go DOWN stream. 
That's not what the problem says. The second equation has a similar 
error.

Again, you don't need to include the "hrs" since the definitions of 
the variables imply that times are in hours.

Now you're on to the next phase of the problem: solving the equations. 
The first thing I do when I see rational expressions like these is to 
multiply each equation through by the product of the denominators. 
This gets rid of the denominators, so you'll have more 
familiar-looking equations. I'll show you on the first equation (but 
remember, you'll have to correct this equation, so you won't get what 
I get here):

   12/(v+c) - 6 = 12/(v-c)

Multiply each side by (v+c)(v-c), the product of the two denominators:

   12(v+c)(v-c)/(v+c) - 6(v+c)(v-c) = 12(v+c)(v-c)/(v-c)

Simplify each fraction:

   12(v-c) - 6(v+c)(v-c) = 12(v+c)

Expand the multiplications using the distributive property, then 
simplify the equation.

Again, this equation is wrong! Correct it and see what you can do.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/26/2000 at 07:19:04
From: Stan
Subject: Re: A Confusing Word Problem

Hi Dr. Rick!

I fixed the wrong equations and tried to do what you told me; this is 
how far I got without any "guidance" (the equations below are
yesterday's fixed equations, and under the line is how I tried to 
simplify them into correct (I think) equations:

   1. [12/(v-c)]-6 hrs = 12/(v+c)

   2. [12/(2v-c)]-1 hr = 12/(2v+c)

-----------

   1.                      12/(v-c) -6 = 12/(v+c)
      12(v+c)(v-c)/(v-c) - 6(v+c)(v-c) = 12(v+c)(v-c)/(v+c)
                 12(v+c) - 6(v+c)(v-c) = 12(v-c)

      [I don't know what to do from here.]

   2.                         12/(2v-c) -1 = 12/(2v+c)
      12(2v+c)(2v-c)/(2v-c) -1(2v+c)(2v-c) = 12(2v+c)(2v-c)/(2v+c)
                   12(2v+c) -1(2v+c)(2v-c) = 12(2v-c)

I don't know if I simplified it correctly:

              24v+12c+(-2v-c)(2v-c) = 24v-12c
     24v+12c - 4v^2 +2vc -2vc + c^2 = 24v-12c
                24v+12c - 4v^2 +c^2 = 24v-12c

     [I don't know what to do from here.]

Now I'm not sure if I did this stuff right. Please correct it if I 
didn't; also, I can't continue #1. How do you distribute in this?

     -6(v+c)(v-c

Please help me  - what do I do next? I feel I'm getting closer to the 
answer, through your help. :)

Stan


Date: 04/26/2000 at 08:30:56
From: Doctor Rick
Subject: Re: A Confusing Word Problem

Hi, Stan. I like your persistence! You did the next step correctly. 
Let's keep going.

You can continue #1 in the same way that you did #2. You correctly 
distributed

     -(2v+c)(2v-c) = (-2v-c)(2v-c)
                   = -2v(2v-c) + -c(2v-c)
                   = -4v^2 + 2vc - 2vc + c^2
                   = -4v^2 + c^2

Now do the same with

     -6(v+c)(v-c)

Here's a useful fact, which you can derive in the same way - it's 
worth remembering: whenever you multiply the sum of two numbers by the 
difference of the same numbers, it's equal to the difference between 
the squares of the numbers:

  (a+b)(a-b) = a^2 - b^2

I would suggest that you expand (v+c)(v-c) first, then distribute the 
-6.

Also, look at what you have for #2:

     24v + 12c - 4v^2 + c^2 = 24v - 12c

There is one term that is found on both sides of the equation. If you 
subtract this term from each side, you will simplify the equation. You 
can simplify further by combining the "c" terms (after bringing them 
to the same side of the equation).

When you're done simplifying, you'll be ready to combine the two 
equations into one equation that only has the variable c (eliminating 
v).

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/26/2000 at 13:02:00
From: Stan
Subject: Re: A Confusing Word Problem

Okay, I did all the stuff you told me  and I hope I applied it right. 
Please help me solve this!
c
   1.                        12/(v-c) -6 = 12/(v+c)

        12(v+c)(v-c)/(v-c) - 6(v+c)(v-c) = 12(v+c)(v-c)/(v+c)

                   12(v+c) - 6(v+c)(v-c) = 12(v-c)

                12v + 12c - 6[v^2 - c^2] = 12v - 12c

                12v + 12c - 6v^2 + 6c^2  = 12v - 12c
                     +12c                       +12c

                12v + 24c - 6v^2 + 6c^2  = 12v  
               -12v                       -12v 

                       24c - 6v^2 + 6c^2 = 0

     That is what I got so far for 1. 

   2.                       12/(2v-c) -1 = 12/(2v+c)

    12(2v+c)(2v-c)/(2v-c) -1(2v+c)(2v-c) = 12(2v+c)(2v-c)/(2v+c)

                 12(2v+c) -1(2v+c)(2v-c) = 12(2v-c)

               24v + 12c + (-2v-c)(2v-c) = 24v - 12c

      24v + 12c - 4v^2 + 2vc - 2vc + c^2 = 24v - 12c

                  24v + 12c - 4v^2 + c^2 = 24v - 12c
                       +12c                     +12c

                  24v + 24c - 4v^2 + c^2 = 24v 
                 -24v                     -24v

                        24c - 4v^2 + c^2 = 0

    That is what I got so far for 2.

I don't know if I should do this step - it looks unnecessary:

SUBTRACT ONE FROM THE OTHER:

      (24c -6v^2 +6c^2 = 0)   <-- 1
     -(24c -4v^2 + c^2 = 0)   <-- 2
     ----------------------
         (-2v^2 - 5c^2 = 0)

I still have two variables.  :(  Is this needed? Tell me what I did 
wrong; I understand that I have to combine like terms and all that 
stuff. But how do I get one variable by itself, especially working 
with squares?

Thanks for your help so far,
Stan


Date: 04/26/2000 at 13:32:33
From: Doctor Rick
Subject: Re: A Confusing Word Problem

Hi! We're getting there ... you've got the two equations correct. Now 
just a little adjustment and you'll have a single equation to solve.

When you subtract the equations, all you eliminate is the 24c term. 
Since the c^2 term is still there, you haven't eliminated c 
altogether.

It will be easier to eliminate v than c. Each equation has just one 
occurrence of v, and it's a v^2 term. If you can get the coefficient 
of v^2 to be the same in each equation, then when you subtract the 
equations, you will eliminate the v^2 term - and get rid of v 
altogether.

Here's how. The first equation has -6v^2; the second equation has 
-4v^2. If you multiply each side of the first equation by 4/6 (or 
2/3), then the v^2 term will be (2/3)(-6)v^2 = -4v^2, the same as the 
second equation.

Try that. Remember that when you multiply the left side of the 
equation by 2/3, you need to apply the distributive property.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/26/2000 at 14:21:42
From: Stan
Subject: Re: A Confusing Word Problem

All right, this is what I did:

COMBINE 2 EQUATIONS:

      16c - 4v^2 + 4c^2 = 0    (after being multiplied by 2/3)
     -24c + 4v^2 -  c^2 = 0    (multiplied by -1 so I can combine)
     --------------------
      -8c - 3c^2        = 0    <-- This is what I get!

Now what do I do with this equation? What does it mean, and how do you 
get rid of this square?

Thanks for your help,
Stan


Date: 04/27/2000 at 07:59:34
From: Doctor Jerry
Subject: Re: A Confusing Word Problem

Hi Stan,

The equation

     -8c - 3c^2 = 0

can be factored and thereby solved for c, like this:

     -c(8 + 3c) = 0

Because the product of two numbers (-c and 8+3c) can be zero only when 
at least one of them is 0, the two solutions are c = 0 and 8+3c = 0,

     8 + 3c = 0
         3c = -8
          c = -8/3.

You can substitute each of these two values into other equations to 
find the corresponding values of other variables.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/27/2000 at 08:35:09
From: Doctor Rick
Subject: Re: A Confusing Word Problem

Hi, Stan.

Since Dr. Jerry wasn't in on the beginning of the problem, he didn't 
know that a negative answer doesn't make sense (the current in the 
river can't flow upstream!). You had one small error when you combined 
the two equations. Look again:

      16c - 4v^2 + 4c^2 = 0
     -24c + 4v^2 -  c^2 = 0
     ----------------------
      -8c        - 3c^2 = 0

You added +4 and -1 and got -3. That's not right.

After you correct this error, you'll have the solution to the problem. 
You can find v by going back to one of the two equations above, 
substituting the value for c, and solving for v. 

The solution c = 0 is spurious; if you solve for v when c = 0, you 
will find v = 0, and this set of values for v and c does not solve the 
original equations. You should check that the other solution does 
solve the original problem.

You've done well! I hope you can use some of what you've learned on 
this problem so you will be able to see what steps to take to solve 
other problems.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/27/2000 at 10:25:00
From: Stan
Subject: Re: A Confusing Word Problem

Okay. I substituted c = 8/3 and this is what I did so far:

     16(8/3) - 4v^2 + 4[(8/3)^2] = 0

        42 2/3 - 4v^2 + 4(7 1/9) = 0

        42 2/3 - 4v^2 + (256/9)  = 0

                 (640/9) - 4v^2 =  0
                         + 4v^2   +4v^2

                           4v^2 = 640/9

                            v^2 = 160/9

                              v = 4.216370214

The number looks weird - there's no fraction equivalent to that (I 
think.) What's this number mean? Is it the mph of the river? Is that 
the ANSWER I'm looking for? I thought it would come to a round number. 
Please check for any mistakes!  :)

Thanks for your help,
Stan


Date: 04/27/2000 at 12:56:33
From: Doctor Rick
Subject: Re: A Confusing Word Problem

Hi, Stan.

You've got the answer: c = 8/3, v = sqrt(160/9). Now go back to the 
definition of the variables, many e-mails ago. How did we define c 
and v? 

Not every correct answer is a nice, round number. In the real world, 
this will rarely happen. I would do one more thing: besides giving the 
decimal answer, I would simplify the square root, so I could give the 
exact answer in a nicer form. Here is one step, to show you what I 
mean:

     sqrt(160/9) = sqrt(160)/sqrt(9)

You can also factor a perfect square out of 160.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/28/2000 at 11:39:14
From: Stan
Subject: Re: [A Confusing Word Problem] Thank you!

Dear Dr. Rick,

I got the problem right! I am so happy, but I know that I couldn't 
have done this without your 3-day help. Thank you so much - I knew I 
could do it, but I needed some guidance. I've learned a lot and thank 
you; I'll try to apply it to the rest of my 8th grade year and time to 
come. Again, thank you and I'm sure that if for some reason I have a 
problem I can ask you for help; you're a nice doctor... I can actually 
UNDERSTAND you!

Thanks. :)
Stan
    
Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Word Problems

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