hfelltennis
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Two ants are at a common point at time t=0, the first ant starts crawling along a straight line at the rate of 4 ft/min. Two minutes later, the second ant starts crawling in a direction perpendicular to that of the first, at a rate of 5 ft/min. How fast is the distance between them changing when the first insect has traveled 12 feet?
Thank you!

Sagot :

Ok, so I'll be quantizing time here to get you the most accurate result I can give you.

What you have to know is that:

[tex]D=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } [/tex]

Also:

t=time snapshot (in minutes)

x=distance first ant has travelled

y=distance second ant has travelled (perpendicular to the first ant)

D=distance between both ants

--------------------------------------

Now, when:

t=0, x=0, y=0, D=0

t=1, x=4, y=0, D=4

t=2, x=8, y=0, D=8

t=3, x=12, y=5, D=13

Now we quantize time and find the tangent line (slope) between two points on the distance vs time graph at t=3 and t=3.000001, knowing that distance is represented by feet and time is represented in minutes.

When t=3.000001, x=12.000004, y=5.000005 D=13.00000562...

Now we find the slope on the distance vs time graph between the points (3,13) and (3.000001, 13.00000562).

[tex]m=\frac { 13.00000562-13 }{ 3.000001-3 } \\ \\ =\frac { 281 }{ 50 } [/tex]

So, at the point (3,13) on the distance vs time graph, the two ants are moving away from each other at approximately 5.62 feet per minute.

Notations you may require to understand problem in the attachment below.
View image Аноним
AL2006

I looked at this just before going to bed, and it ruined my night, worrying
about it. Then, this morning, when I actually set pencil to paper, it turned
out not to be such a vicious limbic twister.

You DO need to know how to take a chain derivative, though.  I must say,
I'm surprised to see this much Calculus at the middle school level.  Maybe
it's MY problem, but I just can't see a way to do it without this derivative,
so I'll just show you how I did it:

The ants begin together, and crawl out along the legs of a right triangle.

-- The first one moves at 4 ft/min, so after 't' minutes, he's 4t-ft from
the vertex.  The question concerns what happens when this one has
covered 12-ft.  That will happen in (12/4) = 3 minutes after he starts out.

-- The second one starts out 2-min later, and moves at 5-ft/min. So,
after 't' minutes, he's  5(t-2) ft  from the vertex.

-- The distance between them is the hypotenuse of the right triangle.
The square of the distance is the sum of the squares of the legs:

            D² = (4t)² + [ 5(t-2) ]²

We're actually going to need to know how far apart they are after 3 minutes,
so let's do that now.

           D² = (4 x 3)² + (5 x 1)² = (12)² + (5)² = 144 + 25 = 169

           D = √169 = 13-ft apart, after the first  guy has covered 12 feet.
     We will need this shortly.

OK.  Now, let's simplify the equation for the distance between them,
(as we get ready to differentiate it).

             D² = (4t)² + [ 5(t-2) ]²

             D² = 16t² + (25) (t² - 4t + 4)

             D² = 16t² + 25t² - 100t + 100

             D² = 41t² - 100t + 100

The question asks "How fast is the distance between them changing . . . ? ",
and we know that this is the derivative of 'D' with respect to 't' .  It's easy to
take the derivative of the right side of the equation, but what about the derivative
of ' D² ' ?   That's the part that I'm not so sure about in middle school, so I'll just
give it to you:

     (Derivative of D²) is [ (2D) times (derivative of D) ] .

Now we can take the derivative of each side of the equation:

     (2D) x (derivative of D) = 82 t - 100 .

                derivative of D = (82 t - 100) / (2D)

That's how fast the distance between them is changing at any time 't' .

After 3 minutes, we calculated 'D' earlier.  It's 13-ft

               derivative of D = [ (82 x 3) - 100 ] / (2 x 13)

                                       =     146               /      26  

                                       =      5.615  feet per second.

Are you confident ?  Do you trust me?
I'm not confident at all, and I don't trust myself.
The best I can suggest is for you to go through this and look for my mistakes,
or better yet, sit down with your teacher and go through it for mistakes.
That'll help you think about it and understand it, even if my work is wrong.


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