An inchworm ran into a log while on his way to the raspberry patch. The diameter of the log is 32 cm. How far did the inchworm t
ravel while on the log?
2 answers:
Maybe I'm overthinking this one, but here's what I think it's talking about.
If I'm wrong, then I ought to at least get a few points for my talent at making
easy things difficult, and inventing obstacles to place in my own path.
-- The worm is 1 inch long.
-- The outside of the log is a cylinder. Its cross-section is a
perfect circle with a circumference of 32-cm.
-- The axis (length) of the log is perpendicular (across) the path
that leads to raspberry nirvana.
-- The ground is hard. The log contacts the ground along a line,
and doesn't sink into it at all.
-- The worm sees the log ahead of him. He continues crawling, until
he is directly under a point on the log that's 1-inch above him.
He then stands up to his full height, sticks his front legs to the log,
hoists himself up onto the bark, and starts to walk up and over it.
-- When he reaches a point on the other side of the log that's exactly 1-inch
above the ground, he hooks his sticky back feet to it, drops straight down to
the ground, and continues on his quest.
-- The question is: What's the length of the part of the log's circumference
that he traveled between the two points that are exactly 1-inch off the ground ?
I thought I was going to be able to be able to talk through this, but I can't.
I need a picture. Please see the attached picture.
Here comes the worm, heading from left to right.
He sees the log in front of him.
He doesn't bother going around it ... he knows he'll be able to get over it.
When he gets under the log, he starts standing straight up, trying to
grab onto the bark. But he can't reach it. He's too short, only 1 inch.
Finally, when he gets to point 'F', the bark is only 1" above him,
so he can hook on and haul himself up to point 'A'.
He continues on ... up, around, and over the log.
Eventually it dawns on him that the log won't last forever, and he'll
soon need to get down to the ground. As he comes down the right
side of the log, he starts looking down. It's too high. He can't reach
the ground, and he's afraid to jump.
Then he reaches point 'B'. It's exactly 1-inch above the ground, and
he leaves the log and gets down.
What was the length of the path he followed on the log ... the long way,
over the top from 'A' to 'B' ?
Here's what I did:
Draw radii from the center of the log to 'A' and 'B' .
Each of them is 16 cm long (1/2 of the diameter).
Draw the radius from the center of the log to the ground (' E ').
It's 16 cm all the way.
Point 'D' is 1 inch = 2.54 cm above the ground, so the
vertical leg of each little right triangle is (16 - 2.54) = 13.46 cm.
There are two similar right triangles, back to back, inside the log.
They are 'CAD' on the left, and 'CBD' on the right.
I want to know the size of the angles at the top of each triangle.
(One will be enough, since they're equal angles.)
For each of those angles, the side adjacent to it is 13.46 cm.
And the hypotenuse of each right triangle is a radius, so it's 16 cm.
The cosine of those angles is (adjacent/hypotenuse) = 13.46/16 = 0.84125 .
Each angle is 32.73 degrees.
Both of them put together add up to 65.45 degrees .
The full circumference of the log is (pi)(D) = 32pi cm.
The short arc between 'A' and 'B' is (65.45/360) of the full circumference.
The rest of the circumference is the distance that the worm crawled along it.
That's (1 - 65.45/360) times (32 pi) = (0.818) x (32 pi) = <em>82.25 cm</em> .
Having already wasted enough time on this one in search of 5 points,
and then gone back through the whole thing to make corrections for
the customary worm crawling over the metric log, I'm not going to bother
looking for a way to check it.
That's my answer, and I'm sticking to it.
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Answer:
f(r) = (x -1)(x -4+√7)(x -4-√7)
Step-by-step explanation:
The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.
The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.
Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:
r^2 -8r +9 = (r^2 -8r +16) +9 -16 = (r -4)^2 -7
This is zero when ...
(r -4)^2 = 7
r -4 = ±√7
r = 4±√7
Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...
f(r) = (r -1)(r -4 -√7)(r -4 +√7)
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<em>Comment on the graph</em>
I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)
