SOLUTION: Yolanda finds a pretty (logarithmic) spiral seashell creature at the beach. Ever the scientist, she measures the distances of the first two loops of its spiral from the center and

Question 289938: Yolanda finds a pretty (logarithmic) spiral seashell creature at the beach. Ever the scientist, she measures the distances of the first two loops of its spiral from the center and finds that they are 2 centimeters and 4 centimeters. The shell is not yet big enough to have a third loop, but she predicts that after a few years it will be at a distance of 8 centimeters along the same line. Maximilian disagrees, saying that the distance will be 6 centimeters. Who is right?

Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
"Yolanda finds a pretty (logarithmic) spiral seashell creature at the beach. Ever the scientist, she measures the distances of the first two loops of its spiral from the center and finds that they are 2 centimeters and 4 centimeters. The shell is not yet big enough to have a third loop, but she predicts that after a few years it will be at a distance of 8 centimeters along the same line. Maximilian disagrees, saying that the distance will be 6 centimeters. Who is right?"

a and b are arbitrary positive real constants
looked on Internet for an example:
"Help with logarithmic spirals?
I'm not certain how to do this?
(from: http://answers.yahoo.com/question/index?qid=20080929235934AA4DffJ)
A fossil ammonite has the form of a logarithmic spiral. Yolanda, now a paleontologist, measures the distances of loops #2 and #4 along a radial line from the center of the spiral. She finds that the distances are 8 centimeters and 18 centimeters. The fossil is in poor condition, and she cannot measure the distance of loop #3 from the center. But she can calculate it. What is the distance in centimeters? Do not include units in your answer.
the equation for a logarithmic spiral is
r=a exp[b theta] where r is radial distance, a, and b are constants that set the dimensions of the sprial, and theta is angle displaced along the spiral
we have that r=8 for the second sprial where theta = 4 pi
and that r=18 for the fourth sprial where theta = 8 pi
write the equations for these spirals and take the ratio:
18/8 = a e^(8 pi b)/ae^(4pib)
9/4 = e^(4 pi b)
take logs of both sides to get that b=ln(9/4)/4 pi = 0.0646
so for the third sprial where theta = 6 pi
r=ae^(6 pi b) = a e^1.216
take ratio of this with equation for second spiral, and get:
r/8 = e^1.216/e^0.811
r=8 e^1.216/e^0.811 = 12"
solving:
r = 2 cm for 1st loop --> theta = 2pi (2pi per revolution)
r = 4 cm for 2nd loop --> theta = 4pi
r = ? for 3rd loop --> theta = 6pi

a and b are arbitrary positive real constants
2 = ae^(2*pi*b) (first loop)
4 = ae^(4*pi*b) (second loop)
4/2 = ae^(4*pi*b)/ae^(2*pi*b) (ratio)
2 = e^(4*pi*b)/e^(2*pi*b)
2 = e^(2*pi*b) (log rule: e^x = y --> ln y = x)
ln 2 = 2*pi*b
(ln 2)/(2*pi) = b = approx. 0.1103
third loop: r = ae^(0.1103*6*pi) = ae^(2.0794)
2nd loop: 4 = ae^(0.1103*4*pi) = ae^(1.3863) (take ratio)
r/4 = ae^(2.0794)/ ae^(1.3863)
r = 4 * e^(2.0794)/ e^(1.3863)
e^(2.0794)/ e^(1.3863) = approx. 8/4 = 2
r = 4 * 2 = 8 cm
It is not 6 cm, so Yolanda is right, and Maximilian is wrong.

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