Circumference of a 123-year-old Tree
Date: 08/01/99 at 21:02:06 From: Aubrey Lepker Subject: Circumference and Radius I am very confused can you please help? Here is the math question I am working on. If a tree grows a 1/4-inch-thick ring per year for the first 60 years, a 1/3-inch-thick ring per year for the next 80 years, and a 1/2-inch-thick ring per year for the next 200 years, what would be the circumference of a 123-year-old tree? I do not know how to find the circumference. Can you please assist? The second question (using the same information as above) is: What would be the radius of a tree with 310 rings? I do not know the formula for figuring this out either. Thank you for any assistance you can provide. Sincerely, Aubrey Lepker
Date: 08/01/99 at 22:01:54 From: Doctor Peterson Subject: Re: Circumference and Radius Hi, Aubrey. There are two tricks you need to keep in mind to solve this. First, the easy part: if you know the radius of a circle, you just have to multiply it by 2 pi (about 6.28) to get the circumference: C = 2*pi*r Second, when something is defined "piecewise" like this, you have to work it out one piece at a time. You have one rate of growth for years 0 to 60, another for years 61 to 140 (the next 80 years), and another for subsequent years. To find the circumference at 123 years, first find the radius, then use the formula to find the circumference. To find the radius, find what it is at 60 years, then add to that the increase in the radius over the next 123 - 60 = 63 years. Basically, this is the sum of two simple rate problems. The second question is really just the same, except it will require all three rates. Remember that 310 rings means 310 years old. Its radius will be the sum of the amount grown in the first 60 years, in the next 80 years, and in the last 310-140 = 170 years, each at a different rate. If that's not enough to help you solve it, write back and tell where you still have trouble, and I'll see what I can do. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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