Recently in Victoria, the tutor noticed a sequoia tree in front of the Parliament Buildings.  The plaque said it grows, on average, 1 foot per year. Is that exponential growth?

I’ve introduced exponential growth in a couple of contexts: see my articles here and here. Its basic attribute is that with exponential growth, the quantity increases by a fixed percentage every year, rather than a fixed amount.

Growth of a foot per year is not exponential, by that description. However, natural growth processes almost always are. How can we reconcile the situation?

The answer is that growth is increase in mass, not height. Mass, in the case of the tree, is directly connected with its volume. So the question is, if its height increases by one foot per year, is its volume growing by a fixed percentage per year?

For today’s post, we’ll define the (mathematical) parameters of the situation and get some initial numbers.

Let’s make the simplifying assumption that the tree is a cone. I saw it in the dark, but I’d say it might be 10 feet across (radius=5). The plaque says it’s 100 feet high. The formula for volume of a cone is

V=πr^2h/3

where r is the radius, while h is the height.

Right now, the tree’s volume is approximately π(5^2)100/3=2618ft^3.

Since its diameter is about 1/10 its height, we’ll imagine the tree’s diameter grows 1/10 of a foot per year, so its proportion of diameter to height persists.

Next year, the tree should be 101 ft tall and 10.1 ft across (radius 5.05 ft). Its new volume will be

V=π5.05^2(101)/3=2697ft^3

In absolute terms, the tree’s growth, this year to next, will be (2697-2618)=79ft^3. Its percentage growth will be 79/2618=3%.

If its growth is 3 percent every year, the tree is growing exponentially. Next post, we’ll investigate whether we can confirm it:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.