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anonymous
 5 years ago
The rate of growth of a tree is calcuated ltaewaccording to the formua : dh/dt = 2+ 1/2t raised to the power of a half (1/2) where h is the height of the tree in metres and t is time in years.
a) How much does the tree grow between the first and the fourth year?
b) How much does the tree grow between the fourth and ninth year?
anonymous
 5 years ago
The rate of growth of a tree is calcuated ltaewaccording to the formua : dh/dt = 2+ 1/2t raised to the power of a half (1/2) where h is the height of the tree in metres and t is time in years. a) How much does the tree grow between the first and the fourth year? b) How much does the tree grow between the fourth and ninth year?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is your 't' raised to the power of 1/2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to integrate this to find h as a function of t, then find the heights at various times and just do what the question asks you to do. I'll integrate it for you if you're unsure.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks you can intergrate as i am unsure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Before I do that, I need you to verify your expression for the derivative of h with respect to t:\[\frac{dh}{dt}=2+\frac{1}{2t^{1/2}}\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0H=2t+4t^1/2+C, since you are wanting the change in heighth I don't thin C is important and letting C=0 at first year 2+4 = 6 at 4th year 8+8= 16 for 10 ft of growth. for 9th yr. 18=12=30 for an additional 14 ft of growth. This is more or less a swag and I would wait for lokisan for the solution, but, does this come close to what you think it would be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would say, since when you integrate, you end up with a constant, that constant disappears when you're dealing with differences (as in a definite integral), and since we'd be calculating height between two years, we can just write,\[h(t_2)h(t_1)=\int\limits_{t_1}^{t_2}2+\frac{1}{2}t^{1/2}dt=2t+t^{1/2}_{t_1}^{t_2}=2t_2+{t_2}^{1/2}2t_1{t_1}^{1/2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, between first and fourth year:\[h(4)h(1)=2.4+\sqrt{4}2.1\sqrt{1}=103=7\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Between fourth and ninth,\[h(9)h(4)=2.9+32.42=2110 = 11\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Well the swag was close, but no cigar!lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hehe, it was your constant. You were right about it not being important, though.
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